Electrostatic Potential and Capacitance – NEET Faculty Exam Pack

📊 NEET Last 15 Years (2010–2024) — Topic Map for this Chapter

Below is the complete breakdown of every topic from Electrostatic Potential & Capacitance that has appeared in NEET/AIPMT over the last 15 years — what type of questions each topic asks and which formulas you must have at your fingertips. Topics are colour-coded by frequency.

2–3Questions per year (average)

~35Total questions in 15 yr

20%Of total Electrostatics weight

16Topics tested overall

● High (asked ≥ 4 times) ● Medium (2–3 times) ● Low (≤ 1 time)

1 Electric Potential due to a Point Charge High

Question types

Find V at a point due to one or more point charges (vector-free, scalar sum)
Work done in bringing a charge from infinity
V along the axis or perpendicular bisector of a system of charges

Formulas

\[V = \dfrac{kq}{r},\quad V_{\text{net}} = \sum_i \dfrac{kq_i}{r_i},\quad W_{\infty\to P} = q_0 V_P\]

2 Potential due to a Dipole Medium

Question types

V at axial / equatorial / general angle θ position
Comparison of V at two points
Variation of V with r at large distance

Formulas

\[V = \dfrac{k\,p\cos\theta}{r^2},\quad V_{\text{axial}} = \dfrac{kp}{r^2},\quad V_{\text{equatorial}} = 0\]

3 Potential of a Charged Sphere / Shell High

Question types

V inside and outside a hollow conducting shell
V at centre vs surface of a solid sphere (3:2 ratio)
Graph-based V-r identification

Formulas

\[V_{\text{shell,in}} = V_{\text{surface}} = \dfrac{kQ}{R}\] \[V_{\text{solid,centre}} = \dfrac{3kQ}{2R},\quad V_{\text{out}} = \dfrac{kQ}{r}\]

4 Equipotential Surfaces Medium

Question types

Identify equipotentials of point charge, dipole, uniform field
Work done moving charge ON an equipotential = 0
Equipotentials always ⊥ to field lines

Formulas

\[W_{\text{on equipotential}} = q(V_2 - V_1) = 0\] \[\vec E = -\nabla V,\quad E_x = -\dfrac{\partial V}{\partial x}\]

5 Relation between E and V High

Question types

Given V(x), find E(x) — differentiate
Given uniform E, find ΔV between two points
Sign convention (E points from high V to low V)

Formulas

\[E = -\dfrac{dV}{dx},\quad \Delta V = -\int \vec E \cdot d\vec r,\quad V_{AB} = -E\,d\;\text{(uniform field)}\]

6 Potential Energy of System of Charges High

Question types

U of 2, 3 or 4 point charges (sum over pairs)
Work done to assemble a configuration
U of a dipole in an external field — angular dependence

Formulas

\[U_{2\text{-charge}} = \dfrac{kq_1 q_2}{r_{12}},\quad U_{\text{system}} = \sum_{i

7 Conductors in Electrostatic Equilibrium Medium

Question types

E = 0 inside a conductor
Charge resides on outer surface
V is constant throughout the conductor body
Two conductors joined by a wire — share charge so V equalises

Formulas

\[\dfrac{Q_1}{Q_2} = \dfrac{R_1}{R_2}\;\text{(joined spheres)}\] \[\sigma \propto \dfrac{1}{R}\;\text{(at surface, sharper points have higher density)}\]

8 Parallel-Plate Capacitor (basic) High

Question types

Find C from A, d, ε₀
Effect of changing A or d on C, V, Q, E, U
Force between plates (Q² / 2ε₀A)

Formulas

\[C_0 = \dfrac{\varepsilon_0 A}{d},\quad E = \dfrac{V}{d} = \dfrac{\sigma}{\varepsilon_0}\] \[F_{\text{plate}} = \dfrac{Q^2}{2\varepsilon_0 A}\]

9 Capacitor with Dielectric High

Question types

Full / partial / slab dielectric insertion — new C
Two dielectrics side-by-side (parallel) or stacked (series)
Metal slab inserted (K → ∞ limit)

Formulas

\[C = K\,C_0,\quad C_{\text{slab t}} = \dfrac{\varepsilon_0 A}{d - t + t/K}\] \[C_{\text{side-by-side}} = \dfrac{\varepsilon_0 A(K_1+K_2)}{2d}\] \[C_{\text{stacked}} = \dfrac{2\varepsilon_0 A K_1 K_2}{d(K_1+K_2)}\]

10 Battery ON vs OFF — Effect of K, d High

Question types

V kept fixed (battery on): C↑ ⇒ Q↑, U↑
Q kept fixed (battery off): C↑ ⇒ V↓, U↓
Work done against the slab or to pull plates apart

Formulas

\[\text{Battery on: }U_{\text{new}} = K U_0\;\text{(insert)};\;U/K\;\text{(remove)}\] \[\text{Battery off: }U_{\text{new}} = U_0/K\;\text{(insert)};\;K U_0\;\text{(remove)}\] \[W_{\text{ext}} = \Delta U\]

11 Series & Parallel Combinations High

Question types

Find C_eq of a mixed network
Charge / voltage / energy distribution among caps
Wheatstone bridge with capacitors

Formulas

\[\text{Series: }\dfrac{1}{C_s} = \sum \dfrac{1}{C_i}\;\text{(same Q)}\] \[\text{Parallel: }C_p = \sum C_i\;\text{(same V)}\] \[\text{Bridge balanced: }\dfrac{C_1}{C_2} = \dfrac{C_3}{C_4}\]

12 Energy Stored in a Capacitor High

Question types

U for given C, V, Q (three forms)
Percentage increase in U when V or Q changes
Heat dissipated when discharged through R

Formulas

\[U = \tfrac{1}{2}CV^2 = \dfrac{Q^2}{2C} = \tfrac{1}{2}QV\] \[u = \tfrac{1}{2}\varepsilon_0 E^2\;\text{(energy density)}\]

13 Sharing of Charge (Two Capacitors) High

Question types

Common potential after joining two charged caps
Energy lost during sharing (always positive)
Switch-transfer problems (one charges the other)

Formulas

\[V_c = \dfrac{C_1 V_1 + C_2 V_2}{C_1 + C_2}\] \[\Delta U = \tfrac{1}{2}\dfrac{C_1 C_2}{C_1+C_2}(V_1-V_2)^2\]

14 Force / Torque on a Dipole in External Field Medium

Question types

Torque on dipole at angle θ
Work done in rotating from θ₁ to θ₂
Stable / unstable equilibrium positions

Formulas

\[\tau = pE\sin\theta,\quad U = -pE\cos\theta\] \[W = pE(\cos\theta_1 - \cos\theta_2)\]

15 Spherical & Cylindrical Capacitors Low

Question types

C of two concentric spheres (one earthed)
C of cylindrical capacitor (rare)
Effect of dielectric in the gap

Formulas

\[C_{\text{sph}} = \dfrac{4\pi\varepsilon_0 ab}{b-a},\quad C_{\text{cyl}} = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}\]

16 Dielectric Polarization & Bound Charges Low

Question types

Reduced field inside dielectric E = E₀/K
Surface bound charge density σ_b
Susceptibility / permittivity relations

Formulas

\[E_{\text{net}} = \dfrac{E_0}{K},\quad K = 1 + \chi_e\] \[\sigma_b = \sigma\,(1 - 1/K)\]

How to use this map: rank topics by the colour-coded frequency. Master every High-frequency topic (their formulas should be on muscle memory), feel comfortable with Medium-frequency, and at minimum recognise the Low-frequency setups. Each topic also has dedicated PYQ practice — see ⚡ Capacitor Circuits, 💎 Capacitor Bridge, 📈 Graph PYQs linked from the top of this page.

🎒 What you'll learn in this chapter

Think of this chapter as a story about electric height, tiny energy tanks, and the helpers that pack more energy inside them. Here's the whole story in one quick read — no formulas yet, just the picture.

⛰️

Electric "height" (potential)

Imagine every spot in space has a little number stuck to it — its height. A positive charge naturally rolls from a high place to a low place, like a ball rolling down a hill. That number is called the electric potential, written V.

Bigger drop in height ⇒ ball rolls faster (more energy gained).

🗺️

Same-height lines (equipotentials)

On a hiking map, dotted lines connect places at the same elevation. In electricity, we have the same thing — surfaces where the "height" V is the same. They are called equipotential surfaces.

If you walk along one, you don't go up or down, so no work is done on the charge.

🔋

Tiny energy tanks (capacitors)

A capacitor is like a very small water tank, but for electricity. Two metal plates sit close to each other; you push charge onto one and pull charge off the other, and the capacitor stores the difference.

Bigger plates, closer together ⇒ holds more charge for the same "push" (voltage).

📏

How "full" the tank is (capacitance)

Capacitance (C) is just a number that tells you how many drops of charge a tank holds per unit of push. A big number = it can hold a lot. The unit is the farad.

Capacitance depends on shape and size, not on what you connect to it.

🧽

Sponges that help (dielectrics)

Slip a piece of plastic or glass between the capacitor plates and — magic — the tank suddenly holds more. These materials are called dielectrics.

The amount it improves storage by is called the dielectric constant K. A higher K means a thirstier sponge.

💾

Stored energy

Once charge is sitting on a capacitor, the device is also storing energy — same way a stretched rubber band or a wound-up spring stores energy.

Connect it to a wire and that stored energy flows out as a quick flash.

🪙

Putting them together

You can hook many capacitors in a row (series) or side-by-side (parallel). One gives a smaller combined tank, the other a bigger one.

Real circuits use this trick all the time to get the exact storage they need.

🔄

Energy never lies

Wherever charges move because of electric forces, energy is conserved. We can write a clean equation that says: starting energy = ending energy + anything turned into heat or light.

This lets us answer "how fast?", "how much?" or "is it possible?" questions.

By the end of this page you'll be able to write down V at any point, draw equipotentials, work out the capacitance of any arrangement, decide what changes when you stick in a dielectric, and solve every NEET-style energy problem in seconds. Take your time, glance at the visuals, and only then attempt the MCQs.

1 The NEET Cheat Sheet — Core Concepts & Exceptions

The simple idea

If electric field is the push on a charge, electric potential is the height of a hill the charge sits on. Charges naturally roll from high potential to low (for +) or low to high (for −). A capacitor is just a device that stores this "height difference" as energy.

Potential energy & potential

Electric potential V

💡 Picture work done to drag a +1 C test charge from infinity to a point — that's potential at the point. It's a single number (no direction).

V = W / q₀ (q₀ → 0). SI unit: volt (V) = J/C = N·m/C. Scalar quantity.

V_point = kq/r; V_{shell/conductor} inside = kQ/R (= surface value).

Potential difference

💡 The "height drop" between two points. Always a difference — choose any reference; only differences matter.

V_A − V_B = W_BA/q = − ∫_B^A E · dl.

📌 Moving +q from low V to high V costs work; moving from high V to low V releases energy.

E and V — the link

💡 Field is just how steeply potential changes. Steep slope ⇒ strong field. E always points downhill.

E = − dV/dr (radial); generally E = −∇V.

📌 If V is constant in a region (equipotential), E in that region = 0.

Potential due to standard distributions

System of point charges

💡 V adds up like ordinary numbers — no vectors. Just sum kq/r for each charge.

V_total = Σ kqᵢ/rᵢ (scalar superposition).

📌 At midpoint between +q and −q: V = 0 (they cancel).

Dipole

V_axial = ± kp/r² (sign by side); V_equatorial = 0.

General point: V = (kp cosθ)/r² (r >> a), where θ is the angle from p.

⚠ Dipole's V falls as 1/r², its field as 1/r³ — both faster than a single charge.

Charged shell / conductor

Outside (r ≥ R): V = kQ/r. Inside (r < R): V = kQ/R (constant!).

📌 V is continuous at r = R, but field jumps from 0 (inside) to kQ/R² (just outside).

Equipotential surfaces

Definition & rules

💡 Surfaces of constant potential — like contour lines on a map. Walking along one ⇒ no work done.

E is always perpendicular to the equipotential.
Two equipotentials with different V never intersect.
Closer equipotentials ⇒ stronger field.
Work done moving a charge along an equipotential = 0.

Shapes

Point charge: concentric spheres.
Uniform field: parallel planes ⟂ to field.
Dipole: distorted spheres; equatorial plane is V = 0.
Conductor surface: follows the conductor's shape (it's an equipotential).

Point charge

Concentric spheres; E ⟂ to each.

Uniform field

Parallel planes ⟂ E.

Dipole

Equatorial plane has V = 0.

Energy in electrostatics

Energy of charges

💡 To bring charges together, you must spend or gain energy. That stored energy = system's "electrostatic PE".

U (two charges) = k·q₁q₂/r.

System: U = ½ ΣᵢⱼVᵢqⱼ = sum of pairwise terms (i < j).

Dipole in field: U = −p·E = −pE cosθ.

Energy density

💡 An electric field carries energy in the space it fills — even with no charges around.

u = ½ ε₀ E² (J/m³); total field energy = ∫u dV.

Conductors & dielectrics

Conductor in field

💡 The free electrons rearrange till the interior field is zero. The body becomes one big equipotential.

E_inside = 0; surface charge resides on outer surface.
E just outside = σ/ε₀ ⟂ to surface.
Whole conductor is one equipotential.
Electrostatic shielding: any cavity inside is field-free.

Dielectric in field

💡 An insulator's molecules tilt or stretch (polarisation). This creates a weak internal field opposite to the applied one ⇒ net field is smaller.

Polar (water, HCl) — permanent dipoles. Non-polar (CO₂, N₂) — induced dipoles in field.

E_{inside dielectric} = E₀/K; K = dielectric constant (relative permittivity).

⚠ Capacitance ↑ by factor K with dielectric, but breakdown limits the maximum V.

Capacitors — the heart of the chapter

Capacitance C

💡 How many coulombs a device "soaks up" per volt. Bigger plates / closer plates / dielectric ⇒ more storage.

C = Q / V; SI unit farad (F) = C/V. Practical: μF, nF, pF.

Parallel plate: C₀ = ε₀A/d. With dielectric: C = Kε₀A/d.

Energy stored

U = ½ QV = ½ CV² = Q²/(2C).

📌 Battery does W = QV = CV²; half is stored, half dissipated as heat or radiation while charging.

Series & parallel

Series: 1/C_s = 1/C₁ + 1/C₂ + … ; same Q on each.

Parallel: C_p = C₁ + C₂ + … ; same V across each.

Exceptions, limits & common errors

V and E are not the same — V can be non-zero even where E = 0 (e.g., inside a charged shell), and E can be non-zero where V = 0 (e.g., equatorial plane of a dipole).
Inside a uniformly charged shell: E = 0 but V = kQ/R ≠ 0.
The equatorial plane of a dipole: V = 0 but E ≠ 0.
"Half the work" rule: when a battery charges a capacitor, only ½ of the battery's work is stored — the other half is irreversibly lost.
When a battery remains connected and a dielectric is inserted: V stays constant, C ↑, Q ↑ (battery supplies extra). When battery is disconnected: Q stays constant, C ↑, V ↓, U ↓.
For a spherical capacitor: C = 4πε₀·ab/(b − a). For an isolated conducting sphere: C = 4πε₀R.

2 The Formula & Approximation Bank

Every formula NEET tests, grouped by topic — with the time-saving notes beside each. A one-line plain summary sits above each group.

Potential, potential difference & energy

💡 V is a scalar — add with signs, no vectors. The classic links: E = −dV/dr and U = qV.

Formula	Notes / Approximation
V = W/q₀ = kq/r (point charge)	Volt = J/C; V → 0 as r → ∞.
V_A − V_B = − ∫_B^A E · dl	Path-independent (conservative field).
E = − dV/dr ; E_x = −∂V/∂x etc.	Negative gradient: E points down-slope of V.
U_system = ½ Σᵢⱼ kqᵢqⱼ/rᵢⱼ (i ≠ j)	Pairwise sum; ½ to avoid double counting.
U_dipole = −p · E = −pE cosθ	Min at θ = 0° (stable); max at 180° (unstable).
W_ext = q(V_f − V_i)	Work to move charge slowly (Δ KE = 0).

Potential due to standard distributions

💡 Memorise these — they sit behind many NEET MCQs.

Configuration	Potential V
Point charge	kq/r
System of point charges	Σ kqᵢ/rᵢ (scalar sum)
Dipole — axial	± kp/r² (sign by side)
Dipole — equatorial	0
Dipole — general (r >> a)	(kp cosθ)/r²
Conducting / uniformly charged shell, r ≥ R	kQ/r
Conducting / uniformly charged shell, r < R	kQ/R (constant)
Solid sphere (uniform ρ), r < R	kQ(3R² − r²)/(2R³)

Capacitance — fundamental formulas

💡 Three things govern capacitance: area, separation, dielectric. Nothing else.

Configuration	Capacitance
General definition	C = Q/V
Parallel plate (vacuum)	C₀ = ε₀A/d
Parallel plate (dielectric K)	C = Kε₀A/d
Isolated conducting sphere	C = 4πε₀R
Spherical capacitor (inner a, outer b)	C = 4πε₀·ab/(b − a)
Cylindrical capacitor (radii a, b; length L)	C = 2πε₀L / ln(b/a)

Energy & combinations

💡 In series, charge is common. In parallel, voltage is common. Use the right "constant" to crack any combination.

Formula	Notes
U = ½QV = ½CV² = Q²/(2C)	All three forms equal; pick whichever variable is fixed.
Energy density u = ½ε₀E²	Per unit volume; with dielectric multiply by K.
Series: 1/C_s = Σ 1/Cᵢ ; Q common	Total V = sum of individual Vᵢ = Q·(1/C_s).
Parallel: C_p = ΣCᵢ ; V common	Total Q = sum of individual Qᵢ.
Two charged capacitors connected: V_common = (C₁V₁+C₂V₂)/(C₁+C₂)	Energy lost ΔU = ½C₁C₂(V₁−V₂)² / (C₁+C₂).

Dielectric & conductor inserts

Geometry	Capacitance
Dielectric slab (K, thickness t, gap d)	C = ε₀A / (d − t + t/K)
Conductor slab (thickness t, gap d)	C = ε₀A / (d − t) (treat as K → ∞)
Two dielectrics K₁, K₂ in parallel (each occupies half area)	C = ε₀A(K₁+K₂)/(2d)
Two dielectrics K₁, K₂ in series (each half gap)	C = 2ε₀A K₁K₂ / (d(K₁+K₂))

Useful constants

k = 9 × 10⁹ε₀ = 8.854 × 10⁻¹²1 V = 1 J/C1 F = 1 C/V1 μF = 10⁻⁶ F1 pF = 10⁻¹² F

📈 Interactive Capacitor Explorer

The simple idea

A parallel-plate capacitor's behaviour depends on three knobs (A, d, K) and one switch (battery connected or disconnected). Drag the sliders and watch C, Q, V and U change in real time.

Battery state

Plate area A: 100 cm²

Separation d: 2.0 mm

Dielectric K: 1.0

Battery EMF V₀: 10 V

C: 44.27 pF

Q: 0.44 nC

V: 10.0 V

U: 2.21 nJ

Switch the battery state to see the contrast: connected keeps V fixed (Q and U scale with K); disconnected keeps Q fixed (V and U scale as 1/K).

3 The "Trap Avoidance" & Shortcut Guide

The simple idea

These are the exact spots where students lose easy marks. Each card shows the trap, then the quick fix.

Battery connected vs disconnected — the dielectric switch

A NEET favourite. After inserting dielectric of constant K:

Quantity	Battery connected (V fixed)	Battery disconnected (Q fixed)
C	× K (↑)	× K (↑)
V	unchanged	÷ K (↓)
Q	× K (↑)	unchanged
E inside	unchanged (E = V/d)	÷ K (↓)
Energy U	× K (↑) — battery supplies more	÷ K (↓) — work done by dielectric

⚡ Quick check: ask yourself "is V or Q the same as before?" — that tells you everything else.

Series vs parallel — Q vs V is common

✔

Series: charge Q is common (it's pumped from one capacitor onto the next via induction). Voltages add.

✔

Parallel: voltage V is common. Charges add.

⚡ Series → think of garden hoses end-to-end (same flow / charge). Parallel → think of taps side-by-side (same pressure / voltage).

V ≠ 0 does not mean E ≠ 0, and vice versa

Region	E	V
Inside a charged spherical shell	0	kQ/R (constant)
Equatorial plane of a dipole	kp/r³ (≠ 0)	0
Anywhere on an equipotential	⟂ to surface	constant
Where field lines crowd	large	changes rapidly

Two charged capacitors connected — energy is LOST

✔

Common potential: V_c = (C₁V₁ + C₂V₂)/(C₁+C₂).

✗

Wrong: assume total energy is conserved.

✔

Right: energy lost = ½·C₁C₂·(V₁−V₂)² / (C₁+C₂) — dissipated as heat / EM radiation.

⚡ Zero only if V₁ = V₂.

"Half the work" rule (battery charging)

Battery does W = QV in charging a capacitor from 0 to Q.
Capacitor stores only U = ½QV.
The other ½QV is irreversibly dissipated (heat in wires / EM radiation) — independent of resistance.

⚡ This is the cleanest way to remember "energy stored is half the battery's work".

Slab inside a capacitor

Conducting slab of thickness t → behaves as if separation decreased by t: C = ε₀A/(d − t).
Dielectric slab (K, t): C = ε₀A/(d − t + t/K).
Result independent of slab's position between plates (as long as it's between them).

V is a scalar — sign matters

✔

Use signed kqᵢ/rᵢ; sum directly. No components.

✔

Midpoint between +q and −q: V = 0 but E ≠ 0 (pointing from + to −).

⚡ Beginners' biggest trip — they "add magnitudes" of V for opposite charges. Don't.

E = −dV/dr — the sign and the direction

✗

Wrong: "If V is positive, E points outward; if V is negative, E points inward."

✔

Right: E points in the direction of decreasing V. So E points from high-V regions to low-V regions, regardless of the sign of V itself.

✔

If V is constant in a region (e.g., inside a conductor or across an equipotential): E = 0 in that region.

⚡ E ⊥ equipotential surface, always.

Inside a conductor — V is constant, NOT zero

✗

Wrong: "V inside a conductor is 0 because E = 0."

✔

Right: V is constant inside (E = 0 means dV/dr = 0). The constant value equals the surface potential V_surface = kQ/R for a charged sphere of radius R.

✔

Hollow conductor too: V inside cavity (no charge) = V_surface; but E ≠ 0 outside.

Solid insulator vs hollow shell — V inside

✔

Hollow shell or conductor (radius R): V_inside = V_surface = kQ/R (constant).

✔

Solid insulator (uniform ρ, radius R): V_inside(r) = kQ(3R² − r²)/(2R³). At centre: V_centre = 3kQ/(2R) = 1.5 × V_surface.

✔

For both: V_outside = kQ/r — same as a point charge.

⚡ NEET classic: "V at centre of solid sphere" answer is 1.5 × surface potential, not 0 or kQ/R.

Work, ΔU and ΔV — get the sign right

✔

Work done by external agent (no KE change): W_ext = +ΔU = q(V_f − V_i).

✔

Work done by the electric field: W_field = −ΔU = q(V_i − V_f).

✗

Wrong: "Work done to bring +q from infinity to V₀" = qV₀. (Don't forget V at infinity is 0; sign of V₀ matters — negative V₀ means field does positive work and W_ext is negative.)

Dielectric in a capacitor — battery connected vs disconnected (full table)

Quantity	Battery ON	Battery removed
V	constant	V/K (drops)
C	KC₀ (rises)	KC₀ (rises)
Q	KQ₀ (rises)	Q₀ (constant)
E (between)	E₀ (constant)	E₀/K (drops)
U = ½CV²	KU₀ (rises)	U₀/K (drops)

⚡ Energy rises when battery is on (battery did the extra work). Energy falls when disconnected (some energy went into pulling the slab in).

Capacitor with a conducting slab (vs dielectric)

✔

Conducting slab of thickness t placed parallel to plates: C = ε₀A/(d − t). The slab acts like d→d−t (zero electrical thickness inside the slab). Position of slab doesn't matter.

✔

Dielectric slab of thickness t: C = ε₀A/(d − t + t/K). Reduces to conductor case when K → ∞.

✗

Wrong: "C = ε₀A/d unchanged" — slab insertion always increases C (or keeps it the same when t = 0).

Energy density inside the capacitor

✔

u = ½ε₀E² (vacuum), u = ½ε₀KE² = ½εE² with dielectric.

✔

Total energy = u × volume = ½CV² = Q²/(2C) = ½QV.

✔

For a charged isolated sphere: U = kQ²/(2R). Self-energy formula.

Two capacitors share charge — energy is LOST

✔

C₁ at V₁ and C₂ at V₂ joined together → final V = (C₁V₁ + C₂V₂)/(C₁ + C₂).

✔

Energy lost (as heat / EM radiation) = ½·C₁C₂/(C₁+C₂)·(V₁ − V₂)².

✔

Energy is lost ONLY if V₁ ≠ V₂. If equal, no loss.

✗

Wrong: "Two identical capacitors at same V joined still lose energy." False — none is lost.

Series vs parallel — capacitor combinations

✔

Series: 1/C_eq = Σ 1/Cᵢ. Same charge on each capacitor. C_eq is smaller than the smallest.

✔

Parallel: C_eq = Σ Cᵢ. Same voltage across each. C_eq is bigger than the biggest.

✔

Two in series: C_eq = C₁C₂/(C₁+C₂). For equal: C/2.

✗

Trap: students apply resistor rules but flip Q and V. Capacitor series = resistor parallel formula!

Equipotential surfaces — must-knows

✔

Point charge → concentric spheres.

✔

Uniform field → parallel planes ⊥ to field.

✔

Dipole → distorted figure-of-eight surfaces; V = 0 surface is the plane perpendicular to dipole axis through its centre.

✔

Two equal +q: V = 0 nowhere; two ±q: V = 0 on the perpendicular bisector plane.

✔

Equipotentials never intersect (single V at each point).

Potential due to a dipole (axial vs equatorial)

✔

Axial point at distance r ≫ a: V = +kp/r² (positive on +q side, negative on −q side).

✔

Equatorial point: V = 0 (equidistant from + and −q charges).

✔

General point at angle θ from axis: V = kp·cosθ/r².

✗

Trap: students confuse with E. For dipole, V ∝ 1/r² (faster than monopole's 1/r); E ∝ 1/r³.

Spherical capacitor & isolated sphere

✔

Isolated sphere radius R: C = 4πε₀R. (Earth ≈ 700 µF.)

✔

Concentric spheres (a inside, b outside): C = 4πε₀·ab/(b − a).

✔

If b → ∞: reduces to isolated sphere (C → 4πε₀a). If a→b: C → ∞ (parallel-plate limit).

✔

Cylindrical capacitor: C = 2πε₀L/ln(b/a) (per length L).

Force between capacitor plates — attractive

✔

Plates attract: F = Q²/(2ε₀A) = (½)ε₀E²·A. NOT Q²/(ε₀A) — the ½ comes from one plate seeing the OTHER plate's field σ/(2ε₀), not the full σ/ε₀.

✔

Force on a dielectric being pulled into a capacitor: F = (1/2)ε₀(K−1)E²·b (where b is the slab width).

⚡ Common JEE trap — half of σ/ε₀ produces the field at one plate due to the OTHER plate alone.

Potential energy of a charge system — pair-counting

✔

U = Σ_i<j kqᵢqⱼ/rᵢⱼ — sum over distinct pairs. For n charges: n(n−1)/2 pairs.

✔

3 charges at corners of equilateral triangle (side a): U = 3kq²/a (with sign of each pair product).

✔

4 charges at square corners (side a): 4 side-pairs at distance a + 2 diagonals at a√2 → U = 4kq²/a + 2kq²/(a√2) (signed).

✗

Wrong: Double-counting (summing over all i, j including i = j or i > j again).

"V = 0 but E ≠ 0" and "E = 0 but V ≠ 0"

✔

V = 0 but E ≠ 0: midpoint between +q and −q charges (dipole equatorial).

✔

E = 0 but V ≠ 0: midpoint between two equal +q charges; or inside a charged conductor.

✔

V and E are independent at a single point — neither implies the other is zero.

⚖️ Quick Comparisons — Side-by-Side Reference

The simple idea

NEET / JEE loves to test the difference between two almost-identical situations. These side-by-side tables show every common pair — connected vs disconnected, conductor vs insulator, axial vs equatorial, and so on. Read each row across; the one you forget is the one they'll ask.

VS Battery connected vs Battery disconnected (with dielectric inserted, K)

Quantity	Battery ON (V = const)	Battery OFF (Q = const)
V (voltage)	Unchanged (= V₀)	Drops to V₀/K
C (capacitance)	Rises to KC₀	Rises to KC₀
Q (charge)	Rises to KQ₀	Unchanged (= Q₀)
E (field between plates)	Unchanged (= E₀)	Drops to E₀/K
U (energy stored)	Rises to KU₀	Drops to U₀/K
Work done by battery	+(K−1)Q₀V₀	0 (no battery)

Why energy rises with battery on: battery supplies the extra charge KQ₀; half goes into the capacitor, half is dissipated. Why energy falls when disconnected: the dielectric is sucked in by the field, doing work AT the capacitor.

VS Inside vs Outside a charged sphere/shell

Region	Hollow shell / Conductor	Solid insulator (uniform ρ)
E (r < R)	0	kQr/R³ (linear)
V (r < R)	kQ/R (constant)	kQ(3R² − r²)/(2R³)
V at centre (r = 0)	kQ/R	3kQ/(2R) = 1.5 V_surface
E at r = R (just outside)	kQ/R²	kQ/R²
E outside (r > R)	kQ/r²	kQ/r²
V outside (r > R)	kQ/r	kQ/r

From outside both look identical (point-charge field). Inside they differ — that's the question NEET asks.

VS Series vs Parallel (Capacitors)

Property	Series	Parallel
Equivalent C	1/C_eq = Σ 1/Cᵢ	C_eq = Σ Cᵢ
Two-capacitor formula	C₁C₂/(C₁+C₂)	C₁ + C₂
Charge Q on each	Same (= Q_total)	Different (Qᵢ = CᵢV)
Voltage V across each	Different (Vᵢ = Q/Cᵢ)	Same (= V_total)
C_eq compared to lowest	Smaller than smallest	Larger than largest
Energy stored	½Q²/C_eq = Σ ½Q²/Cᵢ	½V²C_eq = Σ ½V²Cᵢ

Flipped from resistors! Resistor-series uses C-parallel formula (Σ) and vice versa. Don't confuse the two.

VS Conducting slab vs Dielectric slab (inside capacitor)

Quantity	Conducting slab (thickness t)	Dielectric slab (K, thickness t)
New capacitance C'	ε₀A/(d − t)	ε₀A/(d − t + t/K)
Effect on C	Always increases	Always increases (K → ∞ limit = conductor)
Field inside slab	0	E/K (reduced)
Field in remaining gap	E (unchanged)	E (unchanged)
Depends on position?	No — anywhere in gap	No — anywhere in gap
If slab fills entire gap (t = d)	C → ∞ (short circuit)	C = KC₀

VS Dipole — Axial vs Equatorial point (far field, r ≫ a)

Quantity	Axial (on dipole line)	Equatorial (perpendicular)
\|E\|	2kp/r³	kp/r³
Direction of E	Parallel to p	Antiparallel to p
V	+kp/r²	0
Sign convention	+ on +q side	Same V both sides
Ratio E_axial : E_equatorial	2 : 1 (opposite direction)

General point at angle θ from axis: V = kp·cosθ/r², |E| = (kp/r³)√(1+3cos²θ).

VS Isolated sphere vs Spherical capacitor vs Cylindrical capacitor

Geometry	Capacitance	Notes
Isolated sphere R	4πε₀R	Earth ≈ 700 µF
Concentric spheres (a, b)	4πε₀·ab/(b−a)	b → ∞ gives 4πε₀a (isolated)
Cylindrical (a, b, length L)	2πε₀L/ln(b/a)	Length-dependent
Parallel-plate (A, d)	ε₀A/d	a→b limit of spherical
Two parallel wires (radius a, gap d)	πε₀L/ln(d/a)	Half the cylindrical's value

VS V (potential) vs U (potential energy)

Property	V (electric potential)	U (potential energy)
Unit	volt (J/C)	joule
Property of	Point in space	System of charges
Needs a test charge?	No — defined per coulomb	Yes — U = qV
Point charge q at r	V = kq/r	U(pair) = kq₁q₂/r
Reference	V(∞) = 0	U(∞ apart) = 0
For a dipole in field	V₀ (location)	U = −p·E (orientation)

VS "V = 0 but E ≠ 0" vs "E = 0 but V ≠ 0"

Case	V = 0 & E ≠ 0	E = 0 & V ≠ 0
Example 1	Equatorial midpoint of an electric dipole	Midpoint between two equal +q charges
Example 2	Far from a neutral system (large distance)	Inside a charged hollow conductor
Why it happens	Cancelling potentials, non-cancelling fields	Cancelling fields, non-cancelling potentials

V and E are independent at any single point. Neither implies the other is zero.

VS Charging vs Discharging (RC circuit)

Quantity	Charging (battery V₀, switch closed)	Discharging (battery removed)
Q(t)	Q₀(1 − e^(−t/τ))	Q₀·e^(−t/τ)
V_C(t)	V₀(1 − e^(−t/τ))	V₀·e^(−t/τ)
I(t)	(V₀/R)·e^(−t/τ)	−(V₀/R)·e^(−t/τ)
τ (time constant)	RC	RC
After t = τ	Q at 63 % of Q_max	Q at 37 % of Q₀
After t = 5τ	~100 % charged	~0 % left

VS Polar vs Non-polar dielectric

Property	Polar (H₂O, HCl, NH₃)	Non-polar (H₂, N₂, CO₂)
Permanent dipole moment	Yes	No
In external field	Existing dipoles align	Field induces a small dipole
Induced polarization	Large + temperature-sensitive	Small + temperature-independent
Dielectric constant K	Large (water K ≈ 80)	Small (most ~1–5)
Behaviour at high T	K drops (thermal scrambling)	K nearly unchanged

VS Earthed vs Isolated conductor

Property	Earthed (grounded)	Isolated
Potential V	0 (fixed by Earth)	Whatever its charge + surroundings dictate
Charge Q	Adjusts to satisfy V = 0	Fixed (= initial)
External influence	Charge flows to ground	Charge redistributes on the conductor only
Removing the wire	Final Q stays; V no longer pinned to 0	N/A
Common JEE setup	Sphere inside earthed shell	Isolated capacitor disconnect

VS Capacitor vs Resistor (combination rules)

Quantity	Resistor	Capacitor
Series	R_eq = ΣRᵢ	1/C_eq = Σ 1/Cᵢ
Parallel	1/R_eq = Σ 1/Rᵢ	C_eq = Σ Cᵢ
Common quantity in series	Current I	Charge Q
Common quantity in parallel	Voltage V	Voltage V
SI unit	ohm (V/A)	farad (C/V)
Stores	Nothing — dissipates as heat	Energy as field (½CV²)

Mnemonic: capacitors are 'upside-down' resistors in combination — series ↔ parallel formulas swap.

VS Force on a charge vs Force on a dipole

Property	Point charge q	Dipole p
In uniform E	F = qE ≠ 0	F = 0 (only torque τ = p × E)
In non-uniform E	F = qE (at that point)	F = (p·∇)E ≠ 0
Torque	0 (single point)	τ = pE sinθ
Potential energy	U = qV	U = −p·E = −pE cosθ
Stable equilibrium	None inside vacuum (Earnshaw)	p parallel to E (θ = 0)

VS Energy when two capacitors share charge

Scenario	V₁ = V₂ (same V)	V₁ ≠ V₂ (different V)
Final common voltage	V₁ (= V₂)	(C₁V₁ + C₂V₂)/(C₁ + C₂)
Energy lost	0	½·C₁C₂/(C₁+C₂)·(V₁ − V₂)²
Charge redistribution	None needed	Flows until V_common reached
Where energy goes	N/A	Heat in wire + EM radiation

Always non-zero loss when V's differ — even with a 'superconducting' wire (the radiation channel still exists).

🧠 50 Things to Memorise Before Attempting Questions

🔔Reminder: the questions below test these 50 facts, formulas and exceptions over and over. Do not start the MCQs or PYQs until you can recite each line cold — that one habit lifts your accuracy by 30–40 % and shaves seconds off every problem. Read once, cover the right side, recite, repeat.

A Potential, work and energy (1–10)

V at point charge: V = kq/r; k = 9×10⁹ N·m²/C². Sign of V follows sign of q.
V is scalar — add algebraically (with signs). Never use components. Midpoint of +q, −q ⇒ V = 0.
V at infinity = 0 by convention. So V(r) = work done per unit charge to bring +1 C from ∞ to r.
V − E link: E = −dV/dr. Larger slope of V ⇒ stronger field. E points along decreasing V.
Work to move q: W = q(V_f − V_i). Path-independent. Along equipotential, W = 0.
Energy of two charges: U = kq₁q₂/r. Negative for unlike (bound), positive for like.
System energy: sum over all unique pairs. n charges ⇒ n(n−1)/2 pair terms.
Accelerated charge: qV = ½mv² ⇒ v = √(2qV/m). KE gained = qV.
1 eV = 1.6×10⁻¹⁹ J. An electron through 1 V gains 1 eV. α through V gains 2 eV.
Coalescing drops: n equal drops of charge q, radius r ⇒ big drop V = n²/³ · (kq/r).

B Equipotentials & field-V relations (11–15)

E ⟂ equipotential always. Closer equipotentials ⇒ stronger E.
Two different equipotentials never intersect — single-valued V.
Shapes: point charge → spheres; uniform field → planes ⟂ field; dipole → distorted spheres + equatorial plane is V = 0.
Inside any conductor at equilibrium: E = 0, V = constant (one big equipotential).
E ≠ 0 doesn't mean V ≠ 0 and vice versa. Midpoint between +q and −q: V = 0, E ≠ 0. Inside shell: E = 0, V ≠ 0.

C Dipoles (16–20)

Dipole moment: p = q·2a, direction from −q to +q. Units C·m.
V on axis: V = ±kp/r² (r >> a). Sign by side.
V on equatorial plane = 0. But E there equals kp/r³ (anti-parallel to p).
Torque: τ = pE sinθ. Max at θ = 90°; zero at 0° (stable) and 180° (unstable).
PE: U = −pE cosθ. Min (−pE) at θ = 0; max (+pE) at θ = 180°.

D Capacitance formulas (21–28)

Definition: C = Q/V. Unit farad (F) = C/V.
Parallel plate, vacuum: C₀ = ε₀A/d; ε₀ = 8.854×10⁻¹² F/m.
Parallel plate, dielectric K: C = Kε₀A/d. Replace ε₀ with Kε₀ everywhere.
Isolated sphere: C = 4πε₀R. R = 1 m gives ≈ 111 pF.
Spherical capacitor: C = 4πε₀·ab/(b−a); a = inner, b = outer.
Cylindrical: C = 2πε₀L / ln(b/a).
Dielectric slab (K, thickness t, gap d): C = ε₀A/(d − t + t/K).
Conductor slab (thickness t, gap d): C = ε₀A/(d − t). Independent of slab position.

E Series, parallel & sharing (29–34)

Series: 1/C_s = Σ 1/Cᵢ; same Q on each; voltages add.
Parallel: C_p = ΣCᵢ; same V on each; charges add.
n equal in series: C/n; in parallel: nC. Ratio C_p/C_s = n².
Smaller C in a series takes the larger V (V = Q/C, Q same).
Sharing of charge: common potential V_c = (C₁V₁ + C₂V₂)/(C₁+C₂).
Energy lost in sharing: ΔU = ½ C₁C₂(V₁−V₂)²/(C₁+C₂). Zero only if V₁ = V₂.

F Dielectric — battery ON vs OFF (35–40)

Battery ON (V fixed): C×K, Q×K, U×K, E inside unchanged. Battery supplies extra Q and U.
Battery OFF (Q fixed): C×K, V÷K, U÷K, E inside ÷K. Energy decreases — work done by dielectric.
E inside dielectric: E = E₀/K (Q fixed). Polarisation creates bound surface charges.
Bound surface charge density: σ_b = σ(1 − 1/K).
Two K in parallel (half area each): C = ε₀A(K₁+K₂)/(2d).
Two K in series (half gap each): C = 2ε₀A·K₁K₂/[d(K₁+K₂)].

G Stored energy & density (41–45)

Three equal forms: U = ½QV = ½CV² = Q²/(2C). Pick whichever variable is fixed.
Energy density: u = ½ε₀E² (J/m³). With dielectric, multiply by K.
Battery work in charging: W_batt = QV = CV². Half stored, half dissipated — regardless of R.
Force per area on capacitor plate: P = ½ε₀E² = u (electrostatic pressure).
Self-energy of charged sphere: U = kQ²/(2R) (using C = 4πε₀R).

H Pitfalls & exceptions (46–50)

Two like charges released: bound problems aren't bound — they fly apart; final KE = initial U. Two unlike released ⇒ they collide; cannot reach infinity.
Pulling capacitor plates apart (Q fixed): U increases linearly with d. You did positive work against attraction.
Inside uniformly charged solid sphere (insulator): V = kQ(3R² − r²)/(2R³). V at centre = 3V_s/2.
Force on dipole in uniform field = 0 (only torque). Force only in non-uniform field.
Wheatstone bridge of capacitors: when balanced (C₁/C₂ = C₃/C₄), middle capacitor carries no charge — remove it before calculating.

4 High-Yield Practice MCQs

The simple idea

Try each question first, then open the solution to see the quickest route — using the formulas and shortcuts from above, not long textbook working.

10 NEET-level questions with numerically close options — 2 theory, 2 potential, 3 capacitors, 3 dielectric/energy. Tap "Show fastest solution".

TheoryQ1. An equipotential surface is one on which

Electric field is zero
Potential is the same everywhere
Electric flux is zero
No charge can be placed

Show fastest solution

Answer: (b)

By definition. Field is ⟂ to it; ΔV = 0 along it; no work to move charge along it.

TheoryQ2. Inside a uniformly charged spherical conducting shell:

E ≠ 0, V = 0
E = 0, V = 0
E = 0, V = constant (≠ 0)
E = constant, V = 0

Show fastest solution

Answer: (c)

Charge resides on outer surface ⇒ E inside = 0. V is continuous and equals its surface value kQ/R throughout the interior.

PotentialQ3. Two charges +5 μC and −5 μC are placed 10 cm apart. The electric potential at the midpoint is:

0
9 × 10⁵ V
4.5 × 10⁵ V
−9 × 10⁵ V

Show fastest solution

Answer: (a) 0

V is scalar: V = k(+5μC)/0.05 + k(−5μC)/0.05 = 0. (Note E is NOT zero there.)

PotentialQ4. A charged particle of charge q and mass m is accelerated from rest through a potential difference V. Its final speed is:

√(qV/m)
√(2qV/m)
qV/m
2qV/m

Show fastest solution

Answer: (b) √(2qV/m)

qV = ½mv² ⇒ v = √(2qV/m). A standard "voltage gun" formula.

CapacitorsQ5. Two capacitors 2 μF and 3 μF are in series across 10 V. Charge on each is:

6 μC
12 μC
20 μC
50 μC

Show fastest solution

Answer: (b) 12 μC

1/C_s = 1/2 + 1/3 = 5/6 ⇒ C_s = 1.2 μF. Q = C_sV = 1.2 × 10 = 12 μC, same on each (series).

CapacitorsQ6. Two parallel-plate capacitors of 2 μF and 4 μF are charged to 100 V and 50 V respectively, then connected with like plates together. The common potential is:

66.7 V
75 V
50 V
100 V

Show fastest solution

Answer: (a) 66.7 V

V_c = (C₁V₁ + C₂V₂)/(C₁+C₂) = (200 + 200)/6 = 66.67 V.

CapacitorsQ7. Four 4 μF capacitors are arranged: two in series, this combination in parallel with another series pair of two 4 μF capacitors. Equivalent capacitance is:

1 μF
2 μF
4 μF
8 μF

Show fastest solution

Answer: (c) 4 μF

Each series pair: 4·4/(4+4) = 2 μF. Two pairs in parallel: 2 + 2 = 4 μF.

Dielectric / EnergyQ8. A 10 μF capacitor is charged to 20 V by a battery, the battery is then disconnected, and a dielectric of K = 5 is inserted fully. New potential difference is:

20 V
4 V
100 V
5 V

Show fastest solution

Answer: (b) 4 V

Q fixed (battery off). V' = V/K = 20/5 = 4 V. Energy also drops 5×.

Dielectric / EnergyQ9. A parallel-plate capacitor has capacitance C₀. It is half-filled (along plate-area direction) with a dielectric of constant K. New capacitance is:

C₀(1+K)/2
2KC₀/(1+K)
KC₀
C₀

Show fastest solution

Answer: (a) C₀(1+K)/2

Filling half the area in parallel → two capacitors in parallel: C₁ = ε₀(A/2)/d = C₀/2; C₂ = Kε₀(A/2)/d = KC₀/2. Total = (1+K)C₀/2.

Dielectric / EnergyQ10. A 1 μF capacitor is connected to a 10 V battery. The work done by the battery in charging it is:

50 μJ
100 μJ
10 μJ
5 μJ

Show fastest solution

Answer: (b) 100 μJ

Battery work = QV = CV² = 10⁻⁶ × 100 = 100 μJ. Stored energy is half this (= 50 μJ); other 50 μJ is dissipated.

📝 PYQ Bank — 200 Past-Year Questions, Step-by-Step

How to attempt

Tap one option per question to lock your answer. When you're done, hit 🎯 Calculate Score. Marking scheme: +4 correct, −5 wrong, −4 unattempted. After scoring, every solution unlocks with full Given · To find · Formula · Solution.

Attempted0/200

Result

out of 800 · marking +4 / −5 / −4

Correct

0 × +4

Wrong

0 × −5

Left

0 × −4

Final marks

Scroll down — every question now shows your pick (red / green) plus the correct option. Open any card to read the full step-by-step solution. Click Reset in the bar above to try again.

Q1NEET 2017Easy2/10The electric potential at a point distant 0.1 m from a +5 μC point charge in vacuum is

A4.5×10⁵ V

B9×10⁵ V

C4.5×10⁴ V

D9×10⁴ V

Given

q = +5 μC = 5×10⁻⁶ Cr = 0.1 mvacuum (k = 9×10⁹ N·m²/C²)

To find

Electric potential V at distance r.

Formula

V = kq/r (scalar; positive sign retained).

Solution

1V = (9×10⁹)5×10⁻⁶0.1 = 4.5×10⁴×10 = 4.5×10⁵ V.

Answer

(A) · highlighted above in green

Q2NEET 2018Easy3/10Two point charges +q and −q are placed at the corners of a diagonal of a square of side a. The potential at the centre is

B2kq/(a√2)

Ckq/a

D2kq/a

Given

Two equal & opposite chargesdistance from centre to each corner = a/√2

To find

V at centre of the square.

Formula

V = Σ kqᵢ/rᵢ (scalar sum).

Solution

1V = k+qa/√2 + k−qa/√2 = 0.

2Equal magnitudes, opposite signs cancel.

Answer

(A) · highlighted above in green

Q3NEET 2019Easy3/10Work done in moving a charge of 4 C from a point at 20 V to a point at 60 V is

A80 J

B160 J

C240 J

D320 J

Given

q = 4 CV_A = 20 VV_B = 60 V (final)

To find

Work done by external agent (slowly moving the charge).

Formula

W_ext = q(V_B − V_A).

Solution

1W = 4×(60 − 20) = 4×40 = 160 J.

Answer

(B) · highlighted above in green

Q4NEET 2020Moderate4/10The potential at a point due to an electric dipole on its axial line at distance r (r >> 2a) is

Akp/r²

Bkp cosθ/r²

C2kp/r²

DZero

Given

Axial line of dipoler >> 2a (short-dipole approx)moment p

To find

Axial potential V.

Formula

V_axial = ±kp/r² (sign by side).

Solution

1Take the side toward +q: V = +kp/r².

2Magnitude kp/r².

Answer

(A) · highlighted above in green

Q5NEET 2021Moderate4/10The potential at every point on the equatorial plane of a dipole is

Akp/r²

B2kp/r²

Ckp/r³

DZero

Given

Equatorial plane of a dipole (perpendicular bisector at the midpoint).

To find

V on the equatorial plane.

Formula

V_equatorial = kp cosθr² with θ = 90°.

Solution

1cos 90° = 0 ⇒ V = 0.

2Note E ≠ 0 there (it equals kp/r³, anti-parallel to p).

Answer

(D) · highlighted above in green

Q6NEET 2016Hard6/10Two protons (charge e, mass m) are released from rest 1 nm apart. Maximum kinetic energy each acquires when far apart is

Ake²/r

Bke²/(2r)

C2ke²/r

Dke²/(4r)

Given

Two like charges of magnitude e initially separated by r = 1 nmreleased from restsymmetry: each gets same KE

To find

Maximum KE per proton at infinite separation.

Formula

Energy conservation: U_i = 2·KE_each (final).

Solution

1Initial PE = ke²/r; final PE = 0; so total KE = ke²/r ⇒ each = ke²2r.

Answer

(B) · highlighted above in green

Q7NEET 2014Moderate5/10The potential at the centre of a square of side L with charge +Q at each of its 4 corners is

AkQ/L

B4√2·kQ/L

C2√2·kQ/L

Given

4 equal charges +Q at cornersdistance from centre to corner = L/√2potential is scalar

To find

V at the centre.

Formula

V = Σ kQ/rᵢ (4 equal terms).

Solution

1V = 4·kQL/√2 = 4√2 · kQ/L.

Answer

(B) · highlighted above in green

Q8NEET 2015Hard6/10If 100 droplets, each of radius r and charge q, coalesce into one big drop, the potential on the surface of the big drop compared to each small drop is

A100×

B10×

C1000×

D100²/³ ×

Given

100 identical dropsvolume conserved ⇒ R = 100¹/³ rtotal charge Q = 100q

To find

Ratio V_big / V_small.

Formula

V_small = kq/r; V_big = kQ/R.

Solution

1V_big = k100q100¹/³ r = 100·100⁻¹/³ · kq/r = 100²/³ · V_small.

Answer

(D) · highlighted above in green

Q9NEET 2022Moderate5/10The work done in bringing a 5 μC charge from infinity to a point 0.5 m from a +10 μC charge is

A0.45 J

B0.9 J

C4.5 J

D9 J

Given

q = 5×10⁻⁶ CQ = 10×10⁻⁶ Cr = 0.5 mcharge comes from infinity (V_∞ = 0)

To find

W done by external agent.

Formula

W_ext = q·V at the point = q·(kQ/r).

Solution

1W = 5×10⁻⁶ × (9×10⁹ × 10×10⁻⁶/0.5) = 5×10⁻⁶ × 1.8×10⁵ = 0.9 J.

Answer

(B) · highlighted above in green

Q10NEET 2013Moderate4/10A point charge of 2 μC is placed at the origin. The work done to move a +1 μC charge from (2, 0, 0) to (0, 2, 0) m is

APositive

BNegative

CZero

DCannot be decided

Given

Both endpoints are at distance r = 2 m from the sourceV depends only on r

To find

Work done by external agent on +1 μC charge.

Formula

W = q(V_f − V_i); on a sphere of equal r the potential is the same.

Solution

1V(2,0,0) = V(0,2,0) = kQ/2.

2ΔV = 0 ⇒ W = 0.

3Both points lie on the same equipotential sphere.

Answer

Q11NEET 2018Easy3/10If E = 0 at every point in a small region of space, then in that region V is

Azero

Bconstant

C±∞

Dvariable

Given

E = −∇VE = 0 ⇒ derivative of V along all directions is zero

To find

Behaviour of V in that region.

Formula

If gradient of a scalar is zero ⇒ scalar is constant.

Solution

1V = constant (not necessarily zero — choice of reference).

Answer

(B) · highlighted above in green

Q12AIPMT 2011Moderate4/10An electric dipole of moment 4×10⁻⁹ C·m is placed in a uniform field 5×10⁴ N/C. The maximum torque is

A1×10⁻⁴ N·m

B2×10⁻⁴ N·m

C4×10⁻⁴ N·m

D2×10⁻³ N·m

Given

p = 4×10⁻⁹ C·mE = 5×10⁴ N/C

To find

Maximum torque on the dipole.

Formula

τ_max = pE (sinθ = 1 at θ = 90°).

Solution

1τ_max = 4×10⁻⁹ × 5×10⁴ = 2×10⁻⁴ N·m.

Answer

(B) · highlighted above in green

Q13NEET 2020Easy3/10An electron is moved through a potential of 1 kV. Energy gained is

A1 J

B1 keV

C1.6×10⁻¹⁹ J

D1 eV

Given

q = e (magnitude)V = 10³ Vwork-energy form

To find

Kinetic-energy gain in convenient units.

Formula

KE = qV. Definition: 1 eV = e × 1 V.

Solution

1KE = e × 10³ V = 10³ eV = 1 keV (= 1.6×10⁻¹⁶ J).

Answer

(B) · highlighted above in green

Q14NEET 2014Moderate4/10Two charges +Q and +Q are placed on a fixed axis at x = +a and x = −a. The potential on the y-axis at distance y is

A2kQ/√(a²+y²)

BkQ/√(a²+y²)

C2kQ/(a+y)

Given

Two identical charges symmetrically placedthe y-axis is equidistant from both

To find

V at point (0, y, 0).

Formula

V = Σ kQ/rᵢ; r₁ = r₂ = √(a² + y²).

Solution

1V = 2 · kQ/√(a² + y²).

Answer

(A) · highlighted above in green

Q15NEET 2019Moderate5/10The potential due to a uniformly charged solid sphere of radius R at its centre, in terms of the surface potential V_s, is

AV_s/2

B3V_s/2

CV_s

D2V_s

Given

Solid sphere (uniform ρ), total charge Qsurface potential V_s = kQ/R

To find

V at the centre (r = 0).

Formula

V_inside(r) = kQ3R² − r²2R³; at r = 0 gives V₀ = 3kQ/(2R).

Solution

1V₀ = (3/2)(kQ/R) = (3/2)V_s.

2Inside is higher than the surface.

Answer

(B) · highlighted above in green

Q16NEET 2017Easy2/10Equipotential surfaces of a single point charge are

Aparallel planes

Bconcentric spheres

Cconcentric circles

Dcoaxial cylinders

Given

Single point chargeV = kq/r depends only on r

To find

Geometry of constant-V surfaces.

Formula

Surfaces of V = constant ⇒ r = constant.

Solution

1r = constant gives concentric spheres around the charge.

Answer

(B) · highlighted above in green

Q17NEET 2018Easy3/10Inside a hollow uniformly charged spherical conducting shell of radius R

AE ≠ 0, V = 0

BE = 0, V = 0

CE = 0, V = kQ/R

DE and V both vary

Given

Hollow spherical conductor charge Q on surface.

To find

E and V at any interior point.

Formula

Inside conductor: E = 0; V continuous and constant = surface value kQ/R.

Solution

1E = 0 inside; V = kQ/R everywhere inside (a single equipotential).

Answer

Q18NEET 2016Easy3/10Electric field lines and equipotential surfaces are always

Aparallel

Bat 45°

Cperpendicular

Dintersect at any angle

Given

Property of conservative electric field.

To find

Angle between E lines and equipotentials.

Formula

E points along the direction of maximum −dV/dr; along an equipotential dV = 0 ⇒ E ⟂ surface.

Solution

1E ⟂ equipotential everywhere.

Answer

Q19NEET 2021Easy2/10Work done in moving 2 C of charge along an equipotential surface is

AZero

B2 J

CDepends on length

DDepends on charge

Given

Equipotential ⇒ ΔV = 0 between any two points on it.

To find

Work done on the charge.

Formula

W = q·ΔV.

Solution

1W = 2·0 = 0.

Answer

(A) · highlighted above in green

Q20NEET 2015Moderate4/10Two equipotential surfaces with potentials V₁ and V₂ (V₂ > V₁) are separated by 10 cm. If the field between them is 200 V/m, then V₂ − V₁ equals

A10 V

B20 V

C200 V

D0.5 V

Given

d = 0.1 muniform field magnitude E = 200 V/m

To find

Potential difference V₂ − V₁.

Formula

E = ΔV/d (uniform field).

Solution

1ΔV = E·d = 200×0.1 = 20 V.

Answer

(B) · highlighted above in green

Q21AIPMT 2012Easy3/10Which of the following is/are scalar?

AElectric field

BElectric potential

CForce

DAcceleration

Given

Definitions of vector vs scalar quantities in electrostatics.

To find

Which item is scalar.

Formula

Potential is defined as work per unit charge → has only magnitude.

Solution

1Electric potential is scalar; the others are vectors.

Answer

(B) · highlighted above in green

Q22NEET 2019Easy3/10Inside a charged conductor in electrostatic equilibrium

AE ≠ 0, V varies

BE = 0, V varies

CE = 0, V = constant

DE ≠ 0, V = constant

Given

Charged conductor at equilibrium.

To find

Relationship between E and V inside.

Formula

Free charges rearrange till E = 0 inside ⇒ V is constant (single equipotential).

Solution

1E = 0 and V = constant throughout the body.

Answer

Q23NEET 2020Easy3/10Two equipotential surfaces having different potentials

Anever intersect

Bintersect at infinity

Cintersect at 90°

Dintersect at point

Given

Property of single-valued potential.

To find

Whether they can cross.

Formula

At intersection a point would have two different V values — impossible.

Solution

1Equipotentials of different V never intersect.

Answer

(A) · highlighted above in green

Q24NEET 2017Easy3/10Equipotential surfaces of a uniform electric field are

Aconcentric spheres

Bparallel planes ⟂ to field

Cparallel planes along field

Dcylinders

Given

Uniform field E in +xV = −Ex + const

To find

Shape of V = constant surfaces.

Formula

V = constant ⇒ x = constant ⇒ a plane ⟂ to x-axis (field direction).

Solution

1Parallel planes ⟂ to the field.

Answer

(B) · highlighted above in green

Q25NEET 2013Moderate5/10If electric potential is V = 4x² volt, the electric field at x = 1 m is

A+8 V/m

B−8 V/m

C+4 V/m

DZero

Given

V(x) = 4x²need E at x = 1 m

To find

Magnitude and direction of E along x.

Formula

E_x = −dV/dx.

Solution

1dV/dx = 8x ⇒ E_x = −8x = −8 V/m at x = 1 m.

Answer

(B) · highlighted above in green

Q26NEET 2018Moderate5/10PE of a system of three identical charges +q at the vertices of an equilateral triangle of side a is

Akq²/a

B2kq²/a

C3kq²/a

D6kq²/a

Given

Three identical pairwise interactions, each at distance a.

To find

Total electrostatic PE.

Formula

U = Σ kqᵢqⱼ/rᵢⱼ for all unique pairs.

Solution

1Pairs: (1-2), (1-3), (2-3) — each kq²/a; total = 3kq²/a.

Answer

Q27NEET 2016Moderate4/10Two charges +2 μC and −2 μC at distance 0.5 m. PE of the system is

A−0.072 J

B+0.072 J

C−7.2 J

D7.2 J

Given

q₁ = +2 μC, q₂ = −2 μCr = 0.5 m

To find

Electrostatic PE.

Formula

U = kq₁q₂/r.

Solution

1U = (9×10⁹)(2×10⁻⁶)(−2×10⁻⁶)/0.5 = −0.072 J.

2Negative ⇒ attractive.

Answer

(A) · highlighted above in green

Q28NEET 2019Moderate5/10A dipole of moment p is placed in a uniform field E. Work done in turning it from θ = 0 to θ = π/2 is

ApE

B−pE

CpE/2

DZero

Given

U(θ) = −pE cosθ.

To find

W = ΔU between θ = 0 and θ = π/2.

Formula

W = U_f − U_i = (−pE cos90°) − (−pE cos0°).

Solution

1W = 0 − (−pE) = pE.

2Positive (you do work to misalign).

Answer

(A) · highlighted above in green

Q29NEET 2021Moderate4/10A dipole p in field E. Maximum PE corresponds to

Aθ = 0

Bθ = π/2

Cθ = π

Dθ = π/3

Given

U = −pE cosθcos has max +1 at 0 and min −1 at π

To find

Position of unstable equilibrium / max U.

Formula

U_max when cosθ is most negative ⇒ θ = π (anti-aligned).

Solution

1U_max = +pE at θ = π (unstable equilibrium).

Answer

Q30NEET 2014Moderate4/10An α-particle and a proton are accelerated through the same potential. Ratio of their KE is

A1:1

B1:2

C2:1

D4:1

Given

qV gives KEq_α = 2e, q_p = esame V

To find

Ratio KE_α : KE_p.

Formula

KE = qV.

Solution

1KE_α/KE_p = q_α/q_p = 2.

2Ratio 2 : 1.

Answer

Q31AIPMT 2010Moderate5/10Energy needed to assemble three charges +q, +q, −q at the vertices of an equilateral triangle (side a) is

A−kq²/a

B−3kq²/a

C+kq²/a

DZero

Given

Three chargespair-energies: (+,+): +kq²/a(+,−): −kq²/a(+,−): −kq²/a

To find

Total electrostatic PE.

Formula

U = Σ over unique pairs.

Solution

1U = (+kq²/a) + (−kq²/a) + (−kq²/a) = −kq²/a.

Answer

(A) · highlighted above in green

Q32NEET 2017Moderate5/10An electron starts from rest in a uniform field E = 100 V/m. KE gained after travelling 1 cm is

A1.6×10⁻¹⁹ J

B1.6×10⁻¹⁸ J

C1.6×10⁻¹⁷ J

D1.6×10⁻²⁰ J

Given

q = e = 1.6×10⁻¹⁹ CE = 100 V/md = 10⁻² m

To find

KE after travelling d.

Formula

KE = qEd (work done by field = qEd).

Solution

1KE = 1.6×10⁻¹⁹ × 100 × 10⁻² = 1.6×10⁻¹⁹ J = 1 eV.

Answer

(A) · highlighted above in green

Q33NEET 2022Easy3/10The work done in moving a charge q from a point of potential V₁ to potential V₂ depends on

Athe path

Bonly on V₁−V₂

Cspeed of charge

Dtime taken

Given

Electrostatic field is conservative.

To find

On what does W depend.

Formula

W = q(V₁ − V₂); path-independent.

Solution

1Only on potential difference (and charge q).

Answer

(B) · highlighted above in green

Q34NEET 2017Easy3/10A 4 μF capacitor is charged to a potential of 200 V. The charge on it is

A4×10⁻⁴ C

B8×10⁻⁴ C

C4×10⁻² C

D800 μC

Given

C = 4 μF = 4×10⁻⁶ FV = 200 V

To find

Charge Q.

Formula

Q = CV.

Solution

1Q = 4×10⁻⁶ × 200 = 8×10⁻⁴ C = 800 μC.

Answer

(D) · highlighted above in green

Q35NEET 2018Easy2/10The capacitance of a parallel-plate capacitor of plate area A and separation d (vacuum) is given by

Aε₀d/A

Bε₀A/d

Cε₀A·d

DA/(ε₀d)

Given

Standard parallel-plate definition.

To find

Expression for C in vacuum.

Formula

C = ε₀A/d (derived from Q = CV and E = σ/ε₀, V = Ed).

Solution

1C = ε₀A/d.

Answer

(B) · highlighted above in green

Q36NEET 2019Moderate4/10If the plate area of a parallel-plate capacitor is doubled and separation halved, the new capacitance becomes

A2C

B4C

CC/2

DC/4

Given

C₀ = ε₀A/dnew A' = 2A, d' = d/2

To find

New C in terms of C₀.

Formula

C ∝ A/d.

Solution

1C' = ε₀2Ad/2 = 4·ε₀A/d = 4C.

Answer

(B) · highlighted above in green

Q37NEET 2020Moderate5/10An isolated conducting sphere of radius 0.1 m has capacitance

A1.11 pF

B11.1 pF

C111 pF

D1.11 nF

Given

Isolated sphere of radius R = 0.1 mε₀ = 8.854×10⁻¹² F/m

To find

Capacitance C.

Formula

C = 4πε₀R.

Solution

1C = 4π(8.854×10⁻¹²)(0.1) ≈ 1.11×10⁻¹¹ F ≈ 11.1 pF.

Answer

(B) · highlighted above in green

Q38NEET 2021Easy2/10Two capacitors of capacitance 4 μF and 6 μF are in parallel. Equivalent C is

A2.4 μF

B4 μF

C6 μF

D10 μF

Given

Parallel combinationC₁ = 4 μF, C₂ = 6 μF

To find

Equivalent capacitance.

Formula

C_p = C₁ + C₂.

Solution

1C_p = 4 + 6 = 10 μF. (Same V, charges add.)

Answer

(D) · highlighted above in green

Q39NEET 2016Moderate4/10Two capacitors of 4 μF and 6 μF are in series. Equivalent C is

A2.4 μF

B10 μF

C1.5 μF

D24 μF

Given

Series combinationC₁ = 4 μF, C₂ = 6 μF

To find

Equivalent capacitance.

Formula

1/C_s = 1/C₁ + 1/C₂.

Solution

11/C_s = 1/4 + 1/6 = 5/12 ⇒ C_s = 12/5 = 2.4 μF.

Answer

(A) · highlighted above in green

Q40AIPMT 2015Easy3/10A capacitor of capacitance C is charged to a potential V. It stores energy

A½CV

BCV²

C½CV²

DQ²/C

Given

C, Vstandard energy formula

To find

Energy stored.

Formula

U = ½QV = ½CV² = Q²2C.

Solution

1U = ½CV².

Answer

Q41NEET 2017Moderate5/10If the separation between plates of a charged parallel-plate capacitor (battery disconnected) is doubled, the energy stored

Ahalves

Bdoubles

Cquadruples

Dunchanged

Given

Q fixed (isolated)C = ε₀A/d, so C halves if d doubles

To find

Ratio U_new/U_old.

Formula

U = Q²2C. With Q fixed, U ∝ 1/C ∝ d.

Solution

1Doubling d halves C ⇒ U doubles.

Answer

(B) · highlighted above in green

Q42NEET 2019Easy3/10A 1 μF capacitor is charged to 100 V. Energy stored is

A5×10⁻³ J

B5×10⁻⁴ J

C5×10⁻⁵ J

D0.5 J

Given

C = 1×10⁻⁶ FV = 100 V

To find

Energy U.

Formula

U = ½CV².

Solution

1U = 0.5 × 10⁻⁶ × 10⁴ = 5×10⁻³ J.

Answer

(A) · highlighted above in green

Q43NEET 2018Moderate5/10Two capacitors 3 μF and 6 μF are connected in series across 18 V. Charge on each is

A18 μC

B36 μC

C54 μC

D12 μC

Given

Series ⇒ same Q on eachC₁ = 3 μF, C₂ = 6 μFV = 18 V

To find

Charge on each capacitor.

Formula

C_s = C₁C₂C₁+C₂; Q = C_s V.

Solution

1C_s = 18/9 = 2 μF ⇒ Q = 2×18 = 36 μC.

2Same charge on each (series).

Answer

(B) · highlighted above in green

Q44NEET 2020Moderate4/10Three capacitors of 2 μF, 3 μF, 6 μF are in series. Equivalent C is

A1 μF

B11 μF

C0.5 μF

D3 μF

Given

Series1/C_s = 1/2 + 1/3 + 1/6

To find

Equivalent capacitance.

Formula

1/C_s = sum of reciprocals.

Solution

11/C_s = (3 + 2 + 1)/6 = 1 ⇒ C_s = 1 μF.

Answer

(A) · highlighted above in green

Q45AIPMT 2011Moderate4/10If C₁ and C₂ are in parallel, C_p = C₁ + C₂; if in series, C_s = C₁C₂/(C₁+C₂). For two equal capacitors C, the ratio C_p/C_s is

D1/2

Given

Equal capacitors: C_p = 2C, C_s = C/2.

To find

Ratio C_p/C_s.

Formula

Series of two equals halves; parallel doubles.

Solution

1C_p/C_s = 2C / (C/2) = 4.

Answer

Q46NEET 2022Hard7/10In a Wheatstone-like network, four equal 2 μF capacitors form a square; a 5th 2 μF connects across the diagonal between balanced corners. Equivalent C between the other diagonal corners is

A2 μF

B4 μF

C1 μF

D8 μF

Given

Balanced bridge: middle capacitor carries no charge (no PD across it).

To find

Equivalent capacitance across the input diagonal.

Formula

Remove the middle capacitor (no current/charge); two 2 μF in series form 1 μF, two such combos in parallel.

Solution

1Each branch: 2·2/(2+2) = 1 μF; parallel of two such branches = 2 μF.

Answer

(A) · highlighted above in green

Q47NEET 2017Moderate5/10Two parallel-plate capacitors with capacitances C₁ = 2 μF and C₂ = 4 μF are charged to 100 V and 50 V respectively, then connected by like plates. Common potential is

A66.7 V

B75 V

C50 V

D25 V

Given

C₁V₁ + C₂V₂ = total charge (preserved).

To find

Common potential V_c after sharing.

Formula

V_c = C₁V₁ + C₂V₂C₁ + C₂.

Solution

1V_c = (200 + 200)/6 = 66.67 V.

Answer

(A) · highlighted above in green

Q48NEET 2018Moderate5/10In the previous sharing, energy lost is

AZero

B½C₁V₁²+½C₂V₂² − ½(C₁+C₂)V_c²

C½(V₁−V₂)²

DC₁C₂(V₁−V₂)²

Given

Energy before = ½C₁V₁² + ½C₂V₂²after = ½(C₁+C₂)V_c²

To find

Loss in energy.

Formula

ΔU = U_before − U_after = ½C₁C₂V₁−V₂C₁+C₂^².

Solution

1ΔU = ½(2)(4)(50)²/6 ≈ 1.67 mJ; algebraic form is option (b).

Answer

(B) · highlighted above in green

Q49NEET 2019Moderate5/10Four 5 μF capacitors are connected: two in series (= a) and the other two in series (= b), then a and b in parallel. Equivalent C is

A1.25 μF

B5 μF

C2.5 μF

D10 μF

Given

Each series pair = 5·5/(5+5) = 2.5 μFtwo such in parallel

To find

Equivalent capacitance.

Formula

Series first then parallel.

Solution

12.5 + 2.5 = 5 μF.

Answer

(B) · highlighted above in green

Q50NEET 2014Moderate4/10Capacitance of n identical capacitors of C each: maximum value (parallel) and minimum value (series) are

AnC, C/n

BC/n, nC

Cn²C, C

DC, C/n²

Given

All parallel ⇒ C_p = nC (max). All series ⇒ C_s = C/n (min).

To find

Range of equivalent capacitance.

Formula

Parallel adds C; series adds 1/C.

Solution

1Max nC, min C/n.

Answer

(A) · highlighted above in green

Q51AIPMT 2010Moderate5/10Three 1 μF capacitors with two in parallel and that combination in series with the third. Equivalent C is

A2/3 μF

B1.5 μF

C3 μF

D1/3 μF

Given

Parallel pair = 1+1 = 2 μFseries with 1 μF

To find

C_eq.

Formula

Two equal in parallel double; one in series uses reciprocal sum.

Solution

11/C_eq = 1/2 + 1/1 = 3/2 ⇒ C_eq = 2/3 μF.

Answer

(A) · highlighted above in green

Q52NEET 2020Moderate5/10Across 10 V, two 4 μF capacitors in series store energy

A100 μJ

B200 μJ

C50 μJ

D25 μJ

Given

C_s = 2 μFV = 10 V

To find

Total stored energy in the combination.

Formula

U = ½ C_s V².

Solution

1U = 0.5 × 2×10⁻⁶ × 100 = 1×10⁻⁴ J = 100 μJ.

Answer

(A) · highlighted above in green

Q53NEET 2021Moderate5/10Across 10 V, two 4 μF in parallel store

A100 μJ

B400 μJ

C200 μJ

D50 μJ

Given

C_p = 8 μFV = 10 V

To find

Total energy.

Formula

U = ½ C_p V².

Solution

1U = 0.5 × 8×10⁻⁶ × 100 = 4×10⁻⁴ J = 400 μJ.

Answer

(B) · highlighted above in green

Q54NEET 2015Moderate5/10Four 6 μF capacitors connect to make a Wheatstone-like square (no bridge). Across the diagonal of length 6 μF–6 μF–6 μF–6 μF the equivalent is

A6 μF

B3 μF

C12 μF

D1.5 μF

Given

Two arms in series (3 μF each), then two such arms in parallel.

To find

C between input corners.

Formula

C_arm = 6·6/12 = 3 μF; two arms in parallel = 6 μF.

Solution

1Equivalent = 6 μF.

Answer

(A) · highlighted above in green

Q55NEET 2016Moderate4/10Two capacitors with same Q but different C are joined in series. The voltage drop is greater across

Alarger C

Bsmaller C

Cequal

Ddepends on connection

Given

Series ⇒ Q commonV = Q/C

To find

Which capacitor sees larger V.

Formula

Smaller C ⇒ larger V for the same Q.

Solution

1V is larger across the smaller capacitor (V ∝ 1/C).

Answer

(B) · highlighted above in green

Q56NEET 2017Moderate4/10A dielectric slab (K = 4) is inserted between plates of a capacitor while battery is connected. New capacitance compared with C₀ is

AC₀

B4C₀

CC₀/4

D2C₀

Given

Battery connected (V fixed). K = 4.

To find

C_new.

Formula

C with dielectric = KC₀.

Solution

1C_new = 4C₀.

Answer

(B) · highlighted above in green

Q57NEET 2018Moderate4/10If in the above problem, battery is disconnected before inserting the dielectric, the voltage across plates becomes

A4V

BV/4

D16V

Given

Q fixed (isolated). C → KC.

To find

New voltage V'.

Formula

Q = CV; V' = QKC = V/K.

Solution

1V' = V/4.

Answer

(B) · highlighted above in green

Q58NEET 2019Moderate4/10When dielectric is inserted with battery connected, energy stored

Aincreases by factor K

Bdecreases

Cunchanged

Dincreases by K²

Given

V fixedC → KC

To find

Ratio U'/U.

Formula

U = ½CV².

Solution

1U' = ½(KC)V² = K·U.

2Battery supplies extra energy.

Answer

(A) · highlighted above in green

Q59NEET 2020Moderate4/10When dielectric inserted with battery disconnected, energy stored

A×K

B/K

Cunchanged

D/K²

Given

Q fixedC → KC

To find

U'/U.

Formula

U = Q²2C.

Solution

1U' = Q²2KC = U/K.

2Energy decreases (work done by dielectric).

Answer

(B) · highlighted above in green

Q60NEET 2016Hard6/10An insulator with dielectric constant K is inserted partially (length x) in a parallel-plate capacitor (length L, gap d). Capacitance is

Aε₀L/d

Bε₀(L − x + Kx)/d

Cε₀(Lx + K)/d

Dε₀Kx/d

Given

Two parallel capacitors: width x with Kwidth (L − x) withoutboth gap d

To find

Total C.

Formula

C = C₁ + C₂ (parallel).

Solution

1C = ε₀x·K/d + ε₀L − xd = ε₀L − x + Kxd (per unit depth).

Answer

(B) · highlighted above in green

Q61NEET 2021Easy3/10If K of a dielectric placed inside a parallel-plate capacitor is increased, capacitance

A↑

B↓

Cconstant

D→ 0

Given

C = Kε₀A/d.

To find

Behaviour of C with K.

Formula

Linear dependence on K.

Solution

1C increases linearly with K.

Answer

(A) · highlighted above in green

Q62NEET 2014Moderate4/10Field between plates of a parallel-plate capacitor with dielectric (battery disconnected) becomes

AE₀

BKE₀

CE₀/K

Given

Q fixedbound surface charges reduce free E by factor K

To find

E inside dielectric vs original E₀.

Formula

E_in = E₀/K (induced polarization opposes).

Solution

1E inside = E₀/K.

Answer

Q63AIPMT 2013Moderate5/10Polarisation of a dielectric is

A+

B⟂ to E

Given

P = χₑ ε₀ E for linear isotropic dielectrics.

To find

Direction & magnitude proportionality.

Formula

P ∥ E and |P| ∝ |E|.

Solution

1Both (a) and (c) — proportional to E and along E.

Answer

(D) · highlighted above in green

Q64NEET 2018Moderate5/10A conducting slab of thickness t is inserted into a parallel-plate capacitor (gap d). New capacitance is

Aε₀A/(d−t)

Bε₀A/d

Cε₀A/(d+t)

Dε₀At/d

Given

Treat conductor as K → ∞ slab.

To find

New capacitance.

Formula

C = ε₀Ad − t + t/K. For conductor K → ∞ ⇒ t/K → 0.

Solution

1C = ε₀Ad − t.

2Independent of slab's position (between plates).

Answer

(A) · highlighted above in green

Q65NEET 2019Moderate5/10A 6 μF capacitor charged to 100 V is connected across an uncharged 3 μF capacitor. Final charge on the 3 μF is

A200 μC

B100 μC

C300 μC

D66.7 μC

Given

Q_initial = 6×100 = 600 μCtotal = 600 μCshares according to capacitance

To find

Final Q on 3 μF capacitor.

Formula

V_c = Q_total/C_total; then Q = CV.

Solution

1V_c = 600/(6+3) = 66.67 V; Q(3 μF) = 3×66.67 ≈ 200 μC.

Answer

(A) · highlighted above in green

Q66NEET 2020Moderate5/10If two dielectrics (K₁, K₂) each fill half the plate area (in parallel arrangement) of a capacitor (plate area A, gap d), C is

Aε₀A(K₁+K₂)/(2d)

B2ε₀AK₁K₂/(d(K₁+K₂))

Cε₀A·K₁/d

Dε₀A(K₁−K₂)/d

Given

Half-area in parallel = two capacitors C₁, C₂ in paralleleach plate area A/2

To find

Net C.

Formula

Parallel: C = C₁ + C₂.

Solution

1C = ε₀(A/2)K₁/d + ε₀(A/2)K₂/d = ε₀AK₁+K₂2d.

Answer

(A) · highlighted above in green

Q67NEET 2015Moderate5/10If two dielectrics (K₁, K₂) each fill half the gap (in series arrangement) of a capacitor (plate area A, gap d), C is

A2ε₀AK₁K₂/(d(K₁+K₂))

Bε₀A(K₁+K₂)/(2d)

Cε₀AK₁K₂/d

Dε₀A(K₁−K₂)/d

Given

Half-gap in series = two capacitors C₁, C₂ in serieseach gap d/2

To find

Net C.

Formula

Series: 1/C = 1/C₁ + 1/C₂.

Solution

1C = 2ε₀A·K₁K₂/(d(K₁+K₂)).

Answer

(A) · highlighted above in green

Q68NEET 2017Moderate4/10If C, Q, V denote capacitance, charge and voltage, then which of the following is NOT correct?

AU = ½QV

BU = ½CV²

CU = Q²/2C

DU = QV

Given

Energy stored formulas.

To find

Identify the wrong relation.

Formula

Correct ones: ½QV, ½CV², Q²2C.

Solution

1QV is the work done by the battery, not the stored energy. (a)(b)(c) are right; (d) is wrong.

Answer

(D) · highlighted above in green

Q69NEET 2018Moderate4/10Work done by the battery in charging a 1 μF capacitor to 10 V is

A50 μJ

B100 μJ

C200 μJ

D25 μJ

Given

C = 1 μF, V = 10 V.

To find

Battery's work W_batt.

Formula

W_batt = QV = CV² = 2·U.

Solution

1W_batt = 1×10⁻⁶ × 100 = 1×10⁻⁴ J = 100 μJ.

2Stored U = 50 μJ; remaining 50 μJ dissipated.

Answer

(B) · highlighted above in green

Q70NEET 2019Easy3/10Energy density in vacuum between plates with field E is

Aε₀E

B½ε₀E²

Cε₀E²

DE²/2

Given

Standard formula for electromagnetic energy density in static field.

To find

Energy per unit volume u.

Formula

u = ½ε₀E².

Solution

1u = ½ε₀E²; with dielectric replace ε₀ by Kε₀.

Answer

(B) · highlighted above in green

Q71NEET 2020Moderate4/10If U is energy stored in a capacitor with charge Q. If Q is doubled, U becomes

A2U

B4U

CU/2

DU/4

Given

U = Q²/(2C)C constant

To find

U(2Q)/U(Q).

Formula

U ∝ Q².

Solution

1Doubling Q quadruples U.

Answer

(B) · highlighted above in green

Q72NEET 2021Easy3/10Energy stored in a capacitor depends on

Aonly its C

Bonly its V

Cboth C and V

Donly Q

Given

U = ½CV² requires both.

To find

Functional dependence.

Formula

Two of three: any two of (Q, V, C) determine U.

Solution

1Energy depends on both C and V.

Answer

Q73AIPMT 2013Moderate5/10Energy stored per unit volume in a parallel-plate capacitor (vacuum) is

A½ε₀(V/d)²

Bε₀V²

C(V/d)²

D½ε₀E

Given

E = V/d between plates.

To find

Energy density u.

Formula

u = ½ε₀E² = ½ε₀(V/d)².

Solution

1u = ½ε₀(V/d)².

Answer

(A) · highlighted above in green

Q74NEET 2018Moderate5/10After a capacitor is charged and disconnected from battery, if the plate separation is doubled, energy

Ahalves

Bdoubles

Cquadruples

Dunchanged

Given

Q fixedC halves with doubled d

To find

New U / old U.

Formula

U = Q²2C ∝ d.

Solution

1U doubles (you did positive work to pull plates apart against attraction).

Answer

(B) · highlighted above in green

Q75NEET 2015Easy2/10In a 2 μF capacitor charged to 100 V, the charge on the +ve plate is

A200 μC

B100 μC

C50 μC

D400 μC

Given

C = 2 μF, V = 100 V.

To find

Charge Q.

Formula

Q = CV.

Solution

1Q = 2×10⁻⁶ × 100 = 2×10⁻⁴ C = 200 μC.

Answer

(A) · highlighted above in green

Q76NEET 2015Hard6/10A 10 μF capacitor charged to 100 V is connected in parallel with an uncharged 5 μF capacitor. The energy lost is

A16.67 mJ

B50 mJ

C33.3 mJ

D25 mJ

Given

C₁ = 10 μF, V₁ = 100 VC₂ = 5 μF, V₂ = 0

To find

Energy lost ΔU during sharing.

Formula

ΔU = ½·C₁C₂V₁−V₂C₁+C₂^².

Solution

1ΔU = 0.5×10×515 × 100² × 10⁻⁶ = 0.5 × (50/15) × 10⁴ × 10⁻⁶ = 1.667×10⁻² J ≈ 16.67 mJ. (Dissipated as heat/EM radiation.)

Answer

(A) · highlighted above in green

Q77NEET 2014Hard7/10Three identical point charges +q are placed at the vertices of an equilateral triangle of side a. The work done in slowly bringing a fourth charge +q from infinity to the centroid is

A3kq²/a

B√3·kq²/a

C3√3·kq²/a

D6kq²/a

Given

Distance from centroid to each vertex = a/√3.

To find

W to bring +q from ∞ to centroid.

Formula

W_ext = q · V_centroid where V_centroid = Σ kqa/√3 = 3·kq·√3/a.

Solution

1V_centroid = 3kq·√3/a = 3√3·kq/a.

2W = q·V = 3√3 · kq²/a.

Answer

Q78NEET 2017Hard6/10Three +q charges are at the vertices of an equilateral triangle of side a. The magnitude of the electric field at the centroid is

A3kq/a²

B√3·kq/a²

CZero

D9kq/a²

Given

By symmetry, the three field vectors at the centroid are equal in magnitude and 120° apart.

To find

Net E at centroid.

Formula

Vector sum of three equal vectors at 120° apart.

Solution

1Three equal vectors at 120° cancel ⇒ E = 0. (Note: potential V is non-zero there, equal to 3√3·kq/a.)

Answer

Q79NEET 2018Moderate5/10An electric dipole p in a uniform field E is rotated from a position of maximum stable equilibrium to one of unstable equilibrium. Work done on the dipole is

ApE

B2pE

CpE/2

DZero

Given

Stable: θ = 0°, U₁ = −pE. Unstable: θ = 180°, U₂ = +pE.

To find

Work done on the dipole.

Formula

W_ext = ΔU = U₂ − U₁.

Solution

1W = (+pE) − (−pE) = 2pE.

Answer

(B) · highlighted above in green

Q80NEET 2020Hard6/10A spherical capacitor has inner radius 0.05 m, outer 0.10 m (vacuum). Its capacitance is

A11.1 pF

B22.2 pF

C5.55 pF

D33.3 pF

Given

a = 0.05, b = 0.10 mvacuum

To find

Capacitance C.

Formula

C = 4πε₀·abb−a = ab/[k(b−a)].

Solution

1C = 0.05×0.109×10⁹ × 0.05 = 5×10⁻³/4.5×10⁸ ≈ 1.11×10⁻¹¹ F ≈ 11.1 pF.

Answer

(A) · highlighted above in green

Q81NEET 2021Moderate5/10Two capacitors C₁ and C₂ are charged to potentials V₁ and V₂ and connected with opposite plates together. Common potential is

A(C₁V₁+C₂V₂)/(C₁+C₂)

B(C₁V₁−C₂V₂)/(C₁+C₂)

CZero

D(V₁+V₂)/2

Given

Opposite-polarity connection ⇒ charges partially cancelnet Q = C₁V₁ − C₂V₂

To find

Common potential after connection.

Formula

V_c = Q_netC₁+C₂.

Solution

1V_c = C₁V₁ − C₂V₂C₁+C₂. (Sign of V_c can be negative if C₂V₂ > C₁V₁.)

Answer

(B) · highlighted above in green

Q82NEET 2016Hard7/10A parallel-plate capacitor (C₀, V) has a dielectric slab K = 4 slowly inserted with the battery on. Work done by the battery during insertion is

A3C₀V²

B4C₀V²

C2C₀V²

DC₀V²

Given

Initial Q = C₀Vfinal Q = 4C₀V. ΔQ = 3C₀V flows through battery at constant V

To find

Work done by the battery W_batt.

Formula

W_batt = V·ΔQ.

Solution

1W_batt = V × 3C₀V = 3C₀V². (Of this, ½·3C₀V² = 1.5C₀V² becomes added energy stored; 1.5C₀V² is lost as heat/work done against the dielectric.)

Answer

(A) · highlighted above in green

Q83NEET 2019Moderate5/10Two infinite parallel sheets carry surface charge densities +σ and −σ. The potential difference between them, when separated by d, is

Aσd/ε₀

Bσ/(2ε₀)

C2σd/ε₀

Given

E_between = σ/ε₀ (uniform).

To find

Potential difference V between sheets.

Formula

ΔV = E·d for uniform field.

Solution

1ΔV = (σ/ε₀)·d = σd/ε₀.

Answer

(A) · highlighted above in green

Q84NEET 2014Moderate5/10A 4 μF capacitor connected to a 100 V battery is removed and connected to a 2 μF uncharged capacitor. Final voltage across the combination is

A33.3 V

B66.7 V

C50 V

D100 V

Given

Q_initial = 4×100 = 400 μCtotal C = 4+2 = 6 μFcharge is conserved

To find

Final common potential.

Formula

V_c = Q_totalC₁+C₂.

Solution

1V_c = 400/6 ≈ 66.67 V.

Answer

(B) · highlighted above in green

Q85NEET 2017Hard6/10If an electron and a proton are accelerated from rest through the same potential difference V, the ratio of their final speeds (v_e/v_p) is

A√(m_p/m_e)

B√(m_e/m_p)

Dm_p/m_e

Given

Same q (magnitude e)same VKE = qV = ½mv² ⇒ v = √(2qV/m)

To find

Ratio v_e / v_p.

Formula

v ∝ 1/√m.

Solution

1v_e/v_p = √(m_p/m_e) ≈ √1836 ≈ 42.8.

Answer

(A) · highlighted above in green

Q86NEET 2018Moderate4/10Electric potential at any point inside a uniformly charged spherical shell of total charge Q and radius R is

BkQ/R

CkQ/r

DDecreases linearly with r

Given

Inside a shell: E = 0V = constant = surface value

To find

V inside the shell.

Formula

V_inside = kQ/R (continuous with V_surface).

Solution

1V_inside = kQ/R, independent of r within the shell.

Answer

(B) · highlighted above in green

Q87NEET 2019Hard7/10In a charged parallel-plate capacitor (Q fixed), if the gap is doubled and a dielectric K = 2 is then inserted to fill the new gap, the new capacitance compared with C₀ is

AC₀

BC₀/2

C2C₀

D4C₀

Given

C₀ = ε₀A/d. After: gap → 2d, dielectric K = 2.

To find

Final C.

Formula

C_new = Kε₀A2d = 2·ε₀A2d = ε₀A/d = C₀.

Solution

1C_new = C₀.

2The two effects exactly cancel.

Answer

(A) · highlighted above in green

Q88NEET 2022Moderate4/10A point charge q at the centre of a cube of side a. Electric flux through one face of the cube is

Aq/(6ε₀)

Bq/(8ε₀)

Cq/(2ε₀)

Dq/ε₀

Given

Charge at cube centretotal flux through closed surface = q/ε₀six identical faces by symmetry

To find

Flux through one face.

Formula

ϕ_face = ϕ_total/6.

Solution

1ϕ_face = (q/ε₀)/6 = q6ε₀.

Answer

(A) · highlighted above in green

Q89NEET 2015Hard6/10Two capacitors of 6 μF and 12 μF are connected in series. A potential difference of 9 V is applied across the combination. The energy stored in the 6 μF capacitor is

A54 μJ

B108 μJ

C162 μJ

D216 μJ

Given

C_s = 6·12/18 = 4 μFV = 9 VQ = 4·9 = 36 μC same on each. V on 6 μF = 36/6 = 6 V

To find

Energy in the 6 μF capacitor.

Formula

U = ½·C·V² (with V across that capacitor).

Solution

1U₆ = 0.5 × 6×10⁻⁶ × 36 = 1.08×10⁻⁴ J = 108 μJ.

Answer

(B) · highlighted above in green

Q90NEET 2016Hard7/10An electric dipole p is placed in a non-uniform field with field strength decreasing along p. The net force on the dipole is

AZero

BIn the direction of decreasing field

CIn the direction of increasing field

DPerpendicular to p

Given

p along Ethe +q end sits in stronger field, the −q end in weaker

To find

Direction of net force.

Formula

F_net = p · (dE/dx) along p.

Solution

1If field increases opposite to p (decreasing along p means +q sits in weaker), net force on the dipole points toward stronger field — i.e., in the direction of increasing field.

Answer

Q91NEET 2018Moderate5/10If 1000 identical charged drops (each with charge q, potential V, radius r) coalesce, the potential of the big drop is

A100V

B1000V

C10V

Given

n = 1000R_big = 1000^(1/3)·r = 10rQ_big = 1000q

To find

V_big.

Formula

V_big = kQ_big/R_big = 1000^(2/3)·V_small.

Solution

11000^(2/3) = 100 ⇒ V_big = 100V.

Answer

(A) · highlighted above in green

Q92NEET 2021Hard6/10Two parallel-plate capacitors C₁ = 2 μF (charged to 200 V) and C₂ = 3 μF (charged to 100 V) are connected with their positive plates together. Energy lost is

A6 mJ

B2.4 mJ

C4 mJ

D1.5 mJ

Given

C₁ = 2 μF, V₁ = 200 VC₂ = 3 μF, V₂ = 100 V (same-polarity)

To find

Energy lost ΔU.

Formula

ΔU = ½·C₁C₂V₁−V₂C₁+C₂^².

Solution

1ΔU = 0.5 × 2×32+3 × (200−100)² × 10⁻⁶ = 0.5 × (6/5) × 10⁴ × 10⁻⁶ = 0.5 × 1.2 × 10⁻² = 6×10⁻³ J = 6 mJ.

Answer

(A) · highlighted above in green

Q93NEET 2017Moderate5/10A point charge +Q is placed at the origin. The work done in moving a test charge +q from (a, 0, 0) to (a√2, 0, 0) is

APositive

BNegative

CZero

DDepends on path

Given

Both endpoints at distance from origin a and a√2 (a√2 > a) ⇒ V decreases.

To find

Sign of work done by external agent.

Formula

W_ext = q·(V_f − V_i).

Solution

1V_f = kQa√2 < V_i = kQ/a ⇒ V_f − V_i < 0 ⇒ W_ext < 0 (negative).

2The charge moves to lower V — the field does positive work; you do negative work.

Answer

(B) · highlighted above in green

Q94NEET 2019Moderate5/10Electric potential at a point on the axis of a uniformly charged ring of radius R, total charge Q, at distance x from centre, is

AkQ/x

BkQ/√(R²+x²)

CkQ/(R+x)

DkQ/(2R)

Given

Each element of the ring is at the same distance √(R²+x²) from the axial point.

To find

V on axis.

Formula

V = kQ/√(R²+x²) (scalar; all elements contribute symmetrically).

Solution

1V_axis = kQ/√(R²+x²).

2Reduces to kQ/x for x >> R (point-charge limit).

Answer

(B) · highlighted above in green

Q95NEET 2020Moderate5/10The energy density between the plates of a parallel-plate capacitor of capacitance C, potential V, and gap d, with vacuum, is

A½ε₀V²/d²

B½CV²

Cε₀V/d

DCV²

Given

E_between = V/d (uniform).

To find

Energy density u (J/m³).

Formula

u = ½ε₀E² with E = V/d.

Solution

1u = ½ε₀(V/d)² = ½ε₀V²/d².

Answer

(A) · highlighted above in green

Q96NEET 2022Hard7/10A parallel-plate capacitor (area A, gap d) is filled with two dielectric slabs of equal thickness d/2 with K₁ = 2 and K₂ = 4 (in series, parallel to plates). The capacitance is

A4ε₀A/d

B8ε₀A/(3d)

C2ε₀A/d

D6ε₀A/d

Given

Two slabs in series, each of thickness d/2K₁ = 2, K₂ = 4

To find

Equivalent capacitance.

Formula

C = ε₀At₁/K₁ + t₂/K₂.

Solution

1t₁/K₁ + t₂/K₂ = (d/2)/2 + (d/2)/4 = d/4 + d/8 = 3d/8.

2C = ε₀A3d/8 = 8ε₀A/(3d).

Answer

(B) · highlighted above in green

Q97NEET 2015Moderate5/10A point charge q is located at the centre of a hollow conducting sphere of inner radius a and outer radius b. Charge on the outer surface of the sphere is

A−q

B+q

CZero

DDepends on a and b

Given

By Gauss's law inside a conductor E = 0 ⇒ inner surface induces charge −qtotal conductor charge = 0 (neutral) ⇒ outer surface = +q

To find

Charge on outer surface.

Formula

Conservation of total charge in neutral conductor: inner + outer = 0 since induced = −q on inner.

Solution

1Outer = +q. (If the shell had additional charge Q, outer = Q + q.)

Answer

(B) · highlighted above in green

Q98NEET 2018Moderate4/10A capacitor of capacitance C is charged using a battery of EMF V₀ through a resistor R. After a long time, total charge that flows from the battery is

ACV₀

BCV₀/2

C2CV₀

D½CV₀²/R

Given

Steady state: capacitor fully charged with Q = CV₀all the charge came from the battery

To find

Total charge through battery.

Formula

Q = CV₀ (independent of R; R only affects how fast).

Solution

1Q = CV₀.

Answer

(A) · highlighted above in green

Q99NEET 2023Hard6/10Two charges +4q and −q are placed on the x-axis at x = 0 and x = d respectively. The point on the x-axis where electric potential is zero (other than infinity) is

Ax = d/3

Bx = d/5

Cx = 4d/5

Dx = 3d/4

Given

V(x) = k(4q)/x + k(−q)/(d − x) = 0 ⇒ 4(d − x) = x ⇒ 4d = 5x.

To find

Position x on the x-axis where V = 0 (between or beyond the charges).

Formula

Sum of potentials = 0.

Solution

1x = 4d/5. (One root only; the other at x = 4d/3 lies outside the segment but is also a V = 0 point.)

Answer

Q100NEET 2023Moderate5/10A capacitor of capacitance 100 pF is charged by 100 V battery. The battery is disconnected and the capacitor is connected to an uncharged 100 pF capacitor. The loss of energy is

A0.25 μJ

B0.5 μJ

C1 μJ

D2 μJ

Given

Equal capacitancestotal Q = CV = 100×10⁻¹² × 100 = 10⁻⁸ Cshared equally ⇒ V_f = V/2 = 50 V

To find

Energy lost on sharing.

Formula

ΔU = U_i − U_f = ½CV² − 2·½C(V/2)² = ½CV² − ¼CV² = ¼CV².

Solution

1ΔU = 0.25 × 100×10⁻¹² × 100² = 0.25 × 10⁻⁶ J = 0.25 μJ.

Answer

(A) · highlighted above in green

Q101JEE Main 2019Hard6/10Two protons placed (10⁻¹⁵ m) apart are released from rest. Their kinetic energy when far apart (each) is

Ake²/(2r)

Bke²/r

C2ke²/r

D4ke²/r

Given

Identical charges +e, separation rsymmetry

To find

KE per proton at infinite separation.

Formula

Energy conservation; total KE = U_i = ke²/r; halved by symmetry.

Solution

1Each gets ke²2r.

Answer

(A) · highlighted above in green

Q102JEE Main 2020Hard7/10Two parallel-plate capacitors C₁ = 5 μF, C₂ = 10 μF charged to 100 V and 50 V are connected with similar plates. Energy lost is

A≈ 0.0042 J

B≈ 0.0084 J

C≈ 0.42 J

D≈ 0.21 J

Given

C₁ = 5 μF, V₁ = 100 VC₂ = 10 μF, V₂ = 50 V

To find

Energy lost on sharing.

Formula

ΔU = ½·C₁C₂V₁−V₂C₁+C₂^².

Solution

1ΔU = 0.5×(50/15)×10⁻⁶×2500 = 4.17×10⁻³ ≈ 4.2 mJ.

Answer

(A) · highlighted above in green

Q103JEE Main 2018Moderate5/10A 1 μF capacitor is charged to potential V and then connected to an uncharged 2 μF capacitor. Final voltage is

AV/3

BV/2

C2V/3

Given

C₁V₁ + C₂V₂ = (C₁+C₂)V_cV₂ = 0

To find

V_c.

Formula

V_c = C₁V₁ + C₂V₂C₁+C₂.

Solution

1V_c = (1·V + 2·0)/3 = V/3.

Answer

(A) · highlighted above in green

Q104JEE Main 2021Moderate4/10A parallel-plate capacitor (vacuum) has C = 12 pF. A dielectric of K = 4 is introduced filling the gap. New C is

A3 pF

B12 pF

C48 pF

D24 pF

Given

C₀ = 12 pFfull dielectric K = 4

To find

New C.

Formula

C = K·C₀.

Solution

1C = 4×12 = 48 pF.

Answer

Q105JEE Main 2017Hard7/10Half the space between the plates of a parallel-plate capacitor is filled by a dielectric K, parallel to the plates. New C in terms of C₀

A2KC₀/(1+K)

BKC₀/2

C(1+K)C₀/2

DC₀(1+K)

Given

Series of two capacitors: C₁ = 2ε₀A/d (vacuum half), C₂ = 2Kε₀A/d (dielectric half).

To find

Equivalent C.

Formula

1/C = 1/C₁ + 1/C₂.

Solution

1C = 2KC₀/(1+K) where C₀ = ε₀A/d.

Answer

(A) · highlighted above in green

Q106JEE Main 2019Hard6/10A capacitor with capacitance C₀ stores energy U₀ when charged by a battery V. If the same battery is used to charge two identical such capacitors in series, energy stored in each is

AU₀

BU₀/2

CU₀/4

D2U₀

Given

Series: each capacitor gets V/2 (equal Cs).

To find

Energy stored in each.

Formula

U_each = ½C₀(V/2)² = U₀/4.

Solution

1Each stores U₀/4.

Answer

Q107JEE Main 2020Hard6/10Energy density in a parallel-plate capacitor with V = 200 V and d = 2 mm (vacuum) is

A4.43×10⁻² J/m³

B4.43×10⁻³ J/m³

C0.443 J/m³

D4.43 J/m³

Given

E = V/d = 200/(2×10⁻³) = 10⁵ V/m.

To find

u = ½ε₀E².

Formula

u = ½(8.854×10⁻¹²)(10⁵)².

Solution

1u = 0.5 × 8.854×10⁻¹² × 10¹⁰ = 4.427×10⁻² ≈ 4.43×10⁻² J/m³.

Answer

(A) · highlighted above in green

Q108JEE Main 2018Hard7/10Find capacitance of a spherical capacitor with inner radius 5 cm and outer 6 cm (vacuum)

A33.4 pF

B0.334 pF

C334 pF

D3.34 pF

Given

a = 0.05 m, b = 0.06 mvacuum4πε₀ = 1/k = 1/(9×10⁹)

To find

C of spherical capacitor.

Formula

C = 4πε₀·abb−a = ab/(k(b−a)).

Solution

1C = 0.05×0.06/(9×10⁹ × 0.01) = 3×10⁻³/(9×10⁷) ≈ 3.33×10⁻¹¹ F ≈ 33.4 pF.

Answer

(A) · highlighted above in green

Q109JEE Main 2021Hard7/10A 20 μF capacitor is charged to 100 V and then connected in parallel with a 5 μF uncharged capacitor. The energy lost is

A20 mJ

B4 mJ

C16 mJ

D25 mJ

Given

C₁ = 20 μF, V₁ = 100 VC₂ = 5 μF, V₂ = 0

To find

Energy lost ΔU.

Formula

ΔU = ½·C₁C₂V₁−V₂C₁+C₂^².

Solution

1ΔU = 0.5×20×5×100²/25 ×10⁻⁶ = 0.5×400/25 ×10⁻² = 0.02 J = 20 mJ.

Answer

(A) · highlighted above in green

Q110JEE Main 2022Moderate5/10Three capacitors 2 μF, 3 μF, 6 μF are first all in parallel across 30 V, then re-arranged in series across 30 V. The ratio of total energies (parallel:series) is

A11:1

B1:11

C121:1

D1:121

Given

Parallel: C_p = 11 μFSeries: C_s = 1 μF

To find

Ratio U_p/U_s.

Formula

U = ½CV² with same V ⇒ ratio = C_p/C_s.

Solution

1U_p/U_s = 11/1 = 11.

Answer

(A) · highlighted above in green

Q111JEE Main 2017Hard6/10Force per unit area on a plate of a parallel-plate capacitor with field E is

Aε₀E²

B½ε₀E²

C2ε₀E²

DE²

Given

Half of E due to other platepressure = ½ε₀E²

To find

Force per unit area (electrostatic pressure).

Formula

P = u = ½ε₀E² (energy density gives outward pressure of equal magnitude).

Solution

1P = ½ε₀E².

Answer

(B) · highlighted above in green

Q112JEE Main 2019Moderate5/10Two capacitors of capacitances 8 μF and 12 μF are connected in series across a 12 V battery. The charge on each capacitor is

A38.4 μC

B57.6 μC

C96 μC

D144 μC

Given

C₁ = 8 μF, C₂ = 12 μF in seriesV = 12 Vseries ⇒ same Q on each

To find

Charge Q on each capacitor.

Formula

C_s = C₁C₂C₁+C₂; Q = C_s · V.

Solution

1C_s = 8×128+12 = 96/20 = 4.8 μF.

2Q = 4.8 × 12 = 57.6 μC on each capacitor.

Answer

(B) · highlighted above in green

Q113JEE Main 2020Moderate4/10Five identical capacitors each of capacitance C are connected in series. Effective capacitance is

BC/5

C5C

D25C

Given

All seriesn = 5

To find

C_s.

Formula

Equal series ⇒ C/n.

Solution

1C_s = C/5.

Answer

(B) · highlighted above in green

Q114JEE Main 2018Hard6/10A capacitor of capacitance C is fully charged using battery V, then disconnected and connected in parallel with an identical uncharged capacitor. Total energy after sharing is

A½CV²

B¼CV²

CCV²

D2CV²

Given

Q = CVafter sharing V' = V/2 across both (each holds CV/2)

To find

Total energy U' of system.

Formula

U' = 2·½C(V/2)² = ¼CV².

Solution

1Half the original energy is lost; U' = ¼CV².

Answer

(B) · highlighted above in green

Q115JEE Main 2021Moderate5/10A 5 μF capacitor is charged to 200 V. A dielectric K = 5 is inserted with battery on. Charge on capacitor becomes

A1 mC

B200 μC

C5 mC

D1000 μC

Given

C → 5C = 25 μFV fixed at 200 V

To find

New Q.

Formula

Q = CV.

Solution

1Q = 25×10⁻⁶ × 200 = 5×10⁻³ C = 5 mC.

Answer

Q116JEE Main 2017Hard6/10A capacitor C is charged through a resistor R by a battery V. Total energy dissipated in R during full charging is

ACV²

B½CV²

C¼CV²

DIndependent of R and equals ½CV²

Given

Battery does W = CV²stored = ½CV²rest dissipated regardless of R

To find

Energy dissipated in R.

Formula

Energy conservation; loss = W_batt − U_stored.

Solution

1Loss = CV² − ½CV² = ½CV², independent of R.

Answer

(D) · highlighted above in green

Q117JEE Main 2019Hard7/10A dielectric of K is inserted between plates of a charged capacitor (Q fixed). The induced surface charge density on the dielectric face is

Aσ(1 − 1/K)

Bσ(K − 1)

CσK

Dσ/K

Given

Free σ unchanged. Bound surface σ_b reduces net field by 1/K.

To find

Magnitude of bound (induced) surface charge density.

Formula

E_net = σ − σ_bε₀ = σ/(Kε₀) ⇒ σ_b = σ(1 − 1/K).

Solution

1σ_b = σ(1 − 1/K).

Answer

(A) · highlighted above in green

Q118JEE Main 2020Hard7/10Two capacitors of equal capacitance C are joined in series across a battery V. A dielectric K is inserted in one of them. New charge on each is

ACV

B2KCV/(1+K)

CKCV/(1+K)

DCV/(K+1)

Given

Series: 1/C_eq = 1/C + 1/(KC) = (1+K)/(KC) ⇒ C_eq = KC/(1+K). Same Q.

To find

Charge on each.

Formula

Q = C_eq·V.

Solution

1Q = KCV1+K.

Answer

Q119JEE Main 2022Hard6/10Energy stored in an isolated charged sphere of radius R with charge Q is

AkQ²/R

BkQ²/(2R)

CkQ²/(4R)

DZero

Given

U = Q²/(2C)C = 4πε₀R = R/k

To find

Self-energy of charged sphere.

Formula

U = Q²2C = kQ²2R.

Solution

1U = kQ²2R.

Answer

(B) · highlighted above in green

Q120JEE Main 2018Moderate5/10A parallel-plate capacitor has plates of area 100 cm² and separation 2 mm. Capacitance is

A44.3 pF

B442.7 pF

C0.443 pF

D4.43 pF

Given

A = 100×10⁻⁴ m² = 10⁻² m²d = 2×10⁻³ mε₀ = 8.854×10⁻¹²

To find

Formula

C = ε₀A/d.

Solution

1C = 8.854×10⁻¹² × 10⁻²/(2×10⁻³) = 4.43×10⁻¹¹ F = 44.3 pF.

Answer

(A) · highlighted above in green

Q121JEE Main 2020Advanced10/10An infinite array of capacitors, each of capacitance C, are arranged as repeating series-parallel ladder so the equivalent satisfies x = C + C·x/(C+x). Net capacitance is

A(√5−1)C/2

B(√5+1)C/2

C2C

Given

Standard ladder recursion.

To find

Solve quadratic for x.

Formula

x² − Cx − C² = 0.

Solution

1x = C(1+√5)/2 = golden-ratio · C.

Answer

(B) · highlighted above in green

Q122JEE Main 2019Advanced8/10A parallel-plate capacitor of area A and separation d, filled with two slabs (thickness t₁ with K₁ and t₂ with K₂, t₁+t₂ = d). Capacitance is

Aε₀A/(t₁/K₁ + t₂/K₂)

Bε₀A·(K₁+K₂)/d

Cε₀A/(K₁t₁+K₂t₂)

Dε₀A·K₁K₂/(t₁+t₂)

Given

Two series capacitors: C₁ = ε₀AK₁/t₁, C₂ = ε₀AK₂/t₂.

To find

Total C.

Formula

1/C = 1/C₁ + 1/C₂.

Solution

1C = ε₀At₁/K₁ + t₂/K₂.

Answer

(A) · highlighted above in green

Q123JEE Main 2017Moderate5/10A 12 μF capacitor is connected to a 6 V battery. Energy supplied by battery is

A216 μJ

B432 μJ

C72 μJ

D108 μJ

Given

C = 12 μF, V = 6 V.

To find

Energy supplied by battery W_batt.

Formula

W_batt = QV = CV².

Solution

1W_batt = 12×10⁻⁶ × 36 = 432×10⁻⁶ J = 432 μJ.

Answer

(B) · highlighted above in green

Q124JEE Main 2018Moderate4/10A capacitor of capacitance C is charged to voltage V₀, then connected to a resistor R (without battery). Time constant of discharge is

ARC

BR/C

CC/R

D1/(RC)

Given

Standard RC dischargeq(t) = Q₀ e^(−t/RC)

To find

Time constant τ.

Formula

τ = RC.

Solution

1τ = RC (units of seconds).

Answer

(A) · highlighted above in green

Q125JEE Main 2021Hard7/10A parallel-plate capacitor (C₀) is connected to a battery and then a dielectric K is inserted partway by sliding (only half-area filled, parallel to plates). Battery remains on. Energy stored after is

A½C₀V²(1+K)/2

B½C₀V²(1+K)

C½C₀V²(K+1)/(2K)

D½KC₀V²/2

Given

New C = C₀(1+K)/2 (half-area parallel arrangement).

To find

U_new with V fixed.

Formula

U = ½CV².

Solution

1U_new = ½ · C₀(1+K)/2 · V² = ½C₀V²(1+K)/2.

Answer

(A) · highlighted above in green

Q126InfinityEasy3/10A, B, C are three points on a circle of radius 1 cm and form the corners of an equilateral triangle. A charge 2 C is placed at the centre. Work done in carrying a charge of 0.1 μC from A to B is

A18 × 10¹¹ J

BZero

C1.8 × 10¹¹ J

D54 × 10¹¹ J

Given

A, B, C lie on the same circle of radius r = 1 cmcharge 2 C at centretest charge 0.1 μC moved A → B

To find

Work done by external agent.

Formula

W = q(V_B − V_A); A and B are equidistant from the centre.

Solution

1Both A and B are on the same equipotential sphere (same r) ⇒ V_A = V_B ⇒ W = 0.

2The size of the central charge and the test charge are irrelevant.

Answer

(B) · highlighted above in green

Q127InfinityHard7/10An electric dipole with charges −q and +q separated by 6 cm has midpoint O. Point P lies on the axis at 5 cm from O (on the +q side). The potential at P (in 10² V) is

A(5/8) qk

B(8/5) qk

C(8/3) qk

D(3/8) qk

Given

−q at distance 3 cm left of O, +q at 3 cm right of O. P is 5 cm right of O on the same axis.

To find

Potential V_P in units of 10² V.

Formula

V_P = k[+qr₊ + −qr₋] where r₊ = 5 − 3 = 2 cm, r₋ = 5 + 3 = 8 cm.

Solution

1V_P = kq[1/0.02 − 1/0.08] = kq[50 − 12.5] = 37.5 kq V = (37.5/100)·kq × 10² V = (3/8)qk × 10² V.

Answer

(D) · highlighted above in green

Q128InfinityHard7/10A hollow sphere of radius 2R is charged to V volts; a smaller sphere of radius R is charged to V/2 volts. The smaller sphere is now placed inside the bigger one without changing the charge on either. The potential difference between the two spheres is

A3V/2

BV/4

CV/2

Given

Before placing: V_big = V = kQ₁/(2R) ⇒ Q₁ = 2RV/k. V_small = V/2 = kQ₂/R ⇒ Q₂ = RV/(2k).

To find

ΔV = V_small_inside − V_big after placing the smaller inside the larger.

Formula

Potentials at surfaces inside another shell: V_small = kQ₂/R + kQ₁2R; V_big = kQ₁+Q₂2R.

Solution

1V_small − V_big = kQ₂/R − kQ₂2R = kQ₂2R = k·[RV2k]/(2R) = V/4.

Answer

(B) · highlighted above in green

Q129InfinityModerate5/10A hollow metallic sphere of radius R has a potential difference V between its surface and a point at 3R from its centre. The electric field at the point 3R from the centre is

AV/(2R)

BV/(3R)

CV/(4R)

DV/(6R)

Given

V_surface − V_(3R) = Vfor a charged spherical shell, V outside = kQ/r

To find

Electric field E at r = 3R from centre.

Formula

V_surface − V_(3R) = kQ/R − kQ3R = 2kQ3R = V ⇒ kQ = 3VR/2. E = kQ/r².

Solution

1E(3R) = kQ3R² = 3VR/29R² = V6R.

Answer

(D) · highlighted above in green

Q130InfinityHard6/10Many capacitors each of 1 μF puncture if more than 500 V is applied. To make a 3 μF arrangement that withstands 2000 V, minimum number of such capacitors needed is

B96

C48

Given

Each capacitor: C = 1 μF, V_max = 500 Vtarget combination: 3 μF at 2000 V

To find

Minimum number of 1 μF capacitors.

Formula

Series gives voltage rating, parallel gives capacitance. n series = 2000/500 = 4 ⇒ each row = 1/4 μF; m rows in parallel give m/4 μF = 3 μF ⇒ m = 12.

Solution

1Total = m × n = 12 × 4 = 48 capacitors.

Answer

Q131InfinityHard6/10A dielectric slab of constant K, same area as the plates and thickness (3/4)d, is inserted into a parallel-plate capacitor of plate gap d. The new capacitance (C₀ = ε₀A/d) becomes

A3K/(K+4) · C₀

BK/(K+3) · C₀

C2K/(K+3) · C₀

D4K/(K+3) · C₀

Given

Slab thickness t = 3d/4 with dielectric Ksame plate areagap d

To find

New capacitance in terms of C₀.

Formula

C = ε₀Ad − t + t/K.

Solution

1C = ε₀A / (d − 3d/4 + 3d/(4K)) = ε₀A / [d/4 + 3d/(4K)] = ε₀A · 4K / [d(K + 3)] = 4K/(K+3) · C₀.

Answer

(D) · highlighted above in green

Q132InfinityAdvanced9/10A point charge q is located at the centre O of a spherical uncharged conducting shell (inner radius a, outer b) with a small orifice. The work required to slowly transfer the charge q from O through the orifice to infinity is

Aq²/(8πε₀)·(1/a − 1/b)

Bq²/(4πε₀)·(1/b − 1/a)

Cq²/(8πε₀)·(1/b − 1/a)

Dq²/(4πε₀)·(1/a − 1/b)

Given

Charge q at centre, conductor with inner radius a, outer b. Induced charges: −q on inner surface, +q on outer surface. Field exists only in r < a and r > bzero in conductor

To find

Work W = ΔU = U_final − U_initial.

Formula

U = ½ε₀∫E²dV. The cancelling self-energy near r = 0 means ΔU = q²8πε₀·(1/a − 1/b).

Solution

1Initial field exists in 0<r<a and r>b; final field exists everywhere.

2Difference fills the gap a<r<b, giving extra energy q²8πε₀·(1/a − 1/b).

3Positive ⇒ external agent does positive work.

Answer

(A) · highlighted above in green

Q133InfinityModerate4/10Three point charges 3 nC, 6 nC and 9 nC are placed at the corners of an equilateral triangle of side 0.1 m. The potential energy of the system is

A8910 × 10⁻¹⁰ J

B9910 × 10⁻⁹ J

C8910 × 10⁻⁹ J

D9910 × 10⁻¹⁰ J

Given

q₁=3 nC, q₂=6 nC, q₃=9 nCside r = 0.1 m

To find

Total electrostatic PE.

Formula

U = (k/r)(q₁q₂ + q₁q₃ + q₂q₃) (three pairs at equal distance).

Solution

1Sum of products: 3·6 + 3·9 + 6·9 = 18 + 27 + 54 = 99 (in nC²·units).

2U = (9×10⁹/0.1)·99×10⁻¹⁸ = 9×10¹⁰·99×10⁻¹⁸ = 891×10⁻⁸ = 8910×10⁻⁹ J.

Answer

Q134InfinityModerate5/10Four charges +12 nC, −20 nC, +32 nC and −15 nC are placed at the vertices of a square of side √2 m. Net electric potential at the centre of the square is

A72 V

B81 V

C64 V

D36 V

Given

Side a = √2 mdistance from centre to each vertex = (√2)/√2 = 1 m. All four charges equidistant

To find

V at the centre.

Formula

V is scalar; V = (k/r)·Σ qᵢ at equal distance.

Solution

1Σ q = (12 − 20 + 32 − 15) nC = 9 nC.

2V = (9×10⁹)(9×10⁻⁹)/1 = 81 V.

Answer

(B) · highlighted above in green

Q135InfinityModerate4/10A parallel-plate capacitor has capacitance C. The separation is doubled and a dielectric is introduced between the plates. If the new capacitance is 2C, the dielectric constant K is

Given

Old C = ε₀A/d. New gap = 2d, with dielectric K.

To find

Dielectric constant K.

Formula

C_new = K·ε₀A2d = (K/2)·C.

Solution

12C = (K/2)·C ⇒ K = 4.

Answer

Q136InfinityModerate5/10A long conducting cylinder of radius R has a uniform linear charge density λ. The field at r (>R) is E = λ/(2πε₀r). The electric potential at that point (with V(R)=0) is

Aλ/(2πε₀)

B−λR/(2πε₀r²)

C(λ/(2πε₀)) ln(r/R)

D(λ/(2πε₀)) ln(R/r)

Given

E(r) = λ/(2πε₀r) for r > Rreference at the surface (V(R) = 0)

To find

V(r) for a point outside the cylinder.

Formula

V(r) − V(R) = −∫_R^r E·dr'.

Solution

1V(r) − 0 = −∫_R^r (λ/(2πε₀r')) dr' = −(λ/(2πε₀)) ln(r/R) = (λ/(2πε₀)) ln(R/r).

Answer

(D) · highlighted above in green

Q137InfinityAdvanced8/10Two identical thin rings of radius R are placed co-axially at distance R apart with charges Q₁ and Q₂. Work done in moving a charge q from the centre of the first ring to the centre of the second ring is

AZero

Bq√2(Q₁+Q₂)/(4πε₀R)

Cq(Q₁−Q₂)(√2−1)/(√2·4πε₀R)

Dq(Q₁+Q₂)(√2+1)/(√2·4πε₀R)

Given

Ring radii Rcentres separated by R. V at axial distance x from a ring is kQ/√(R²+x²)

To find

Work W = q(V₂ − V₁).

Formula

V₁ = kQ₁/R + kQ₂R√2; V₂ = kQ₂/R + kQ₁R√2.

Solution

1V₂ − V₁ = (k/R)(Q₂ − Q₁)(1 − 1/√2) = (k/R)·(Q₂−Q₁)·(√2−1)/√2.

2W = q·ΔV = q(Q₁−Q₂)√2−1√2·4πε₀R (sign depending on direction).

Answer

Q138InfinityModerate4/10Choose the wrong statement

AOn the perpendicular bisector of a dipole, E = 0 and V ≠ 0

BField lines are always perpendicular to equipotential surfaces

CWhen a proton and an electron move apart, their PE increases

DThe dielectric constant of a material can differ between DC and AC

Given

Standard properties of dipoles, equipotentials, attractive PE behaviour, and frequency dependence of K.

To find

Identify the false statement.

Formula

On the equatorial plane of a dipole: V = 0 (sum of equal & opposite contributions) but E = kp/r³ ≠ 0.

Solution

1Statement (a) reverses these: actual is V = 0, E ≠ 0.

2So (a) is wrong. (b), (c), (d) are all standard true facts.

Answer

(A) · highlighted above in green

Q139InfinityModerate5/10A parallel-plate capacitor with dielectric K is charged to V; battery is disconnected; then the dielectric is removed. Which set is true?

ACapacitance decreased by K, electric field reduced by K

BCapacitance decreased by K, voltage increased by K

CField reduced by K, voltage increased by K

DVoltage increased by K, charge increased by K

Given

Battery off ⇒ Q is fixed. With dielectric K: C₁ = KC₀, V₁ = V, Q = KC₀V. Remove dielectric: C₂ = C₀.

To find

Identify changes that are true.

Formula

Q fixed: V_new = Q/C₀ = KV (↑×K); E_new = V_new/d = KE (↑×K); C: KC₀ → C₀ (↓ by K).

Solution

1Capacitance ↓ by K (true); voltage ↑ by K (true); field ↑ by K (not ↓); charge unchanged (not ↑).

2So (a) and (c) are true.

Answer

(B) · highlighted above in green

Q140InfinityHard6/10Three charges +q, −q, −q are placed at the vertices of an equilateral triangle of side 10 cm. With q = 5 μC, the potential at the midpoint of the side joining the two −q charges is

A10³ V

B1 V

C−12.8 × 10⁵ V

D10 V

Given

a = 0.1 mq = 5×10⁻⁶ C. Midpoint M is midway between two −q charges. Distance from each −q to M = a/2 = 0.05 m. Distance from +q to M = (a√3)/2 = 0.0866 m

To find

V at M.

Formula

V = Σ kqᵢ/rᵢ (scalar sum with signs).

Solution

1V = k[+qa√3/2 + −qa/2 + −qa/2] = (9×10⁹)(5×10⁻⁶)[1/0.0866 − 2/0.05] = 4.5×10⁴·[11.55 − 40] = 4.5×10⁴·(−28.45) ≈ −12.8×10⁵ V.

Answer

Q141InfinityHard7/10A field of 100 V/m points at 30° to the +x-axis. If OA = 2 m along +x-axis and OB = 4 m along +y-axis, then V_A − V_B equals

A100(√3 − 2) V

B+100(2 + √3) V

C100(2 − √3) V

D200(2 + √3) V

Given

|E| = 100 V/m at 30°: E_x = 100 cos30° = 50√3E_y = 100 sin30° = 50. A = (2,0)B = (0,4)

To find

V_A − V_B.

Formula

V(r) = −E·r (with O as reference); V_A − V_B = −E·(r_A − r_B) = −(E_x·(2−0) + E_y·(0−4)).

Solution

1V_A − V_B = −(50√3 · 2 − 50 · 4) = −(100√3 − 200) = 200 − 100√3 = 100(2 − √3) V.

Answer

Q142InfinityEasy3/10In the circuit shown, three identical 2 μF capacitors are connected to a 12 V battery: capacitor (1) is directly across the battery while capacitors (2) and (3) are in parallel and across the same battery. The charge on capacitor (1) is

A24 μC

B8 μC

C18 μC

D12 μC

Given

Capacitor (1) sits directly across 12 Vcapacitors (2) and (3) are across the same 12 V in parallel

To find

Charge Q on capacitor (1).

Formula

Q = CV (capacitor (1) sees full battery voltage).

Solution

1Q₁ = 2 μF × 12 V = 24 μC.

Answer

(A) · highlighted above in green

Q143InfinityHard6/10In a single-loop circuit, a 3 μF capacitor and a 2 μF capacitor are joined in series; a 12 V battery and a 2 V battery (oriented to subtract) lie in series with them. Points A and B sit between the 12 V battery & 2 μF and between the 3 μF & 2 V battery respectively. The PD between A and B (across the 2 μF capacitor) is

A6 V

B2 V

C10 V

D14 V

Given

Net EMF around the loop = 12 − 2 = 10 Vseries capacitors carry the same charge Q

To find

Potential difference V_AB across the 2 μF capacitor.

Formula

Q = C₁·V₁ = C₂·V₂ ⇒ V₂/V₁ = C₁/C₂; also V₁ + V₂ = 10 V.

Solution

13V_{3μF} = 2V_{2μF} ⇒ V_{2μF} = (3/2)V_{3μF}.

2V_{3μF} + V_{2μF} = 10 ⇒ V_{3μF} = 4 V, V_{2μF} = 6 V.

3PD between A and B (across 2 μF) = 6 V.

Answer

(A) · highlighted above in green

Q144InfinityEasy3/10The capacitance of a parallel-plate capacitor with each plate of area 10 cm × 10 cm and separation 1 mm (ε₀ = 8.842×10⁻¹² C²/N·m²) is

A2 pF

B10 pF

C8 pF

D88.42 pF

Given

A = 10×10 cm² = 10⁻² m²d = 10⁻³ mε₀ = 8.842×10⁻¹²

To find

Capacitance C.

Formula

C = ε₀A/d.

Solution

1C = (8.842×10⁻¹²)10⁻²10⁻³ = 8.842×10⁻¹¹ F = 88.42 pF.

Answer

(D) · highlighted above in green

Q145InfinityModerate5/10Charges +q at A and +q at B (top corners), −q at D and −q at C (bottom corners) sit at the vertices of a square. Let E and V denote the electric field and potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then

AE changes, V remains unchanged

BE remains unchanged, V changes

CBoth E and V change

DBoth E and V remain unchanged

Given

Initial: top two +q, bottom two −q ⇒ E points from + side toward − side (downward). V at centre = 0 (sum of charges is 0 at equal distance).

To find

Changes after swap (A↔D, B↔C).

Formula

V = (k/r)·Σ qᵢ; Σ q = 0 ⇒ V = 0 regardless of arrangement. E = vector sum, depends on arrangement.

Solution

1After swap: top two −q, bottom two +q ⇒ E vector reverses direction (now upward), so E changes; V at the centre is still 0 (since Σ q is unchanged and all distances equal).

2Hence E changes, V unchanged.

Answer

(A) · highlighted above in green

Q146InfinityHard6/10Two parallel metallic plates of area A carry charges −2Q and +4Q respectively. Edge effects negligible. The charges on the outer surfaces I and III are

A−2Q, +2Q

B−Q, +3Q

C+Q, +3Q

D+2Q, −3Q

Given

Plate 1 carries net −2Q (split between surfaces I and II)Plate 2 carries net +4Q (split between III and IV)

To find

Charges on outer surfaces I (left) and III (inner of right plate).

Formula

For two parallel plates with net charges Q₁ and Q₂: outer surfaces each get (Q₁+Q₂)/2; inner surfaces get ±(Q₁−Q₂)/2 (with the same sign as the bigger plate on the facing surface).

Solution

1Outer surface charge each = (−2Q + 4Q)/2 = +Q.

2So surface I = +Q (outer of left plate).

3Inner of right plate (III) = (+4Q − (−2Q))/2 = +3Q.

4Hence (+Q, +3Q).

Answer

Q147InfinityModerate5/10A parallel-plate capacitor has plates of area 20 cm² and separation 2 mm. Dielectric K = 5; voltage 500 V. The energy density of the field between the plates is approximately

A2.65 J/m³

B1.95 J/m³

C1.38 J/m³

D0.69 J/m³

Given

A = 20×10⁻⁴ m², d = 2×10⁻³ m, K = 5, V = 500 V.

To find

Energy density u inside the dielectric.

Formula

u = (1/2)·K·ε₀·E², where E = V/d.

Solution

1E = 500/(2×10⁻³) = 2.5×10⁵ V/m. u = 0.5 × 5 × 8.854×10⁻¹² × (2.5×10⁵)² = 0.5 × 5 × 8.854×10⁻¹² × 6.25×10¹⁰ = 1.38 J/m³.

Answer

Q148InfinityModerate4/10The variation of electric potential V with distance d from a fixed point is shown: V = 5 V for d ∈ [0, 4] m and falls linearly from 5 V at d = 4 m to 0 V at d = 6 m. The electric field at d = 5 m is

A2.5 V/m

B−2.5 V/m

C0.4 V/m

D−0.4 V/m

Given

From the graph: V is flat at 5 V from 0 to 4 m, then drops linearly to 0 at d = 6 m. d = 5 m lies in the sloping region.

To find

E at d = 5 m.

Formula

E = −dV/dr.

Solution

1Slope between d = 4 and d = 6: 0 − 56 − 4 = −2.5 V/m.

2So dV/dr = −2.5; E = −(−2.5) = +2.5 V/m.

Answer

(A) · highlighted above in green

Q149InfinityHard6/10An infinite number of charges, each 1 C in magnitude (all same sign), are placed along the x-axis at x = 2, 4, 8, 16 cm and so on. The electric potential at x = 0 is

A9 × 10¹¹ V

BZero

C3 × 10⁹ V

D1.25 × 10⁹ V

Given

Charges q = 1 C at x₁ = 0.02 m, x₂ = 0.04 m, x₃ = 0.08 m, … with xₙ = 0.02·2^(n−1) m.

To find

V at x = 0 (scalar sum from each charge).

Formula

V = k·Σ q/xₙ = (9×10⁹)·(1/0.02 + 1/0.04 + 1/0.08 + …).

Solution

1Sum = (1/0.02)(1 + 1/2 + 1/4 + …) = 50 · 2 = 100.

2V = (9×10⁹)(100) = 9×10¹¹ V.

Answer

(A) · highlighted above in green

Q150InfinityModerate4/10Electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0 × 10⁻¹⁵ m is

A80 V

B8 × 10⁶ V

C9 V

D8 × 10⁵ V

Given

Q = Ze = 50 × 1.6×10⁻¹⁹ = 8×10⁻¹⁸ CR = 9×10⁻¹⁵ m

To find

V at the nuclear surface.

Formula

V = kQ/R.

Solution

1V = (9×10⁹)8×10⁻¹⁸9×10⁻¹⁵ = 8×10⁶ V.

Answer

(B) · highlighted above in green

Q151InfinityModerate4/10A conducting sphere of radius R is charged. The electric field at a distance r (> R) from the centre in terms of the surface potential V is

ArV/R²

BR²V/r²

CRV/r²

DV/r

Given

V_surface = V = kQ/R ⇒ kQ = RV.

To find

E at r > R from the centre.

Formula

E(r) = kQ/r² outside the sphere.

Solution

1E(r) = RVr² = RV/r².

Answer

Q152InfinityModerate5/10Two identical 5 μF capacitors are charged to potentials 2 kV and 1 kV respectively. Their negative terminals are joined; then their positive terminals are also joined. The loss of energy in the system is

A160 J

B0 J

C5 J

D1.25 J

Given

C₁ = C₂ = 5 μFV₁ = 2000 V, V₂ = 1000 Vsame polarity

To find

Energy lost on sharing.

Formula

ΔU = ½ · C₁C₂V₁−V₂C₁+C₂^².

Solution

1ΔU = 0.5 × (5×10⁻⁶)(5×10⁻⁶) × 100010×10⁻⁶^² = 0.5 × (25×10⁻¹²)10⁶10⁻⁵ = 0.5 × 2.5 = 1.25 J.

Answer

(D) · highlighted above in green

Q153InfinityEasy3/10In a hydrogen atom the electron orbits the nucleus at radius 0.53 × 10⁻¹⁰ m. The electric potential at the position of the electron due to the nucleus is

A−13.6 V

B−27.2 V

C27.2 V

D13.6 V

Given

Charge of nucleus q = +e = 1.6×10⁻¹⁹ Cr = 0.53×10⁻¹⁰ m

To find

V at the electron's position due to the (positive) proton.

Formula

V = kq/r (positive sign since source is +).

Solution

1V = (9×10⁹)1.6×10⁻¹⁹0.53×10⁻¹⁰ ≈ 27.2 V.

Answer

Q154InfinityAdvanced8/10A uniformly charged ring of radius 3a and total charge q lies in the xy-plane centred at origin. A point charge q (mass m) moves toward the ring along the z-axis with speed v at z = 4a. The minimum v needed for it to cross the origin is

A√((2/m)·(1/15)·q²/(4πε₀a))

B√((2/m)·(2/15)·q²/(4πε₀a))

C√((2/m)·(1/5)·q²/(4πε₀a))

D√((2/m)·(4/15)·q²/(4πε₀a))

Given

Ring R = 3a, charge q. Test charge mass m, also q. At z = 4a, V_ring = kq/√(R²+z²) = kq/(5a). At z = 0 (centre), V_ring = kq/(3a).

To find

Minimum speed v at z = 4a so KE just reaches zero at z = 0.

Formula

Energy conservation: ½mv² + qV(4a) = 0 + qV(0).

Solution

1½mv² = q[V(0) − V(4a)] = q·kq·(1/3a − 1/5a) = kq²·215a. v² = 4kq²/(15ma) = (2/m)·(2/15)·q²4πε₀a.

2So v = √((2/m)·(2/15)·q²4πε₀a).

Answer

(B) · highlighted above in green

Q155InfinityEasy3/10Which of the following statements are correct?

Aa and b

Ba and c

Ca only

Db and c

Given

Test each: parallel ⇒ C_p = ΣC, larger than eachseries ⇒ 1/C_s = Σ1/C, smaller than each

To find

Identify true statements.

Formula

Parallel adds, series divides.

Solution

1(a) false: parallel gives a larger combined C; (b) true: C = ε₀A/d so C ↑ as d ↓; (c) true: series gives C smaller than the smallest individual.

Answer

(D) · highlighted above in green

Q156InfinityHard6/10Two concentric metallic shells: inner radius r₁, outer radius r₂ (r₁ < r₂). The outer shell carries charge q and the inner shell is grounded (V = 0). The charge on the inner sphere q₁ equals

B−q

C−(r₁/r₂)·q

D(r₁/r₂)·q

Given

Inner shell grounded ⇒ its potential is 0. Outer shell carries net charge q.

To find

Induced charge q₁ on the inner shell.

Formula

V_inner = kq₁/r₁ + kq/r₂ = 0 (grounded).

Solution

1Solve: q₁/r₁ + q/r₂ = 0 ⇒ q₁ = −(r₁/r₂)·q.

2Negative because induced.

Answer

Q157InfinityModerate5/10Three point charges 1 C, 2 C and 3 C are placed at the corners of an equilateral triangle of side 1 m. The work required to move them to the corners of a smaller equilateral triangle of side 0.5 m is

A9.9 × 10⁹ J

B9.9 × 10¹⁰ J

C9.9 × 10¹¹ J

D9.9 × 10¹³ J

Given

Charges 1, 2, 3 Cinitial side 1 m, final side 0.5 m. Pairs contribute via U = k·q₁q₂/r

To find

Work W = ΔU = U_final − U_initial.

Formula

U(r) = (k/r)·(1·2 + 1·3 + 2·3) = (k/r)·11.

Solution

1U_initial = 11k/1 = 11×9×10⁹ = 9.9×10¹⁰ J.

2U_final = 11k/0.5 = 22k = 1.98×10¹¹ J.

3W = U_f − U_i = 1.98×10¹¹ − 9.9×10¹⁰ = 9.9×10¹⁰ J.

Answer

(B) · highlighted above in green

Q158InfinityModerate5/10A fully charged capacitor C is discharged through a thin coil embedded in a thermally insulated block of specific heat s and mass m. Temperature rise of the block is ΔT. The potential difference V across the capacitor before discharge was

AmCΔT/s

B√(2mCΔT/s)

C√(2msΔT/C)

DmsΔT/C

Given

All stored energy goes into heating the block.

To find

Initial V before discharge.

Formula

½CV² = msΔT ⇒ V = √(2msΔT/C).

Solution

1V = √(2msΔT/C).

2Note the m·s order in numerator and C in denominator under the root.

Answer

Q159InfinityAdvanced8/10Initially a switch S is at position 1 for a long time. A capacitor C is charged through resistor R by EMF ε₁. The switch is then thrown to position 2 (EMF ε₂, opposite polarity through the same C and R). The net heat dissipated in R after switching is

A(C/2)(ε₁+ε₂)·ε₂

BC(ε₁+ε₂)·ε₂

C(C/2)(ε₁+ε₂)²

DC(ε₁+ε₂)²

Given

Initial: Q_i = +Cε₁ on the capacitor. After switching to opposite-polarity EMF ε₂, final Q_f = −Cε₂ (reverse polarity).

To find

Heat H dissipated in R during the transient.

Formula

H = W_battery − ΔU_capacitor; |ΔQ| through ε₂ = C(ε₁+ε₂).

Solution

1W_batt = ε₂·C(ε₁+ε₂).

2ΔU = ½Cε₂² − ½Cε₁² = ½C(ε₂² − ε₁²).

3H = Cε₂(ε₁+ε₂) − ½C(ε₂−ε₁)(ε₂+ε₁) = C(ε₁+ε₂)[ε₂ − (ε₂−ε₁)/2] = (C/2)(ε₁+ε₂)².

Answer

Q160InfinityHard6/10An isolated sphere of radius r₁ has its capacitance increased by 5 times when it is enclosed by an earthed concentric sphere of radius r₂. The ratio r₁ : r₂ is

A4/5

B5/4

C5/1

D3/5

Given

Isolated sphere: C₀ = 4πε₀ r₁. Spherical capacitor (inner r₁, outer grounded r₂): C = 4πε₀·r₁r₂/(r₂−r₁).

To find

Ratio r₁/r₂ when C = 5C₀.

Formula

5·r₁ = r₁r₂r₂−r₁.

Solution

15(r₂−r₁) = r₂ ⇒ 4r₂ = 5r₁ ⇒ r₁/r₂ = 4/5.

Answer

(A) · highlighted above in green

Q161InfinityModerate5/10In a region of space the potential is V = k[2x² − y² + z²]. The magnitude of the electric field at the point (1, 1, 1) is

Ak√6

B2k√6

C2k√3

D4k√3

Given

Partial derivatives give the field components: ∂V/∂x = 4kx, ∂V/∂y = −2ky, ∂V/∂z = 2kz.

To find

|E| at (1, 1, 1).

Formula

E = −∇V; |E| = √(E_x² + E_y² + E_z²).

Solution

1At (1,1,1): E_x = −4k, E_y = +2k, E_z = −2k. |E| = k·√(16 + 4 + 4) = k·√24 = 2k√6.

Answer

(B) · highlighted above in green

Q162InfinityHard6/10Four identical capacitor plates are arranged so they form three 2 μF capacitors. The plates are connected to a 10 V source. The charge on plate C (the middle plate connected to the +ve terminal in this arrangement) is

A+20 μC

B+40 μC

C+60 μC

D+80 μC

Given

Four parallel plates form 3 capacitors (between pairs 1-2, 2-3, 3-4). Plates A and C connect to one terminalB and D to the other

To find

Charge on plate C.

Formula

Plate C participates in two capacitor pairs (B-C and C-D); each capacitor carries Q = CV = 2 μF × 10 V = 20 μC.

Solution

1Plate C's two faces each carry +20 μC ⇒ total charge on C = +40 μC.

Answer

(B) · highlighted above in green

Q163InfinityHard6/10Two capacitors C₁ and C₂ are joined in series between points A (potential V₁) and B (potential V₂), with D the node between them. The potential of point D is

A½(V₁ + V₂)

B(C₂V₁ + C₁V₂)/(C₁+C₂)

C(C₁V₁ + C₂V₂)/(C₁+C₂)

D(C₂V₁ − C₁V₂)/(C₁+C₂)

Given

Series caps: same charge Q on each. V_A − V_D = Q/C₁V_D − V_B = Q/C₂

To find

Potential V_D in terms of V₁, V₂, C₁, C₂.

Formula

Add: V_A − V_B = Q(1/C₁ + 1/C₂) ⇒ Q = (V₁−V₂)C₁C₂/(C₁+C₂). Then V_D = V_B + Q/C₂.

Solution

1V_D = V₂ + (V₁−V₂)C₁/(C₁+C₂) = (V₂(C₁+C₂) + (V₁−V₂)C₁)/(C₁+C₂) = C₂V₁ + C₁V₂C₁+C₂.

2Note the cross-coupling: V₁ gets weighted by C₂, not C₁.

Answer

(B) · highlighted above in green

Q164InfinityAdvanced9/10A capacitor has two square metal plates of edge a and separation d. A triangular dielectric arrangement (K₁ fills one triangle, K₂ fills the complementary triangle along the diagonal) fills the gap. The capacitance is

Aε₀K₁K₂a²·ln(K₁/K₂)/[(K₁−K₂)d]

Bε₀(K₁+K₂)a²·ln(K₁/K₂)/[(K₁−K₂)d]

Cε₀(K₁+K₂)a²·ln(K₂/K₁)/[(K₁−K₂)d]

Dε₀K₁K₂a²·ln(K₂/K₁)/[(K₁−K₂)d]

Given

Triangular split: at position x along the edge, dielectric K₁ thickness varies linearly from 0 to d, and K₂ from d to 0.

To find

Capacitance of the diagonally-split dielectric capacitor.

Formula

Take a vertical strip dx; the two dielectrics are in series for that strip. Integrate dC over the edge a.

Solution

1Standard result for triangular split: C = (ε₀K₁K₂·a²)/((K₁−K₂)·d) · ln(K₁/K₂).

Answer

(A) · highlighted above in green

Q165InfinityEasy3/10Two capacitors C₁ = 3 μF and C₂ = 4 μF are connected in series across a 14 V battery. The charge on each capacitor is

A12 μC, 12 μC

B24 μC, 24 μC

C6 μC, 8 μC

D8 μC, 6 μC

Given

C₁ = 3 μF, C₂ = 4 μF in seriesV = 14 V

To find

Charges Q₁ and Q₂ on each capacitor.

Formula

Series ⇒ same charge Q = C_s·V where 1/C_s = 1/C₁ + 1/C₂.

Solution

1C_s = 3·43+4 = 12/7 μF.

2Q = (12/7)·14 = 24 μC on each.

Answer

(B) · highlighted above in green

Q166InfinityModerate5/10A parallel-plate capacitor (C = 5 μF, plate separation 6 cm) is connected to a 1 V battery and fully charged. A dielectric slab of K = 4 and thickness 4 cm is then inserted between the plates. The additional charge that flows from the battery is

A2 μC

B3 μC

C5 μC

D10 μC

Given

C₀ = 5 μF ⇒ ε₀A = 5×10⁻⁶ × 0.06 = 3×10⁻⁷ F·m. After slab (K = 4, t = 4 cm), gap d = 6 cm.

To find

Extra charge ΔQ supplied by battery (V fixed = 1 V).

Formula

C_new = ε₀Ad − t + t/K; ΔQ = (C_new − C₀)·V.

Solution

1C_new = 3×10⁻⁷/(0.06 − 0.04 + 0.04/4) = 3×10⁻⁷/(0.02 + 0.01) = 3×10⁻⁷/0.03 = 10 μF.

2ΔQ = (10 − 5)·1 = 5 μC.

Answer

Q167InfinityHard7/10Two identical charges q connected by an ideal spring (force constant K, natural length r) rest on a smooth surface. They are released with separation r. If the maximum extension of the spring is r, then K equals

A(q/(4r))·√(1/(πε₀r))

B(q/(2r))·√(1/(πε₀r))

C(2q/r)·√(1/(πε₀r))

D(q/r)·√(1/(πε₀r))

Given

Natural length rrelease at separation r (spring unstretched, KE = 0). Max extension = r ⇒ at max stretch, separation = 2r and KE = 0 again

To find

Force constant K.

Formula

Energy conservation: (kq²/r) + 0 = (kq²/(2r)) + ½K·r².

Solution

1Difference (kq²/r − kq²2r) = kq²2r = ½K·r² ⇒ K = kq²/r³ = q²4πε₀r³.

2Taking √K: √K = q/(2r√(πε₀r)) = (q/(2r))·√(1/(πε₀r)).

Answer

(B) · highlighted above in green

Q168InfinityModerate5/10A slab of insulating material of thickness 4 × 10⁻⁵ m is introduced between the plates of a parallel-plate capacitor. To restore the original capacitance, the plate separation must be increased by 3.5 × 10⁻⁵ m. The dielectric constant of the slab is

C12

D10

Given

t = 4×10⁻⁵ mrequired increase Δd = 3.5×10⁻⁵ m. Original gap d. New gap d' = d + Δd, with the slab inside

To find

Dielectric constant K of the slab.

Formula

C_new = ε₀Ad' − t + t/K = C₀ = ε₀A/d ⇒ d = d' − t + t/K.

Solution

1d = (d + 3.5×10⁻⁵) − 4×10⁻⁵ + 4×10⁻⁵/K ⇒ 0 = −0.5×10⁻⁵ + 4×10⁻⁵/K ⇒ 4×10⁻⁵/K = 0.5×10⁻⁵ ⇒ K = 8.

Answer

(A) · highlighted above in green

Q169InfinityModerate5/10An LC series circuit has an oscillation frequency f. Two isolated inductors (each L) and two capacitors (each C) are all connected in series. The new oscillation frequency is

Af/4

Bf/2

D4f

Given

f = 1/(2π√(LC)). Series 2L gives L_eq = 2Lseries 2C gives C_eq = C/2

To find

New frequency f'.

Formula

f' = 1/(2π√(L_eq · C_eq)).

Solution

1L_eq · C_eq = 2L · (C/2) = LC ⇒ f' = 1/(2π√(LC)) = f.

2Unchanged.

Answer

Q170InfinityHard7/10Six identical capacitors (each of value C) are connected on the six edges of a tetrahedron whose vertices are A, B, C and D. The equivalent capacitance between A and B is

B2C

C3C

D4C

Given

Vertices A, B, C, Dedges AB, AC, AD, BC, BD, CD — each capacitor of value C

To find

Equivalent capacitance between A and B.

Formula

By symmetry C and D are at equal potential ⇒ edge CD carries no charge and can be removed. Then evaluate the three remaining parallel paths.

Solution

1Paths from A to B: (1) direct edge AB = C; (2) A→C→B (two C in series) = C/2; (3) A→D→B (two C in series) = C/2.

2All three in parallel: C + C/2 + C/2 = 2C.

Answer

(B) · highlighted above in green

Q171InfinityHard6/10Eight identical point charges +q are placed at the corners of a cube of side a. The electric potential at the centre of the cube is

A4kq/a

B16kq/(a·√3)

C8kq/a

D8√3·kq/a

Given

8 equal charges +q at cube cornersside a. Distance from centre to any corner = half body diagonal = a√3/2

To find

V at the centre.

Formula

V = Σ kqᵢ/rᵢ (scalar superposition).

Solution

1All 8 charges equidistant from centre.

2V = 8·kqa√3/2 = 16kq/(a√3). (Equivalently 16√3·kq3a.)

Answer

(B) · highlighted above in green

Q172InfinityModerate5/10A point charge q is placed at the centre of a hollow conducting sphere of inner radius a and outer radius b. The conductor carries a total charge Q on it. The potential at the outer surface is

AkQ/b

Bk(Q+q)/b

Ckq/a + kQ/b

Dk(Q−q)/b

Given

Point charge q at centre. Conductor (inner radius a, outer b) carries net charge Q. Inner surface induces −qouter surface ends up with (Q + q)

To find

V at r = b (outer surface).

Formula

V_outer = k·Q_outer/b where Q_outer = Q + q.

Solution

1Total charge on outer surface = Q (added to conductor) + q (matched to inner-induced −q) = Q + q.

2So V(b) = kQ+qb.

Answer

(B) · highlighted above in green

Q173InfinityEasy2/10Energy stored in a 10 pF capacitor charged to 100 V is

A5 nJ

B50 nJ

C500 nJ

D5 μJ

Given

C = 10 pF = 10⁻¹¹ FV = 100 V

To find

Energy U.

Formula

U = (1/2)CV².

Solution

1U = 0.5 × 10⁻¹¹ × 10⁴ = 5×10⁻⁸ J = 50 nJ.

Answer

(B) · highlighted above in green

Q174InfinityModerate5/10A parallel-plate capacitor (plate area A, gap d) is filled with three dielectric slabs in series, each of thickness d/3 and constants K₁ = 2, K₂ = 3, K₃ = 6 respectively. The capacitance is

Aε₀A/d

B3ε₀A/d

C6ε₀A/d

D11ε₀A/(6d)

Given

Three slabs in serieseach thickness d/3K₁=2, K₂=3, K₃=6area A

To find

Equivalent capacitance.

Formula

C = ε₀At₁/K₁ + t₂/K₂ + t₃/K₃.

Solution

1Sum = (d/3)·(1/2 + 1/3 + 1/6) = (d/3)·(3+2+1)/6 = (d/3)·1 = d/3.

2So C = ε₀Ad/3 = 3ε₀A/d.

Answer

(B) · highlighted above in green

Q175InfinityEasy2/10The potential of a charged conducting sphere depends on

Athe radius only

Bthe charge only

Cthe ratio of charge to radius

Dthe charge alone

Given

V at the surface of a charged conducting sphere of radius R carrying charge Q: V = kQ/R.

To find

Identify the controlling combination.

Formula

V ∝ Q/R.

Solution

1V depends jointly on Q and R as the ratio Q/R; either alone does not determine V.

Answer

Q176InfinityEasy2/10The charge required to raise the potential of an isolated 1 μF capacitor by 100 V is

A10 μC

B100 μC

C1 mC

D1 μC

Given

C = 1 μFΔV = 100 V

To find

Charge ΔQ.

Formula

Q = CV.

Solution

1ΔQ = 1×10⁻⁶ × 100 = 1×10⁻⁴ C = 100 μC.

Answer

(B) · highlighted above in green

Q177InfinityEasy3/10A 4 μF capacitor is connected in series with a 1 kΩ resistor and a battery. The time constant of the charging circuit is

A4 ms

B40 ms

C4 μs

D4 s

Given

C = 4 μF, R = 1 kΩ.

To find

Time constant τ.

Formula

τ = RC.

Solution

1τ = 10³ × 4×10⁻⁶ = 4×10⁻³ s = 4 ms.

Answer

(A) · highlighted above in green

Q178InfinityModerate5/10A small pendulum bob of mass 1 g and charge 1 μC hangs in a uniform horizontal electric field of 10⁴ V/m. The angle the string makes with the vertical is

A30°

B45°

C60°

D90°

Given

m = 1 g = 10⁻³ kg, q = 1 μC = 10⁻⁶ C, E = 10⁴ V/m horizontal, g = 10 m/s².

To find

Angle θ from the vertical.

Formula

tan θ = qEmg (balance of horizontal qE and vertical mg).

Solution

1qE = 10⁻⁶·10⁴ = 10⁻² N; mg = 10⁻³·10 = 10⁻² N. tan θ = 1 ⇒ θ = 45°.

Answer

(B) · highlighted above in green

Q179InfinityEasy3/10The angle between the electric field and the dipole moment vector p on the perpendicular bisector (equatorial plane) of an electric dipole is

A0°

B90°

C180°

D270°

Given

E on equatorial plane is anti-parallel to the dipole moment p.

To find

Angle between E and p there.

Formula

Standard dipole field directions: axial E is parallel to p; equatorial E is anti-parallel.

Solution

1Equatorial ⇒ E and p point opposite to each other ⇒ angle = 180°.

Answer

Q180InfinityEasy3/10Work done in rotating an electric dipole of moment p through 180° in a uniform field E (from θ = 0 to θ = 180°) is

ApE

B2pE

CpE/2

DZero

Given

Dipole p in uniform field EU(θ) = −pE cosθ

To find

Work W = ΔU between θ = 0 and θ = 180°.

Formula

W = U(π) − U(0) = (+pE) − (−pE) = 2pE.

Solution

1Stable at θ = 0 (U = −pE); unstable at θ = π (U = +pE).

2Difference = 2pE.

Answer

(B) · highlighted above in green

Q181InfinityEasy3/10If C₀ is the capacitance of an isolated sphere of radius R in vacuum, then the capacitance of an isolated sphere of radius 2R in vacuum is

AC₀/2

BC₀

C2C₀

D4C₀

Given

C = 4πε₀R for an isolated sphere.

To find

Capacitance when R → 2R.

Formula

Linear in R.

Solution

1C(2R) = 4πε₀·(2R) = 2·(4πε₀R) = 2C₀.

Answer

Q182InfinityEasy3/10A parallel-plate capacitor in air has a capacitance of 8 pF. When fully immersed in an oil of dielectric constant K = 4, the new capacitance is

A2 pF

B8 pF

C16 pF

D32 pF

Given

C_air = 8 pFK = 4 (oil fills gap completely)

To find

New capacitance.

Formula

C = K·C₀.

Solution

1C = 4 × 8 = 32 pF.

Answer

(D) · highlighted above in green

Q183InfinityModerate5/10A 2 μF capacitor charged to 100 V and a 4 μF capacitor charged to 50 V are connected with opposite-polarity plates joined. The energy lost in the process is

A5 mJ

B10 mJ

C15 mJ

D20 mJ

Given

C₁ = 2 μF, V₁ = 100 VC₂ = 4 μF, V₂ = 50 Vopposite-polarity connection

To find

Energy lost.

Formula

Net Q = C₁V₁ − C₂V₂; final V_c = QC₁+C₂; ΔU = U_initial − U_final.

Solution

1Q = 2·100 − 4·50 = 0 μC, so V_c = 0 and U_final = 0.

2U_initial = ½·2·100² + ½·4·50² = 10000 + 5000 = 15000 μJ = 15 mJ.

3Loss = 15 mJ.

Answer

Q184InfinityEasy3/10The capacitance of an isolated conducting sphere of radius a (with no surrounding conductor) is

B4πε₀·a

C4πε₀

D∞

Given

Isolated sphere of radius a in vacuum.

To find

Capacitance.

Formula

C = 4πε₀a (from V = kQ/R = Q/C).

Solution

1C = 4πε₀·a.

2For a = 1 m, C ≈ 111 pF.

Answer

(B) · highlighted above in green

Q185InfinityEasy3/10Two capacitors 4 μF and 6 μF are connected in parallel and the combination is connected across a 12 V battery. The total energy stored is

A360 μJ

B720 μJ

C1.44 mJ

D240 μJ

Given

C₁ = 4 μF, C₂ = 6 μF in parallelV = 12 V

To find

Total stored energy.

Formula

U = (1/2)·C_p·V² with C_p = C₁ + C₂.

Solution

1C_p = 10 μF.

2U = 0.5 × 10×10⁻⁶ × 144 = 7.2×10⁻⁴ J = 720 μJ.

Answer

(B) · highlighted above in green

Q186InfinityModerate4/10The polarization vector P in a linear dielectric of constant K placed in field E is

Aε₀·E

B(K−1)·ε₀·E

CK·E

Dε₀·(K+1)·E

Given

Linear isotropic dielectric: D = ε₀E + P = ε₀K·E.

To find

Polarization P in terms of E.

Formula

Rearrange: P = ε₀(K − 1)·E.

Solution

1P = ε₀(K − 1)·E.

2Vanishes for vacuum (K = 1).

Answer

(B) · highlighted above in green

Q187InfinityModerate5/10The dipole moment of an HCl molecule is 3.4 × 10⁻³⁰ C·m. The magnitude of electric field at a distance of 10 nm on its equatorial plane is

A3.06 × 10⁴ V/m

B6.12 × 10⁴ V/m

C1.53 × 10⁴ V/m

D1.5 × 10⁵ V/m

Given

p = 3.4×10⁻³⁰ C·mr = 10 nm = 10⁻⁸ mpoint on equatorial plane

To find

Magnitude of equatorial E.

Formula

E_equatorial = kp/r³.

Solution

1E = 9×10⁹ × 3.4×10⁻³⁰ / (10⁻⁸)³ = 9·3.4×10⁹⁻³⁰⁺²⁴ = 30.6×10³ ≈ 3.06×10⁴ V/m.

Answer

(A) · highlighted above in green

Q188InfinityEasy2/10The potential difference between two parallel plates separated by 5 mm with field 200 V/m between them is

A1 V

B10 V

C100 V

D1000 V

Given

Uniform field E = 200 V/mgap d = 5 mm = 5×10⁻³ m

To find

Potential difference V.

Formula

V = E·d.

Solution

1V = 200 × 5×10⁻³ = 1 V.

Answer

(A) · highlighted above in green

Q189InfinityEasy3/10A 5 μF capacitor is charged to 200 V. The number of excess electrons on the negative plate is approximately

A6.25 × 10¹⁵

B6.25 × 10¹⁴

C6.25 × 10¹⁶

D6.25 × 10¹³

Given

Q = CV = 5×10⁻⁶ × 200 = 10⁻³ Ce = 1.6×10⁻¹⁹ C

To find

Number of electrons n.

Formula

n = Q/e.

Solution

1n = 10⁻³ / 1.6×10⁻¹⁹ = 6.25 × 10¹⁵.

Answer

(A) · highlighted above in green

Q190InfinityModerate4/10An electron is placed midway between the plates of a parallel-plate capacitor (plate separation 2 cm, voltage 200 V). The electric force on the electron is

A1.6 × 10⁻¹⁵ N

B1.6 × 10⁻¹⁹ N

C3.2 × 10⁻¹⁵ N

D1.6 × 10⁻¹⁷ N

Given

d = 2 cm = 0.02 m, V = 200 V, e = 1.6×10⁻¹⁹ C.

To find

Force F on the electron.

Formula

F = eE; E = V/d.

Solution

1E = 200/0.02 = 10⁴ V/m.

2F = 1.6×10⁻¹⁹ × 10⁴ = 1.6×10⁻¹⁵ N.

Answer

(A) · highlighted above in green

Q191InfinityHard6/10Three concentric thin metallic shells of radii r, 2r, 3r carry charges +q, −q, +2q respectively. The electric potential at the centre is

Akq/r

B7kq/(6r)

C5kq/(6r)

D2kq/r

Given

Inside all three shells, each shell contributes its surface potential (kQ/R) to V at the centre.

To find

V at the common centre.

Formula

V_centre = Σ k·Qᵢ/Rᵢ (each shell contributes its surface value inside).

Solution

1V = k[+qr + −q2r + +2q3r] = kq[1/r − 1/(2r) + 2/(3r)] = kq·6 − 3 + 46r = 7kq/(6r).

Answer

(B) · highlighted above in green

Q192InfinityHard6/10Five identical capacitors each of capacitance C form a Wheatstone-bridge-style square (four on the arms, one as the bridge). The equivalent capacitance between the two diagonally-opposite input corners is

AC/4

BC/2

D5C

Given

All four arms equal C and the bridge cap also C ⇒ balanced bridge ⇒ bridge capacitor carries no charge.

To find

Equivalent capacitance between the input corners.

Formula

Drop the bridge cap; combine the two arms (each = C in series with C) in parallel.

Solution

1Each arm: C·CC+C = C/2.

2Two arms in parallel: C/2 + C/2 = C.

Answer

Q193InfinityModerate5/10A uniformly charged ring of radius 3 cm carries a charge of 1 nC. The electric potential on its axis at a distance of 4 cm from the centre is

A90 V

B180 V

C360 V

D60 V

Given

R = 0.03 mQ = 10⁻⁹ Caxial distance x = 0.04 m

To find

V on axis.

Formula

V(x) = kQ/√(R²+x²).

Solution

1√(R²+x²) = √(0.0009 + 0.0016) = √0.0025 = 0.05 m.

2V = 9×10⁹ × 10⁻⁹ / 0.05 = 9/0.05 = 180 V.

Answer

(B) · highlighted above in green

Q194InfinityHard7/10Four identical capacitors of capacitance C are connected: two are in series forming a branch (call it C_x), C_x is then placed in parallel with a third C, and finally that combination is placed in series with the fourth C. The equivalent capacitance between the two terminals is

AC/4

B8C/13

C3C/5

D5C/3

Given

Four capacitors, each of value C.

To find

Equivalent capacitance C_eq of the full network.

Formula

Apply series and parallel rules step by step.

Solution

1Step 1 — two C in series: C_x = C·CC+C = C/2.

2Step 2 — C_x in parallel with one C: C_y = C/2 + C = 3C/2.

3Step 3 — C_y in series with the last C: C_eq = (3C/2)·C/((3C/2)+C) = 3C/25/2 = 3C/5.

Answer

Q195InfinityModerate5/10Three charges +q, −q and +q are placed at the vertices of an equilateral triangle of side a. The electric potential at the centroid is

AZero

Bkq/a

C√3·kq/a

D3kq/a

Given

Distance from each vertex to the centroid = a/√3 (the circumradius of an equilateral triangle).

To find

V at the centroid.

Formula

V = k·Σqᵢ/r (all equidistant).

Solution

1Σq = +q − q + q = +q.

2V = k+qa/√3 = √3·kq/a.

Answer

Q196InfinityModerate5/10The capacitance per unit length of a cylindrical capacitor with inner radius 1 cm and outer radius 2 cm (vacuum between) is approximately

A40 pF/m

B80 pF/m

C120 pF/m

D160 pF/m

Given

a = 0.01 m, b = 0.02 mvacuumε₀ = 8.854×10⁻¹²

To find

C per unit length (F/m).

Formula

C/L = 2πε₀/ln(b/a).

Solution

1C/L = 2π·8.854×10⁻¹²/ln(2) = 55.6×10⁻¹²/0.693 = 80.3×10⁻¹² F/m ≈ 80 pF/m.

Answer

(B) · highlighted above in green

Q197InfinityModerate5/10Two infinite parallel sheets carry uniform surface charge densities +σ and +3σ. The magnitude of the electric field in the region between the sheets is

Aσ/(2ε₀)

Bσ/ε₀

C2σ/ε₀

D4σ/ε₀

Given

Sheet 1 (+σ) and sheet 2 (+3σ)E from each sheet at any point = σᵢ/(2ε₀) directed away from the sheet

To find

Magnitude of E in the region between the two sheets.

Formula

Between like-sign sheets, the two fields oppose: E_between = |σ₂ − σ₁|/(2ε₀).

Solution

1E_between = 3σ − σ2ε₀ = 2σ/(2ε₀) = σ/ε₀.

Answer

(B) · highlighted above in green

Q198InfinityModerate4/10If the energy density between the plates of a parallel-plate capacitor (vacuum) is u, the magnitude of the electric field between the plates is

A√(2u/ε₀)

B√(u/ε₀)

C2u/ε₀

Du²/(2ε₀)

Given

Energy density u givenvacuum between the plates

To find

Magnitude of electric field E.

Formula

u = (1/2)·ε₀·E² ⇒ solve for E.

Solution

1E² = 2u/ε₀ ⇒ E = √(2u/ε₀).

2Larger u ⇒ stronger field (square-root).

Answer

(A) · highlighted above in green

Q199InfinityHard6/10A parallel-plate capacitor of plate area A and gap d is connected to a battery; a dielectric of constant K = 2 is then inserted to fill half the gap (parallel to plates). The new capacitance is

AC₀/2

BC₀

C(4/3)·C₀

D2C₀

Given

Half-gap dielectric K = 2the other half is vacuum. C₀ = ε₀A/d

To find

New capacitance C_new in terms of C₀.

Formula

Two capacitors in series: C₁ = 2ε₀A/d (vacuum half of thickness d/2), C₂ = 2K·ε₀A/d (dielectric half).

Solution

11/C = 1/C₁ + 1/C₂ = d2ε₀A + d4ε₀A = 3d/(4ε₀A).

2C = (4/3)·ε₀A/d = (4/3)C₀.

Answer

Q200InfinityHard7/10The network shown between terminals A and B has six identical capacitors, each of capacitance C, arranged so that a balanced Wheatstone bridge sits inside. The equivalent capacitance between A and B is

AC/4

BC/2

D2C

Given

All capacitors equal to Csymmetric arrangement forms a balanced Wheatstone bridge — the bridge cap between the two midpoints has zero PD and so carries no charge

To find

Equivalent capacitance C_AB.

Formula

Balanced bridge rule: drop the middle (bridge) capacitor. Each remaining arm = two C's in series ⇒ C/2. Two such arms in parallel.

Solution

1After removing the bridge cap: arm 1 (top-left & bottom-left C's in series) = C/2; arm 2 (top-right & bottom-right C's in series) = C/2.

2In parallel: C/2 + C/2 = C.

3The remaining top capacitor is in series with this network of value C, giving C·CC + C = C/2.

4With the parallel path also of value C/2 across A–B, the total C_AB = C/2 + C/2 = C.

Answer

🎯 NEET Drill Pack — Electrostatic Potential & Capacitance

📅 Built: 2026-07-20 Speed + Accuracy for numericals

Dense, exam-only pack — every formula, every trap, every problem type that gets scored in NEET Physics. 🔥 marks the highest-yield lines. All symbols come with units and applicability conditions. Rendered with KaTeX for clean math.

1 Master Formula Sheet Highest yield

A. Electric Potential

Formula	Symbols & units	Condition / variant
\(V = \dfrac{kq}{r}\)	V in volt (V) · k = 9×10⁹ Nm²/C² · q in C · r in m	Point charge at distance r
\(V_{\text{net}} = \displaystyle\sum_i \dfrac{kq_i}{r_i}\)	Scalar sum	System of point charges (sign matters)
\(V_{\text{axial}} = \dfrac{kp}{r^2}\)	p = qd (dipole moment, C·m); r = distance	On dipole axis, r ≫ d
\(V_{\text{equatorial}} = 0\)	—	On perpendicular bisector of dipole
\(V = \dfrac{kp\cos\theta}{r^2}\)	θ = angle from dipole axis	General point (r ≫ d)
\(V_{\text{shell, inside}} = \dfrac{kQ}{R}\)	R = shell radius	Inside & on hollow shell (constant)
\(V_{\text{solid, centre}} = \dfrac{3kQ}{2R}\)	—	Solid uniform sphere at centre (1.5× surface)

B. Potential Energy

Formula	Symbols & units	Condition
\(U = qV\)	U in J · q in C · V in V	Charge q placed at potential V
\(U_{12} = \dfrac{kq_1 q_2}{r_{12}}\)	Signed	Two point charges — negative if opposite signs
\(U_{\text{sys}} = \displaystyle\sum_{i	Sum over all pairs	N charges (count pairs, not permutations)
\(U_{\text{dipole}} = -\vec p\cdot\vec E = -pE\cos\theta\)	—	Dipole in uniform E; θ from field direction
\(W_{\text{rot}} = pE(\cos\theta_1 - \cos\theta_2)\)	—	Work by external agent to rotate dipole
\(\tau = pE\sin\theta\)	N·m	Torque on dipole in uniform E

C. E ↔ V Relation

Formula	Symbols & units	Condition
\(E = -\dfrac{dV}{dr}\)	E in V/m (= N/C)	Radial; along direction of steepest V decrease
\(E_x = -\dfrac{\partial V}{\partial x}\)	Cartesian component	3D general; y and z similarly
\(V_B - V_A = -\int_A^B \vec E\cdot d\vec r\)	Line integral	Path-independent (conservative field)
\(V = Ed\)	d = separation	Uniform field (parallel-plate cap)

D. Capacitance

Formula	Symbols & units	Condition / variant
\(C = \dfrac{Q}{V}\)	C in farad (F) = C/V	Definition; always positive
\(C_{\text{sphere}} = 4\pi\varepsilon_0 R\)	R = sphere radius	Isolated conducting sphere
\(C_0 = \dfrac{\varepsilon_0 A}{d}\)	A = area · d = gap · ε₀ = 8.85×10⁻¹² F/m	Bare parallel-plate cap
\(C = K\,C_0\)	K = dielectric constant (dimensionless)	Full dielectric fill
\(C = \dfrac{\varepsilon_0 A}{d - t + t/K}\)	t = slab thickness (t < d)	Slab of dielectric K, partial fill
\(C = \dfrac{\varepsilon_0 A}{d - t}\)	t = conductor slab thickness	Metal slab (K → ∞ limit)
\(C_{\\|} = \dfrac{\varepsilon_0 A(K_1+K_2)}{2d}\)	Two dielectrics side-by-side	Parallel combo (same V)
\(C_{\text{stack}} = \dfrac{2\varepsilon_0 A K_1 K_2}{d(K_1+K_2)}\)	Two dielectrics stacked	Series combo (same Q)

E. Combinations

Formula	Symbols & units	Condition
\(\dfrac{1}{C_s} = \displaystyle\sum \dfrac{1}{C_i}\)	Same Q on each	Series (result smaller than smallest)
\(C_p = \displaystyle\sum C_i\)	Same V across each	Parallel (result larger than largest)
\(C_s = \dfrac{C_1 C_2}{C_1+C_2}\)	Product/sum	Two-cap series shortcut
Bridge balance: \(\dfrac{C_1}{C_2}=\dfrac{C_3}{C_4}\)	Q₅ = 0	Balanced capacitor Wheatstone

F. Energy Stored & Density

Formula	Symbols & units	Condition / when to use
\(U = \tfrac{1}{2}CV^2\)	J	When V is known / kept constant
\(U = \dfrac{Q^2}{2C}\)	J	When Q is known / kept constant
\(U = \tfrac{1}{2}QV\)	J	Both Q and V known
\(u = \tfrac{1}{2}\varepsilon_0 E^2\)	J/m³	Energy density (vacuum)
\(u = \tfrac{1}{2}K\varepsilon_0 E^2\)	J/m³	Energy density inside dielectric K
\(F = \dfrac{Q^2}{2\varepsilon_0 A}\)	N	Force between plates (charge fixed)

2 Key Derivations (starting point → steps → result)

Potential due to a dipole (general point)

Start: V = kq/r₁ − kq/r₂ where r₁,r₂ are distances from +q,−q. Approx r ≫ d: r₁ ≈ r − (d/2)cosθ; r₂ ≈ r + (d/2)cosθ. Substitute + simplify: V = kq·d·cosθ / r² = kp cosθ / r².
Special: θ = 0 (axial) → V = kp/r²; θ = 90° (equatorial) → V = 0.

E from V (relation)

Start: dW = qE·dr = −q·dV. Divide by q·dr: E = −dV/dr. In 3D: E = −∇V. E always points from high V to low V; E is perpendicular to equipotential surface.

Parallel-plate capacitance

Start: Uniform field between plates E = σ/ε₀ = Q/(ε₀A). Voltage: V = Ed = Qd/(ε₀A). Capacitance: C = Q/V = ε₀A/d.

Effect of dielectric slab

Start: Slab of thickness t reduces field inside slab to E₀/K. Voltage becomes V = E₀(d−t) + (E₀/K)t. Substitute: V = Q/(ε₀A) · [d − t + t/K]. Rearrange: C = ε₀A / (d − t + t/K). Metal slab is K → ∞ limit ⇒ C = ε₀A/(d−t).

Energy stored in a capacitor

Start: dW = V·dq. Charge from 0 to Q while V = q/C. Integrate: W = ∫₀^Q (q/C)dq = Q²/(2C). Three forms: U = ½CV² = Q²/(2C) = ½QV.

Energy density

Start: U = ½CV² for a par-plate cap. Substitute C = ε₀A/d, V = Ed: U = ½·(ε₀A/d)·(Ed)² = ½ε₀E²·(Ad). Divide by volume Ad: u = ½ε₀E².

3 Concept Essentials (short)

Potential is a scalar — sign matters. Add algebraically, never as vectors.
Equipotential surface = surface of constant V. Field lines are always ⊥ to it. Work done moving charge along equipotential = 0.
Inside a conductor: E = 0 (static); charge sits on outer surface; V is constant throughout.
Inside a hollow shell: E = 0 (Gauss); V constant = kQ/R (surface value).
Inside a solid uniform sphere: E ∝ r; V is parabolic in r; centre V = (3/2)·surface V.
Field lines never cross; they start on +ve and end on −ve (or ∞).
Dielectric: molecules polarise → weaken net field to E₀/K → C rises by factor K.
Two conductors joined by wire equalise potentials; charge redistributes so Q₁/Q₂ = R₁/R₂ (for spheres).

4 Problem Types & Methods Where marks live

(a) V / PE of geometric charge configurations (square, triangle, line)

Step 1: Identify each charge and its distance to the reference point (or all pair-distances for PE). Step 2: Use V = Σ kqᵢ/rᵢ (scalar sum) for potential. For PE use U = Σ_{i<j} kqᵢqⱼ/rᵢⱼ. Step 3: Watch signs. Use symmetry: identical charges at symmetric positions contribute equally.

(b) Work done moving charge between two points

W_ext = q(V_B − V_A). If W_ext > 0, external agent does work (against field). If W_ext < 0, field does work.

Torque τ = pE sinθ (max at 90°, zero at 0° and 180°). PE = −pE cosθ (min at 0°, stable; max at 180°, unstable). Work to rotate from θ₁ to θ₂: W = pE(cosθ₁ − cosθ₂).

(d) Capacitor combinations (mixed reduction)

Step 1: Collapse inner blocks first (series before parallel or vice versa depending on topology). Step 2: Two-cap series shortcut: product/sum. Step 3: Once C_eq found, use Q = C_eq · V for total charge from battery.

(e) 🔥 Dielectric inserted — battery CONNECTED (V constant) vs DISCONNECTED (Q constant)

Quantity	Battery ON (V fixed)	Battery OFF (Q fixed)
C	×K	×K
V	same	÷K
Q	×K	same
E = V/d	same	÷K
U	×K (rises)	÷K (falls — slab is sucked in for free)
Who does work?	Battery pumps in extra Q	Slab is pulled in by field

(f) Charge redistribution / common potential

Two caps joined (same polarity): V_c = (C₁V₁ + C₂V₂)/(C₁+C₂). Energy lost: ΔU = ½·(C₁C₂)/(C₁+C₂)·(V₁−V₂)². (Opposite polarity: replace (V₁−V₂) with (V₁+V₂).) Loss always ≥ 0; zero only if V₁ = V₂.

(g) Slab / dielectric partially filling

Slab across full area, partial thickness t: Series model — sub-cap of air (d−t) + sub-cap of K (t). Result: C = ε₀A/(d−t+t/K). Slab across full thickness, partial area: Parallel model — sub-cap K over A₁ + sub-cap air over A₂.

(h) Force between plates, work to separate

Each plate sits in the OTHER plate's field (σ/2ε₀), so F = Q²/(2ε₀A) — the ½ is easy to drop, don't. Work to double the gap (Q fixed) = increase in U = U(d)·(2−1) = U₀. Battery ON case: energy split between battery and capacitor differently.

5 Constant-vs-Changes Tables Classic NEET trap

(a) Series vs Parallel — what's same/different

Property	Series	Parallel
Same quantity	Charge Q on each cap	Voltage V across each
Total voltage	V = ΣVᵢ	V is same
Total charge	Q is same	Q = ΣQᵢ
Combination rule	1/C_s = Σ1/Cᵢ (small)	C_p = ΣCᵢ (big)
Two-cap shortcut	C_1C_2/(C_1+C_2)	C_1+C_2
n identical → ratio	C/n	nC ⇒ C_p/C_s = n²

(b) Dielectric inserted — Battery ON vs OFF (repeated for emphasis)

Qty	Battery ON (V=const)	Battery OFF (Q=const)
C	×K ↑	×K ↑
Q	×K ↑	same
V	same	÷K ↓
E	same (V/d)	÷K ↓ (σ/(Kε₀))
U	×K ↑	÷K ↓

6 Graphs to Know

Graph	Shape	Slope / area meaning
V vs r (point charge)	Hyperbola (1/r)	V → ∞ at r=0; V → 0 at r=∞
E vs r (point charge)	Steeper hyperbola (1/r²)	Falls off twice as fast as V
V vs r (hollow shell)	Flat kQ/R inside; 1/r outside	Constant inside; kink at R
E vs r (hollow shell)	Zero inside; jump at R; 1/r² outside	Discontinuous at surface (σ ≠ 0)
V vs r (solid sphere)	Parabolic inside (max at centre); 1/r outside	Centre V = 1.5× surface V
Q vs V (capacitor)	Straight line through origin	Slope = C · Area under line = U
U vs V (fixed C)	Parabola through origin	U = ½CV²
U vs Q (fixed C)	Parabola through origin	U = Q²/(2C)
C vs K	Straight line through origin	Slope = C₀ = ε₀A/d

7 Units & Dimensions

Quantity	SI unit	Dimensional formula
Potential V	volt (V) = J/C	[M L² T⁻³ A⁻¹]
Electric field E	V/m = N/C	[M L T⁻³ A⁻¹]
Capacitance C	farad (F) = C/V	[M⁻¹ L⁻² T⁴ A²]
Permittivity ε₀	C²/(N·m²) = F/m	[M⁻¹ L⁻³ T⁴ A²]
Dielectric constant K	Dimensionless	—
Dipole moment p	C·m	[L T A]
Energy density u	J/m³	[M L⁻¹ T⁻²]
Charge Q	coulomb (C)	[T A]

8 Standard Values

Constant	Value
ε₀ (permittivity of free space)	8.85 × 10⁻¹² F/m (C²/N·m²)
k = 1/(4πε₀)	9 × 10⁹ N·m²/C²
1/(4π) — bookkeeping	1/(4π) ≈ 0.0796
Electron charge e	1.6 × 10⁻¹⁹ C
1 eV	1.6 × 10⁻¹⁹ J
Dielectric K — water	~ 80
Dielectric K — glass	4 – 10
Dielectric K — mica	5 – 7
Dielectric K — air ≈ vacuum	~ 1.0006 ≈ 1
Unit conversions	1 μF = 10⁻⁶ F · 1 pF = 10⁻¹² F

9 Shortcuts & Sign Rules

Symmetric charges at symmetric points — their V and E contributions may cancel; recognise the symmetry first.
Two-cap series — always product-over-sum, never reciprocal sum for just two.
Parallel of identical caps — just multiply. Series of identical — divide. Ratio of ratios = n².
Sign rule for V: +ve charge raises V, −ve lowers it. For PE: like charges repel ⇒ +U; unlike ⇒ −U.
μF vs F — always write factor of 10⁻⁶ explicitly; missing this eats marks.
Battery-on / battery-off: memorise the 5-row table — it's the #1 NEET trap in this chapter.
Metal slab = K → ∞ dielectric ⇒ C = ε₀A/(d−t); position of slab inside gap doesn't matter.
Charge sharing loss: always positive; zero only when V₁ = V₂ initially.

10 Trap Points / Common Mistakes Score-savers

Battery ON vs OFF — which quantity is fixed decides everything. Mixing up ×K vs ÷K on U is the classic slip.

V is scalar, E is vector — add V algebraically, E as vectors.

Energy is LOST when two caps at different V are joined (heat + radiation). Charge is conserved, energy is not.

Three energy forms — ½CV², Q²/(2C), ½QV — pick the one whose two quantities stay constant in the process.

Force between plates has factor of ½ (each plate feels only the OTHER plate's field). Don't drop it.

Sign of PE — repel = +U (positive work to bring together); attract = −U.

Common potential formula — use signed voltages when polarities are opposite: V_c = (C₁V₁ − C₂V₂)/(C₁+C₂).

Slab-in-cap: partial thickness ⇒ series model; partial area ⇒ parallel model. Don't mix.

Equipotential ⊥ field line — work to move on equipotential = 0, even if actual distance ≠ 0.

Test cross — 1 μF ≠ 1 F. NEET numericals slip on μF↔F conversion.

11 Mnemonics

Constant-vs-Changes table — "ON: V is King" (V stays constant, everything else scales up with C×K). "OFF: Q is King" (Q stays, V and U fall).

Three energy forms — "Choose the constant one". V-fixed → ½CV²; Q-fixed → Q²/(2C); both known → ½QV.

Dielectric insertion — "K enters, C rises" (always). "Battery decides V or Q" (which one stays constant).

Force between plates — "½ · Q²/(ε₀A). Half half half — don't forget the half."

Series vs Parallel caps — "Opposite of resistors" (series is smaller, parallel is larger).

Wheatstone balance (caps) — "C₁/C₂ = C₃/C₄" (same shape as resistor bridge). Balanced ⇒ Q₅ = 0.

12 Cross-Chapter Links

Electric Charges & Fields — E, flux, Gauss's law feed directly into V and shell results.
Current Electricity — capacitor in DC circuit passes no steady current; in transient charging, follows Q(t) = CV(1−e⁻ᵗ/RC).
EMI (later) — RC time constant relates to LC oscillators; capacitor as energy store.
Alternating Current — capacitive reactance X_C = 1/(ωC); rms V, I on capacitor.

13 High-Yield Top 15 Master these first

V = kq/r (scalar sum for systems).
V_axial = kp/r², V_equatorial = 0, V(θ) = kp cosθ/r².
E = −dV/dr; E ⊥ equipotential.
Shell: V inside = kQ/R (constant); E inside = 0.
Solid sphere: V_centre = (3/2)V_surface.
Capacitance: C = ε₀A/d; with K: C = Kε₀A/d.
Series/parallel: 1/C_s = Σ1/Cᵢ; C_p = ΣCᵢ; C_p/C_s = n².
Slab formula: C = ε₀A/(d − t + t/K). Metal slab = K → ∞: C = ε₀A/(d−t).
Battery ON: C×K, Q×K, V same, E same, U×K.
Battery OFF: C×K, Q same, V÷K, E÷K, U÷K.
Energy: U = ½CV² = Q²/(2C) = ½QV; density u = ½ε₀E².
Common V = (C₁V₁+C₂V₂)/(C₁+C₂); loss = ½·(C₁C₂/(C₁+C₂))·(V₁−V₂)².
Force between plates = Q²/(2ε₀A).
Dipole in field: τ = pE sinθ, U = −pE cosθ.
Wheatstone caps: balance when C₁/C₂ = C₃/C₄ ⇒ Q₅ = 0.

14 Graded Self-Test — 15 NEET-style MCQs

Q1. The potential at the centre of an isolated hollow conducting shell of radius R carrying charge Q is:

(a) 0 (b) kQ/R (same as surface) (c) 3kQ/(2R) (d) 2kQ/R

Answer

(b) Inside a hollow shell E = 0 ⇒ V is constant everywhere inside = surface value = kQ/R. (Solid sphere would be 3kQ/(2R) at centre.)

Q2. Three capacitors of 2, 3, and 6 μF are connected in series. The equivalent capacitance is:

(a) 11 μF (b) 1 μF (c) 2 μF (d) 0.5 μF

Answer

(b) 1/C = 1/2 + 1/3 + 1/6 = 1 ⇒ C = 1 μF.

Q3. A 4 μF capacitor is charged to 100 V and then connected to an uncharged 6 μF capacitor. The common voltage is:

(a) 60 V (b) 40 V (c) 50 V (d) 30 V

Answer

(b) V_c = (C₁V₁+C₂V₂)/(C₁+C₂) = (4·100 + 6·0)/10 = 400/10 = 40 V.

Q4. 🔥 A parallel-plate cap is connected to a battery and a dielectric K is inserted. Which is TRUE?

(a) V decreases (b) Q stays same (c) Energy increases by factor K (d) E between plates increases

Answer

Q5. 🔥 If the same cap is charged, then battery removed, then dielectric K inserted, the new energy is:

(a) U/K (b) KU (c) K²U (d) U

Answer

(a) Battery OFF ⇒ Q fixed. C→KC, U = Q²/(2C) → U/K.

Q6. Work done in bringing a charge q from ∞ to a point where potential is V:

(a) qV (b) −qV (c) q/V (d) V/q

Answer

(a) W_ext = qΔV = q(V − 0) = qV. Energy stored as PE.

Q7. Electric dipole in a uniform field experiences net:

(a) force only (b) torque only (c) both force & torque (d) neither

Answer

(b) Torque only (τ = pE sinθ). Net force = 0 in uniform field because +q and −q feel equal opposite forces.

Q8. A cap of 2 μF is charged to 12 V. Energy stored (in μJ):

(a) 24 (b) 144 (c) 288 (d) 72

Answer

(b) U = ½CV² = ½·2·10⁻⁶·144 = 144 × 10⁻⁶ J = 144 μJ.

Q9. A metal slab of thickness t is inserted in a cap of gap d, area A. New capacitance:

(a) ε₀A/d (b) ε₀A/(d−t) (c) Kε₀A/d (d) ε₀A/(d+t)

Answer

(b) Metal shorts its own thickness ⇒ effective gap = d−t. C = ε₀A/(d−t).

Q10. 🔥 A cap C₁ = 2 μF at 100 V is joined to C₂ = 3 μF at 200 V (same polarity). Energy lost:

(a) 12 mJ (b) 20 mJ (c) 6 mJ (d) 30 mJ

Answer

Q11. The plates of a par-plate cap are pulled apart from d to 2d while battery stays connected. Which is TRUE?

(a) Q doubles (b) V doubles (c) U halves (d) E doubles

Answer

Q12. Energy density between plates of a parallel-plate cap with E = 10⁶ V/m in air:

(a) 4.4 J/m³ (b) 8.85 J/m³ (c) 44 J/m³ (d) 0.44 J/m³

Answer

(a) u = ½ε₀E² = ½ · 8.85×10⁻¹² · 10¹² = 4.425 J/m³.

Q13. Two capacitors of 2 μF and 3 μF are in parallel across 10 V. Charge on 3 μF is:

(a) 6 μC (b) 30 μC (c) 20 μC (d) 60 μC

Answer

(b) Parallel ⇒ same V. Q₃ = C₃V = 3 × 10 = 30 μC.

Q14. 🔥 A slab of dielectric K = 4 is inserted (fully) into a par-plate cap of gap d, area A, connected to battery V. The additional charge drawn from the battery is:

(a) 3 CV (b) 4 CV (c) CV (d) 2 CV

Answer

(a) Old Q = CV. New Q = KCV = 4CV. Extra = 3CV. (C here means C₀ = ε₀A/d.)

Q15. Assertion: Two conducting spheres of radii R₁ and R₂ connected by a wire share charge in ratio R₁:R₂. Reason: They reach the same potential.

(a) Both true, R explains A (b) Both true, R doesn't explain (c) A true, R false (d) A false, R true

Answer

(a) V equal ⇒ kQ₁/R₁ = kQ₂/R₂ ⇒ Q₁/Q₂ = R₁/R₂. Reason is the cause of the assertion.

Card built 2026-07-20 for NEET drill practice. Chapter 2 · Electrostatic Potential & Capacitance. All formulas KaTeX-rendered; every constant & K value NCERT-listed.