🎒 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.
Vpoint = kq/r; Vshell/conductor inside = kQ/R (= surface value).
Potential difference
💡 The "height drop" between two points. Always a difference — choose any reference; only differences matter.
VA − VB = WBA/q = − ∫BA 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.
Vtotal = Σ kqᵢ/rᵢ (scalar superposition).
📌 At midpoint between +q and −q: V = 0 (they cancel).
Dipole
Vaxial = ± kp/r² (sign by side); Vequatorial = 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.
- Einside = 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.
Einside 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/Cs = 1/C₁ + 1/C₂ + … ; same Q on each.
Parallel: Cp = 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 → ∞. |
| VA − VB = − ∫BA E · dl | Path-independent (conservative field). |
| E = − dV/dr ; Ex = −∂V/∂x etc. | Negative gradient: E points down-slope of V. |
| Usystem = ½ Σᵢⱼ kqᵢqⱼ/rᵢⱼ (i ≠ j) | Pairwise sum; ½ to avoid double counting. |
| Udipole = −p · E = −pE cosθ | Min at θ = 0° (stable); max at 180° (unstable). |
| Wext = q(Vf − Vi) | 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/Cs = Σ 1/Cᵢ ; Q common | Total V = sum of individual Vᵢ = Q·(1/Cs). |
| Parallel: Cp = ΣCᵢ ; V common | Total Q = sum of individual Qᵢ. |
| Two charged capacitors connected: Vcommon = (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.
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.
1
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.
2
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).
3
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 |
4
Two charged capacitors connected — energy is LOST
✔
Common potential: Vc = (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.
5
"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".
6
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).
7
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.
🧠 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 = 9×10⁹N·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)/2pair 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) 0V 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 μC1/Cs = 1/2 + 1/3 = 5/6 ⇒ Cs = 1.2 μF. Q = CsV = 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 VVc = (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 μFEach 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 VQ 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)/2Filling 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 μJBattery 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.
Result
+0
out of 800 · marking +4 / −5 / −4
Correct
0 × +4
Wrong
0 × −5
Left
0 × −4
Final marks
0
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 isA4.5×10⁵ VB9×10⁵ VC4.5×10⁴ VD9×10⁴ V
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 isA0B2kq/(a√2)Ckq/aD2kq/a
A0
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 isA80 JB160 JC240 JD320 J
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) isAkp/r²Bkp cosθ/r²C2kp/r²DZero
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 isAkp/r²B2kp/r²Ckp/r³DZero
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 isAke²/rBke²/(2r)C2ke²/rDke²/(4r)
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 isAkQ/LB4√2·kQ/LC2√2·kQ/LD0
AkQ/L
B4√2·kQ/L
C2√2·kQ/L
D0
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 isA100×B10×C1000×D100²/³ ×
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 isA0.45 JB0.9 JC4.5 JD9 J
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 isAPositiveBNegativeCZeroDCannot be decided
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
(C) · highlighted above in green
Q11NEET 2018Easy3/10If E = 0 at every point in a small region of space, then in that region V isAzeroBconstantC±∞Dvariable
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 isA1×10⁻⁴ N·mB2×10⁻⁴ N·mC4×10⁻⁴ N·mD2×10⁻³ N·m
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 isA1 JB1 keVC1.6×10⁻¹⁹ JD1 eV
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 isA2kQ/√(a²+y²)BkQ/√(a²+y²)C2kQ/(a+y)D0
A2kQ/√(a²+y²)
BkQ/√(a²+y²)
C2kQ/(a+y)
D0
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, isAV_s/2B3V_s/2CV_sD2V_s
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 areAparallel planesBconcentric spheresCconcentric circlesDcoaxial cylinders
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 RAE ≠ 0, V = 0BE = 0, V = 0CE = 0, V = kQ/RDE and V both vary
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
(C) · highlighted above in green
Q18NEET 2016Easy3/10Electric field lines and equipotential surfaces are alwaysAparallelBat 45°CperpendicularDintersect at any angle
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
(C) · highlighted above in green
Q19NEET 2021Easy2/10Work done in moving 2 C of charge along an equipotential surface isAZeroB2 JCDepends on lengthDDepends on charge
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 equalsA10 VB20 VC200 VD0.5 V
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 fieldBElectric potentialCForceDAcceleration
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 equilibriumAE ≠ 0, V variesBE = 0, V variesCE = 0, V = constantDE ≠ 0, V = constant
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
(C) · highlighted above in green
Q23NEET 2020Easy3/10Two equipotential surfaces having different potentialsAnever intersectBintersect at infinityCintersect at 90°Dintersect at point
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 areAconcentric spheresBparallel planes ⟂ to fieldCparallel planes along fieldDcylinders
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 isA+8 V/mB−8 V/mC+4 V/mDZero
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 isAkq²/aB2kq²/aC3kq²/aD6kq²/a
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
(C) · highlighted above in green
Q27NEET 2016Moderate4/10Two charges +2 μC and −2 μC at distance 0.5 m. PE of the system isA−0.072 JB+0.072 JC−7.2 JD7.2 J
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 isApEB−pECpE/2DZero
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 toAθ = 0Bθ = π/2Cθ = πDθ = π/3
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
(C) · highlighted above in green
Q30NEET 2014Moderate4/10An α-particle and a proton are accelerated through the same potential. Ratio of their KE isA1:1B1:2C2:1D4:1
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
(C) · highlighted above in green
Q31AIPMT 2010Moderate5/10Energy needed to assemble three charges +q, +q, −q at the vertices of an equilateral triangle (side a) isA−kq²/aB−3kq²/aC+kq²/aDZero
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 isA1.6×10⁻¹⁹ JB1.6×10⁻¹⁸ JC1.6×10⁻¹⁷ JD1.6×10⁻²⁰ J
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 onAthe pathBonly on V₁−V₂Cspeed of chargeDtime taken
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 isA4×10⁻⁴ CB8×10⁻⁴ CC4×10⁻² CD800 μC
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 byAε₀d/ABε₀A/dCε₀A·dDA/(ε₀d)
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 becomesA2CB4CCC/2DC/4
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 capacitanceA1.11 pFB11.1 pFC111 pFD1.11 nF
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 isA2.4 μFB4 μFC6 μFD10 μF
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 isA2.4 μFB10 μFC1.5 μFD24 μF
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 energyA½CVBCV²C½CV²DQ²/C
A½CV
BCV²
C½CV²
DQ²/C
Given
C, Vstandard energy formula
To find
Energy stored.
Formula
U = ½QV = ½CV² = Q²2C.
Solution
1U = ½CV².
Answer
(C) · highlighted above in green
Q41NEET 2017Moderate5/10If the separation between plates of a charged parallel-plate capacitor (battery disconnected) is doubled, the energy storedAhalvesBdoublesCquadruplesDunchanged
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 isA5×10⁻³ JB5×10⁻⁴ JC5×10⁻⁵ JD0.5 J
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 isA18 μCB36 μCC54 μCD12 μC
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 isA1 μFB11 μFC0.5 μFD3 μF
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 isA1B2C4D1/2
A1
B2
C4
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
(C) · highlighted above in green
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 isA2 μFB4 μFC1 μFD8 μF
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 isA66.7 VB75 VC50 VD25 V
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 isAZeroB½C₁V₁²+½C₂V₂² − ½(C₁+C₂)V_c²C½(V₁−V₂)²DC₁C₂(V₁−V₂)²
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+C₂².
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 isA1.25 μFB5 μFC2.5 μFD10 μF
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) areAnC, C/nBC/n, nCCn²C, CDC, C/n²
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 isA2/3 μFB1.5 μFC3 μFD1/3 μF
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 energyA100 μJB200 μJC50 μJD25 μJ
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 storeA100 μJB400 μJC200 μJD50 μJ
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 isA6 μFB3 μFC12 μFD1.5 μF
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 acrossAlarger CBsmaller CCequalDdepends on connection
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 isAC₀B4C₀CC₀/4D2C₀
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 becomesA4VBV/4CVD16V
A4V
BV/4
CV
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 storedAincreases by factor KBdecreasesCunchangedDincreases by K²
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 storedA×KB/KCunchangedD/K²
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 isAε₀L/dBε₀(L − x + Kx)/dCε₀(Lx + K)/dDε₀Kx/d
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, capacitanceA↑B↓CconstantD→ 0
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) becomesAE₀BKE₀CE₀/KD0
AE₀
BKE₀
CE₀/K
D0
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
(C) · highlighted above in green
Q63AIPMT 2013Moderate5/10Polarisation of a dielectric isA+B⟂ to ECD
A+
B⟂ to E
C
D
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 isAε₀A/(d−t)Bε₀A/dCε₀A/(d+t)Dε₀At/d
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 isA200 μCB100 μCC300 μCD66.7 μC
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 isAε₀A(K₁+K₂)/(2d)B2ε₀AK₁K₂/(d(K₁+K₂))Cε₀A·K₁/dDε₀A(K₁−K₂)/d
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 isA2ε₀AK₁K₂/(d(K₁+K₂))Bε₀A(K₁+K₂)/(2d)Cε₀AK₁K₂/dDε₀A(K₁−K₂)/d
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 = ½QVBU = ½CV²CU = Q²/2CDU = QV
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 isA50 μJB100 μJC200 μJD25 μJ
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 isAε₀EB½ε₀E²Cε₀E²DE²/2
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 becomesA2UB4UCU/2DU/4
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 onAonly its CBonly its VCboth C and VDonly Q
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
(C) · highlighted above in green
Q73AIPMT 2013Moderate5/10Energy stored per unit volume in a parallel-plate capacitor (vacuum) isA½ε₀(V/d)²Bε₀V²C(V/d)²D½ε₀E
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, energyAhalvesBdoublesCquadruplesDunchanged
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 isA200 μCB100 μCC50 μCD400 μC
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 isA16.67 mJB50 mJC33.3 mJD25 mJ
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+C₂².
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 isA3kq²/aB√3·kq²/aC3√3·kq²/aD6kq²/a
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
(C) · highlighted above in green
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 isA3kq/a²B√3·kq/a²CZeroD9kq/a²
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
(C) · highlighted above in green
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 isApEB2pECpE/2DZero
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 isA11.1 pFB22.2 pFC5.55 pFD33.3 pF
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 isA(C₁V₁+C₂V₂)/(C₁+C₂)B(C₁V₁−C₂V₂)/(C₁+C₂)CZeroD(V₁+V₂)/2
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 isA3C₀V²B4C₀V²C2C₀V²DC₀V²
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, isAσd/ε₀Bσ/(2ε₀)C2σd/ε₀D0
Aσd/ε₀
Bσ/(2ε₀)
C2σd/ε₀
D0
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 isA33.3 VB66.7 VC50 VD100 V
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) isA√(m_p/m_e)B√(m_e/m_p)C1Dm_p/m_e
A√(m_p/m_e)
B√(m_e/m_p)
C1
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 isA0BkQ/RCkQ/rDDecreases linearly with r
A0
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 isAC₀BC₀/2C2C₀D4C₀
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 isAq/(6ε₀)Bq/(8ε₀)Cq/(2ε₀)Dq/ε₀
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 isA54 μJB108 μJC162 μJD216 μJ
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 isAZeroBIn the direction of decreasing fieldCIn the direction of increasing fieldDPerpendicular to p
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
(C) · highlighted above in green
Q91NEET 2018Moderate5/10If 1000 identical charged drops (each with charge q, potential V, radius r) coalesce, the potential of the big drop isA100VB1000VC10VDV
A100V
B1000V
C10V
DV
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 isA6 mJB2.4 mJC4 mJD1.5 mJ
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+C₂².
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) isAPositiveBNegativeCZeroDDepends on path
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, isAkQ/xBkQ/√(R²+x²)CkQ/(R+x)DkQ/(2R)
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, isA½ε₀V²/d²B½CV²Cε₀V/dDCV²
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 isA4ε₀A/dB8ε₀A/(3d)C2ε₀A/dD6ε₀A/d
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 isA−qB+qCZeroDDepends on a and b
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 isACV₀BCV₀/2C2CV₀D½CV₀²/R
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) isAx = d/3Bx = d/5Cx = 4d/5Dx = 3d/4
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
(C) · highlighted above in green
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 isA0.25 μJB0.5 μJC1 μJD2 μJ
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) isAke²/(2r)Bke²/rC2ke²/rD4ke²/r
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 isA≈ 0.0042 JB≈ 0.0084 JC≈ 0.42 JD≈ 0.21 J
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+C₂².
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 isAV/3BV/2C2V/3DV
AV/3
BV/2
C2V/3
DV
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 isA3 pFB12 pFC48 pFD24 pF
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
(C) · highlighted above in green
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 C₀A2KC₀/(1+K)BKC₀/2C(1+K)C₀/2DC₀(1+K)
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 isAU₀BU₀/2CU₀/4D2U₀
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
(C) · highlighted above in green
Q107JEE Main 2020Hard6/10Energy density in a parallel-plate capacitor with V = 200 V and d = 2 mm (vacuum) isA4.43×10⁻² J/m³B4.43×10⁻³ J/m³C0.443 J/m³D4.43 J/m³
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 pFB0.334 pFC334 pFD3.34 pF
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 isA20 mJB4 mJC16 mJD25 mJ
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+C₂².
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) isA11:1B1:11C121:1D1:121
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 isAε₀E²B½ε₀E²C2ε₀E²DE²
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 isA38.4 μCB57.6 μCC96 μCD144 μC
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 isACBC/5C5CD25C
AC
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 isA½CV²B¼CV²CCV²D2CV²
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 becomesA1 mCB200 μCC5 mCD1000 μC
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
(C) · highlighted above in green
Q116JEE Main 2017Hard6/10A capacitor C is charged through a resistor R by a battery V. Total energy dissipated in R during full charging isACV²B½CV²C¼CV²DIndependent of R and equals ½CV²
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 isAσ(1 − 1/K)Bσ(K − 1)CσKDσ/K
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 isACVB2KCV/(1+K)CKCV/(1+K)DCV/(K+1)
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
(C) · highlighted above in green
Q119JEE Main 2022Hard6/10Energy stored in an isolated charged sphere of radius R with charge Q isAkQ²/RBkQ²/(2R)CkQ²/(4R)DZero
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 isA44.3 pFB442.7 pFC0.443 pFD4.43 pF
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
C.
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 isA(√5−1)C/2B(√5+1)C/2C2CDC
A(√5−1)C/2
B(√5+1)C/2
C2C
DC
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 isAε₀A/(t₁/K₁ + t₂/K₂)Bε₀A·(K₁+K₂)/dCε₀A/(K₁t₁+K₂t₂)Dε₀A·K₁K₂/(t₁+t₂)
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 isA216 μJB432 μJC72 μJD108 μJ
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 isARCBR/CCC/RD1/(RC)
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 isA½C₀V²(1+K)/2B½C₀V²(1+K)C½C₀V²(K+1)/(2K)D½KC₀V²/2
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 isA18 × 10¹¹ JBZeroC1.8 × 10¹¹ JD54 × 10¹¹ J
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) isA(5/8) qkB(8/5) qkC(8/3) qkD(3/8) qk
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 isA3V/2BV/4CV/2DV
A3V/2
BV/4
CV/2
DV
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 isAV/(2R)BV/(3R)CV/(4R)DV/(6R)
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 isA4B96C48D3
A4
B96
C48
D3
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
(C) · highlighted above in green
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) becomesA3K/(K+4) · C₀BK/(K+3) · C₀C2K/(K+3) · C₀D4K/(K+3) · C₀
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 isAq²/(8πε₀)·(1/a − 1/b)Bq²/(4πε₀)·(1/b − 1/a)Cq²/(8πε₀)·(1/b − 1/a)Dq²/(4πε₀)·(1/a − 1/b)
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 isA8910 × 10⁻¹⁰ JB9910 × 10⁻⁹ JC8910 × 10⁻⁹ JD9910 × 10⁻¹⁰ J
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
(C) · highlighted above in green
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 isA72 VB81 VC64 VD36 V
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 isA2B1C4D8
A2
B1
C4
D8
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
(C) · highlighted above in green
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) isAλ/(2πε₀)B−λR/(2πε₀r²)C(λ/(2πε₀)) ln(r/R)D(λ/(2πε₀)) ln(R/r)
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 isAZeroBq√2(Q₁+Q₂)/(4πε₀R)Cq(Q₁−Q₂)(√2−1)/(√2·4πε₀R)Dq(Q₁+Q₂)(√2+1)/(√2·4πε₀R)
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
(C) · highlighted above in green
Q138InfinityModerate4/10Choose the wrong statementAOn the perpendicular bisector of a dipole, E = 0 and V ≠ 0BField lines are always perpendicular to equipotential surfacesCWhen a proton and an electron move apart, their PE increasesDThe dielectric constant of a material can differ between DC and AC
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 KBCapacitance decreased by K, voltage increased by KCField reduced by K, voltage increased by KDVoltage increased by K, charge increased by K
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 isA10³ VB1 VC−12.8 × 10⁵ VD10 V
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
(C) · highlighted above in green
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 equalsA100(√3 − 2) VB+100(2 + √3) VC100(2 − √3) VD200(2 + √3) V
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
(C) · highlighted above in green
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) isA24 μCB8 μCC18 μCD12 μC
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) isA6 VB2 VC10 VD14 V
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) isA2 pFB10 pFC8 pFD88.42 pF
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, thenAE changes, V remains unchangedBE remains unchanged, V changesCBoth E and V changeDBoth E and V remain unchanged
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 areA−2Q, +2QB−Q, +3QC+Q, +3QD+2Q, −3Q
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
(C) · highlighted above in green
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 approximatelyA2.65 J/m³B1.95 J/m³C1.38 J/m³D0.69 J/m³
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
(C) · highlighted above in green
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 isA2.5 V/mB−2.5 V/mC0.4 V/mD−0.4 V/m
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 isA9 × 10¹¹ VBZeroC3 × 10⁹ VD1.25 × 10⁹ V
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 isA80 VB8 × 10⁶ VC9 VD8 × 10⁵ V
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 isArV/R²BR²V/r²CRV/r²DV/r
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
(C) · highlighted above in green
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 isA160 JB0 JC5 JD1.25 J
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+C₂².
Solution
1ΔU = 0.5 × (5×10⁻⁶)(5×10⁻⁶) × 100010×10⁻⁶² = 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 isA−13.6 VB−27.2 VC27.2 VD13.6 V
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
(C) · highlighted above in green
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 isA√((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))
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 bBa and cCa onlyDb and c
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 equalsA0B−qC−(r₁/r₂)·qD(r₁/r₂)·q
A0
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
(C) · highlighted above in green
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 isA9.9 × 10⁹ JB9.9 × 10¹⁰ JC9.9 × 10¹¹ JD9.9 × 10¹³ J
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 wasAmCΔT/sB√(2mCΔT/s)C√(2msΔT/C)DmsΔT/C
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
(C) · highlighted above in green
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 isA(C/2)(ε₁+ε₂)·ε₂BC(ε₁+ε₂)·ε₂C(C/2)(ε₁+ε₂)²DC(ε₁+ε₂)²
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
(C) · highlighted above in green
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 isA4/5B5/4C5/1D3/5
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) isAk√6B2k√6C2k√3D4k√3
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) isA+20 μCB+40 μCC+60 μCD+80 μC
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 isA½(V₁ + V₂)B(C₂V₁ + C₁V₂)/(C₁+C₂)C(C₁V₁ + C₂V₂)/(C₁+C₂)D(C₂V₁ − C₁V₂)/(C₁+C₂)
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 isAε₀K₁K₂a²·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]
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 isA12 μC, 12 μCB24 μC, 24 μCC6 μC, 8 μCD8 μC, 6 μC
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 isA2 μCB3 μCC5 μCD10 μC
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
(C) · highlighted above in green
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 equalsA(q/(4r))·√(1/(πε₀r))B(q/(2r))·√(1/(πε₀r))C(2q/r)·√(1/(πε₀r))D(q/r)·√(1/(πε₀r))
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 isA8B6C12D10
A8
B6
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 isAf/4Bf/2CfD4f
Af/4
Bf/2
Cf
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
(C) · highlighted above in green
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 isACB2CC3CD4C
AC
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 isA4kq/aB16kq/(a·√3)C8kq/aD8√3·kq/a
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 isAkQ/bBk(Q+q)/bCkq/a + kQ/bDk(Q−q)/b
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 isA5 nJB50 nJC500 nJD5 μJ
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 isAε₀A/dB3ε₀A/dC6ε₀A/dD11ε₀A/(6d)
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 onAthe radius onlyBthe charge onlyCthe ratio of charge to radiusDthe charge alone
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
(C) · highlighted above in green
Q176InfinityEasy2/10The charge required to raise the potential of an isolated 1 μF capacitor by 100 V isA10 μCB100 μCC1 mCD1 μC
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 isA4 msB40 msC4 μsD4 s
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 isA30°B45°C60°D90°
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 isA0°B90°C180°D270°
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
(C) · highlighted above in green
Q180InfinityEasy3/10Work done in rotating an electric dipole of moment p through 180° in a uniform field E (from θ = 0 to θ = 180°) isApEB2pECpE/2DZero
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 isAC₀/2BC₀C2C₀D4C₀
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
(C) · highlighted above in green
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 isA2 pFB8 pFC16 pFD32 pF
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 isA5 mJB10 mJC15 mJD20 mJ
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
(C) · highlighted above in green
Q184InfinityEasy3/10The capacitance of an isolated conducting sphere of radius a (with no surrounding conductor) isA0B4πε₀·aC4πε₀D∞
A0
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 isA360 μJB720 μJC1.44 mJD240 μJ
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 isAε₀·EB(K−1)·ε₀·ECK·EDε₀·(K+1)·E
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 isA3.06 × 10⁴ V/mB6.12 × 10⁴ V/mC1.53 × 10⁴ V/mD1.5 × 10⁵ V/m
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 isA1 VB10 VC100 VD1000 V
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 approximatelyA6.25 × 10¹⁵B6.25 × 10¹⁴C6.25 × 10¹⁶D6.25 × 10¹³
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 isA1.6 × 10⁻¹⁵ NB1.6 × 10⁻¹⁹ NC3.2 × 10⁻¹⁵ ND1.6 × 10⁻¹⁷ N
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 isAkq/rB7kq/(6r)C5kq/(6r)D2kq/r
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 isAC/4BC/2CCD5C
AC/4
BC/2
CC
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
(C) · highlighted above in green
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 isA90 VB180 VC360 VD60 V
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 isAC/4B8C/13C3C/5D5C/3
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
(C) · highlighted above in green
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 isAZeroBkq/aC√3·kq/aD3kq/a
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
(C) · highlighted above in green
Q196InfinityModerate5/10The capacitance per unit length of a cylindrical capacitor with inner radius 1 cm and outer radius 2 cm (vacuum between) is approximatelyA40 pF/mB80 pF/mC120 pF/mD160 pF/m
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 isAσ/(2ε₀)Bσ/ε₀C2σ/ε₀D4σ/ε₀
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 isA√(2u/ε₀)B√(u/ε₀)C2u/ε₀Du²/(2ε₀)
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 isAC₀/2BC₀C(4/3)·C₀D2C₀
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
(C) · highlighted above in green
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 isAC/4BC/2CCD2C
AC/4
BC/2
CC
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
(C) · highlighted above in green