← All Physics chapters  ·  Chapter 2 — Electrostatic Potential & Capacitance
NCERT Class 12 · Chapter 2 · Sub-topic

Effect of Dielectrics

A focused sub-topic master guide on the Effect of Dielectrics from NCERT Class 12 Physics Chapter 2 (Electrostatic Potential and Capacitance). Three sections: ① Concepts & NEET-style problems in Given / Asked / Formula / Technique / Shortcut format, ② Exception & instant-answer questions like 'symmetric corner charges → V = 0 at centre, no calculation', and ③ The high-yield extras most students miss. Maths typeset with KaTeX so every formula is crisp and copy-paste-able.

1.0 Concepts (8 ideas) 1.1 NEET-style problems (12) 2.0 Exception / shortcut Qs (15) 3.0 High-yield extras (12) 📚 Full chapter notes
📖 Source for theory: NCERT Class 12 Physics, Chapter 2 (Electrostatic Potential and Capacitance) — leph102.pdf on ncert.nic.in. All explanations, problems and shortcut tips below are original.

1 Effect of Dielectrics — concepts & NEET-style problems

First the eight core concepts you need cold, then twelve NEET-style numerical problems each broken down into Given / Asked / Formula / Technique / Shortcut / Answer.

1.1 What a dielectric does in an electric field

A dielectric is an electrical insulator that gets polarised when placed in an electric field. Its atoms or molecules either rotate (polar molecule) or stretch (non-polar molecule) so that the centres of positive and negative charge no longer coincide. Each molecule then behaves as a tiny dipole. The induced dipole moment per unit volume is the polarisation vector \(\mathbf{P}\).

\[ \mathbf{P} \;=\; \varepsilon_0\,\chi_e\,\mathbf{E} \qquad K \;=\; 1 + \chi_e \]
🎈 Plain EnglishImagine a sponge made of tiny magnets pointing every which way. Stand it in a wind (electric field) — every little magnet swivels to face one way. The sponge as a whole now leans in the wind's direction, even though no single magnet moved.

1.2 Reduction of field inside a dielectric

The induced surface charges on the dielectric produce a field opposite to the external field. The net (effective) field inside the dielectric is therefore SMALLER than the external field by a factor of the dielectric constant \(K\) (also called relative permittivity \(\varepsilon_r\)).

\[ E_{\text{inside}} \;=\; \dfrac{E_0}{K} \qquad V_{\text{inside}} \;=\; \dfrac{V_0}{K} \quad (\text{if } Q \text{ is fixed}) \]
🎈 Plain EnglishThe sponge soaks up some of the wind. By the time the wind reaches your face on the other side, it feels much gentler — divided by K.

1.3 Capacitance with a dielectric — three rules

Full fill (dielectric occupies the entire gap): capacitance simply multiplies by \(K\). Partial fill (slab thickness \(t < d\)): capacitance increases but only modestly. Conductor slab: same formula as dielectric with \(K \to \infty\) — position-independent.

\[ C_0 \;=\; \dfrac{\varepsilon_0\,A}{d} \qquad C_{\text{full}} \;=\; K\,C_0 \] \[ \text{Slab thickness } t:\quad C \;=\; \dfrac{\varepsilon_0\,A}{d \;-\; t \;+\; \dfrac{t}{K}} \] \[ \text{Metal slab } (K \to \infty):\quad C \;=\; \dfrac{\varepsilon_0\,A}{d \;-\; t} \]
🎈 Plain EnglishA dielectric is the sponge between two trays of soda water (the plates). Full fill = soda holds way more bubbles. Slab fill = only part of the gap is sponge — improvement is real but smaller.

1.4 What stays fixed — battery connected vs disconnected

This is the single biggest gotcha in NEET questions. With the battery connected, the voltage \(V\) is held constant. With the battery disconnected, the charge \(Q\) is held constant. Decide which is fixed BEFORE writing any equation.

\[ \textbf{Battery ON: }\; C \to K C_0,\quad Q \to K Q_0,\quad E \to E_0,\quad U \to K U_0 \] \[ \textbf{Battery OFF: }\; C \to K C_0,\quad V \to \dfrac{V_0}{K},\quad E \to \dfrac{E_0}{K},\quad U \to \dfrac{U_0}{K} \]
🎈 Plain EnglishBattery ON = voltage policeman, won't let V change. Battery OFF = charge prisoner, charge can't escape. Whichever role the battery is playing decides which quantity stays put.

1.5 Polar vs non-polar dielectrics

Non-polar: molecules have no permanent dipole (H₂, N₂, O₂, CO₂). Only induced polarisation. Effect is small. Polar: molecules have a permanent dipole (H₂O, HCl, NH₃). Effect is large, but DROPS with rising temperature (thermal scrambling disrupts alignment).

\[ \chi_e^{\text{polar}} \;\propto\; \dfrac{1}{T} \quad (\text{Curie-style}) \] \[ K_{\text{water}} \approx 80 \qquad K_{\text{air}} \approx 1.0006 \]
🎈 Plain EnglishPolar molecules already have a plus and a minus end. They just rotate to line up — easy. Non-polar ones have to stretch to make a tiny dipole — harder, so the effect is much smaller.

1.6 Dielectric strength and breakdown

Every dielectric has a maximum electric field it can withstand before its electrons get torn from atoms and it suddenly conducts. This maximum field is the dielectric strength. NEET-relevant fact: air ≈ \(3 \times 10^{6}\) V/m; mica ≈ \(10^{8}\) V/m.

\[ V_{\max} \;=\; E_{\text{breakdown}} \times d \]
🎈 Plain EnglishEvery sponge has a snapping point. Push too hard and the sponge tears — turning into a path that lets current rip through. That snapping field is the dielectric strength.

1.7 Energy stored — with and without dielectric

Energy density in vacuum is \(\tfrac{1}{2}\varepsilon_0 E^2\). With a linear dielectric of constant \(K\), it becomes \(\tfrac{1}{2}\varepsilon_0 K E^2 = \tfrac{1}{2}\varepsilon E^2\). Inserting with battery ON RAISES U (battery pumps in extra charge). Inserting with battery OFF LOWERS U (the dielectric gets sucked in, doing work on itself).

\[ u \;=\; \tfrac{1}{2}\,\varepsilon_0\,K\,E^{2} \;=\; \tfrac{1}{2}\,\varepsilon\,E^{2} \] \[ U \;=\; \tfrac{1}{2}\,C\,V^{2} \;=\; \dfrac{Q^{2}}{2\,C} \]
🎈 Plain EnglishBattery ON: the dielectric is sucked in AND the battery rushes extra charge in to fill it — energy goes up. Battery OFF: the dielectric falls in, releasing energy of its own — the capacitor's stored energy drops because some went into the dielectric's motion.

1.8 Force on a dielectric slab being inserted

The slab is pulled INTO the capacitor by a real force. NEET classic: a slab partially inside a battery-connected capacitor feels a force \(F = \tfrac{1}{2}V^{2}\,dC/dx\) pointing into the capacitor.

\[ F \;=\; \tfrac{1}{2}\,V^{2}\,\dfrac{dC}{dx} \] \[ F_{\text{parallel plate}} \;=\; \dfrac{\varepsilon_0\,b\,V^{2}\,(K-1)}{2\,d} \]
🎈 Plain EnglishThe capacitor wants to be more capacitor-y. So it pulls a sponge in by itself. The pull only stops when the sponge is fully in (or some other force holds it back).

1.1 NEET-style problems with structured solutions

A parallel-plate capacitor of capacitance C is connected to a battery of EMF V. A dielectric slab of constant K completely fills the gap. Find the new capacitance, charge, voltage and energy.

Given
Initial: capacitance \(C\), voltage \(V\) (battery still connected), no dielectric.
Asked
\(C'\), \(Q'\), \(V'\), \(U'\) after slab insertion.
Formula
\(C' = KC\); battery ON ⇒ \(V\) fixed ⇒ \(Q' = C'V = KCV\); \(U' = \tfrac{1}{2}C'V^2 = K \cdot \tfrac{1}{2}CV^2\).
Technique
Identify the constrained quantity first. Battery connected → V constant. Multiply C by K, then update Q and U from C.
Shortcut
With battery ON: every quantity except V multiplies by K (C, Q, U). V stays put.
Answer
\(C' = KC\), \(V' = V\), \(Q' = KCV\), \(U' = K \cdot \tfrac{1}{2}CV^2\).

A parallel-plate capacitor is charged to potential V and then DISCONNECTED from the battery. A dielectric slab of K is fully inserted. Find new V, Q, and energy.

Given
Initial: \(C\), \(V\), charge \(Q = CV\). Battery DISCONNECTED before slab insertion.
Asked
\(V'\), \(Q'\), \(U'\) after slab insertion.
Formula
\(C' = KC\); battery OFF ⇒ \(Q\) fixed; \(V' = Q/C' = V/K\); \(U' = \tfrac{1}{2}C'V'^2 = U/K\).
Technique
Spot 'disconnected' as the keyword. Q must stay the same. Compute new C, then V and U from the fixed-Q form.
Shortcut
With battery OFF: C goes up by K, every other quantity (V, E, U) goes DOWN by K. Q never changes.
Answer
\(Q' = Q = CV\), \(V' = V/K\), \(U' = U/K\).

A slab of thickness \(t = d/2\) and dielectric constant K is inserted into a parallel-plate capacitor with plate separation d. By what factor does the capacitance change?

Given
Plate gap \(d\), slab thickness \(t = d/2\), dielectric constant K. Initial \(C_0 = \varepsilon_0 A/d\).
Asked
Ratio \(C/C_0\).
Formula
\(C = \dfrac{\varepsilon_0 A}{d - t + t/K}\). Plug \(t = d/2\): \(C = \dfrac{\varepsilon_0 A}{d - d/2 + d/(2K)} = \dfrac{2K\,\varepsilon_0 A}{d(K+1)}\).
Technique
Use the partial-fill master formula. Don't try to derive series-of-two-capacitors each time.
Shortcut
\(C/C_0 = \dfrac{2K}{K+1}\). For \(K = 3\): ratio = 1.5. Notice it lies between 1 and K — never above K.
Answer
\(C = \dfrac{2K}{K+1}\,C_0\). For K = 3: \(C = 1.5 \, C_0\).

In a parallel-plate capacitor with plate area A and gap d, a metallic plate (a conductor) of thickness t < d is inserted parallel to the plates. What is the new capacitance? Does its position in the gap matter?

Given
Plate area \(A\), gap \(d\), inserted slab of thickness \(t\) made of metal (\(K \to \infty\)).
Asked
\(C\) and whether position matters.
Formula
As \(K \to \infty\): \(t/K \to 0\). \(C = \dfrac{\varepsilon_0 A}{d - t}\). Position does NOT matter.
Technique
Treat the metal slab as a dielectric with K = ∞. Drop the t/K term.
Shortcut
A metal slab effectively reduces the gap from d to (d − t), no matter where it's placed. Many NEET trick questions hide this — solve in 2 seconds.
Answer
\(C = \dfrac{\varepsilon_0 A}{d - t}\). Position-independent.

Two dielectrics K₁ and K₂ each fill half the area of a parallel-plate capacitor (split vertically — side by side, both filling full gap d). Find the equivalent capacitance.

Given
Half area A/2 with K₁ filling full d; the other half-area A/2 with K₂ filling full d.
Asked
\(C_{\text{eq}}\).
Formula
Side-by-side (same V across each) ⇒ PARALLEL: \(C_{\text{eq}} = C_1 + C_2 = \dfrac{\varepsilon_0 (A/2)(K_1 + K_2)}{d}\).
Technique
Side-by-side dielectrics = same V across each = parallel combination.
Shortcut
Equivalent K (single replacement) = \((K_1 + K_2)/2\) — average.
Answer
\(C_{\text{eq}} = \dfrac{\varepsilon_0 A (K_1+K_2)}{2 d}\). Equivalent single \(K = (K_1+K_2)/2\).

Two dielectrics K₁ and K₂ are stacked one above the other in the gap of a parallel-plate capacitor, each of thickness d/2. Find the equivalent capacitance.

Given
Two slabs of thickness \(d/2\) each, stacked vertically.
Asked
\(C_{\text{eq}}\).
Formula
Stacked (same Q through each) ⇒ SERIES: \(\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}\). With \(C_1 = \dfrac{2\varepsilon_0 K_1 A}{d}\), \(C_2 = \dfrac{2\varepsilon_0 K_2 A}{d}\): \(C_{\text{eq}} = \dfrac{2\varepsilon_0 A}{d}\cdot\dfrac{K_1 K_2}{K_1+K_2}\).
Technique
Stacked dielectrics = same Q = series combination. Use 1/C = Σ 1/Cᵢ.
Shortcut
Equivalent K (single replacement) = harmonic mean = \(\dfrac{2 K_1 K_2}{K_1 + K_2}\).
Answer
\(C_{\text{eq}} = \dfrac{2\varepsilon_0 A}{d}\cdot\dfrac{K_1 K_2}{K_1+K_2}\). Equivalent single \(K = \dfrac{2K_1K_2}{K_1+K_2}\).

A capacitor of capacitance C is charged to a potential V by a battery, then disconnected. A dielectric slab of constant K is inserted fully. What is the work done by an external agent to insert the slab?

Given
Initial energy \(U_0 = \tfrac{1}{2}CV^2\). Battery DISCONNECTED. K inserted, Q fixed.
Asked
Work done by the external agent.
Formula
\(U_f = U_0/K\). \(W_{\text{ext}} = U_f - U_0 = U_0(1/K - 1) = -U_0 (K-1)/K\).
Technique
\(W_{\text{ext}}\) is NEGATIVE — the slab is sucked in. The external agent does NEGATIVE work (it has to hold the slab back to slip it in slowly).
Shortcut
Sign trap: positive answer = wrong. Always negative because the capacitor pulls the slab in by itself.
Answer
\(W_{\text{ext}} = -\dfrac{1}{2}CV^2 \cdot \dfrac{K-1}{K}\) (negative: slab pulled in).

Same setup as Q7 but the battery is STILL CONNECTED while the slab is inserted. Find work done by the battery, change in stored energy, and work done by the external agent.

Given
Battery V connected throughout. K fully inserted.
Asked
\(W_{\text{bat}}\), \(\Delta U\), \(W_{\text{ext}}\).
Formula
\(\Delta Q = (KC - C)V = (K-1)CV\). \(W_{\text{bat}} = V \cdot \Delta Q = (K-1)CV^2\). \(\Delta U = \tfrac{1}{2}(K-1)CV^2\). Energy conservation: \(W_{\text{bat}} + W_{\text{ext}} = \Delta U\) ⇒ \(W_{\text{ext}} = \Delta U - W_{\text{bat}} = -\tfrac{1}{2}(K-1)CV^2\).
Technique
With battery on, write energy conservation for the whole system: battery work in + external work in = ΔU of capacitor.
Shortcut
Battery does double the energy that ends up stored. Half goes to capacitor energy, half is given back to the agent (or dissipated). External work is again negative.
Answer
\(W_{\text{bat}} = (K-1)CV^2\); \(\Delta U = \tfrac{1}{2}(K-1)CV^2\); \(W_{\text{ext}} = -\tfrac{1}{2}(K-1)CV^2\).

A dielectric slab is being slowly slid INTO a battery-connected parallel-plate capacitor at constant speed v. What is the current that flows in the connecting wire?

Given
Width of plates b, slab moving with speed v at position x inside; battery V constant.
Asked
Current I in the circuit.
Formula
\(C(x) = \dfrac{\varepsilon_0 b}{d}[\,\ell + (K-1)\,x\,]\). \(I = V \dfrac{dC}{dt} = V \dfrac{dC}{dx}\,v = \dfrac{\varepsilon_0 b V (K-1) v}{d}\).
Technique
Current = rate of charge flow = V·dC/dt. Note dx/dt = v (constant).
Shortcut
I is CONSTANT while slab moves (because dC/dx is constant for linear geometry). When slab stops, I = 0 instantly.
Answer
\(I = \dfrac{\varepsilon_0 b V (K-1) v}{d}\), constant during insertion.

In a parallel-plate capacitor of plate gap d, half the gap (thickness d/2) is filled with a dielectric of constant K and the other half is air. Find the ratio of voltage across the dielectric portion to the voltage across the air portion.

Given
Stacked dielectric + air, each of thickness d/2. Same charge Q flows through both.
Asked
\(V_K / V_{\text{air}}\).
Formula
For stacked layers (series): \(V_i = Q/C_i\). \(C_{\text{air}} = \dfrac{2\varepsilon_0 A}{d}\), \(C_K = \dfrac{2K\varepsilon_0 A}{d}\). Ratio \(V_K/V_{\text{air}} = C_{\text{air}}/C_K = 1/K\).
Technique
Series ⇒ same Q. Voltage is inversely proportional to capacitance.
Shortcut
Voltage drops more across the air gap because air has lower K. Bigger K = less voltage.
Answer
\(V_K/V_{\text{air}} = 1/K\). So air carries K times more voltage than the dielectric layer.

A dielectric slab of constant K = 2 is inserted into a capacitor that had energy U₀ before insertion. The capacitor is connected to a battery throughout. Find new energy.

Given
Battery ON. K = 2. Initial energy \(U_0\).
Asked
\(U'\).
Formula
Battery ON ⇒ \(U' = K U_0 = 2 U_0\).
Technique
Battery ON → energy multiplies by K (battery pumped in extra charge).
Shortcut
Just multiply U by K. The extra energy came from the battery doing 2U(K−1)/K... but you don't need to calculate that — energy answer is simply KU₀.
Answer
\(U' = 2 U_0\).

A parallel-plate capacitor with no dielectric has electric field E₀ between the plates. A slab of constant K = 4 is now fully inserted. Find the new field if (a) battery is on, (b) battery is disconnected.

Given
Initial field \(E_0\). K = 4.
Asked
New field in both cases.
Formula
Battery ON: E stays the same (V/d unchanged). Battery OFF: \(E = E_0/K = E_0/4\).
Technique
Battery ON keeps V (and therefore E) constant. Battery OFF keeps Q (and therefore σ on plates) constant, so the field divides by K.
Shortcut
Battery on → E unchanged. Battery off → E divided by K.
Answer
(a) \(E = E_0\) (battery ON). (b) \(E = E_0/4\) (battery OFF).

2 Exception & instant-answer questions

These are the questions where you DON'T need calculation — a symmetry, a sign, a constraint instantly gives the answer. NEET loves to slip these into a 200-question paper to reward students who recognise the trick.

❓ A symmetric arrangement of charges +q, −q, +q, −q lies at the four corners of a square. What is the potential at the centre?

⚡ 0. The four charges add algebraically: net charge = 0, and by symmetry each pair contributes equal and opposite potential at the centre. No calculation needed.
Why: V at a point is a SCALAR sum of kq_i/r_i. All four corners are equidistant from the centre. The signed sum is 0.

❓ A parallel-plate capacitor has a dielectric slab (K) inserted fully. The capacitor was first DISCONNECTED from the battery. By what factor does the surface charge density σ on the plates change?

⚡ It does NOT change. Q is fixed, A is fixed ⇒ σ = Q/A is fixed. The dielectric only changes the FIELD inside, not the charges on the plates.
Why: σ on the conductor plates only depends on free charge. Bound charge appears only at the dielectric surfaces. The plates still hold the same total Q.

❓ In a capacitor, what happens to the FREE charge density on the conductor plates when a dielectric is inserted with the battery CONNECTED?

⚡ It increases by factor K. Battery feeds in extra free charge to maintain V across smaller-K effective gap.
Why: V = σd/(Kε₀). For V fixed, σ must rise by factor K.

❓ A metallic slab of any thickness t < d is placed parallel to the plates of a capacitor at any position in the gap. Does the answer for capacitance depend on slab position?

⚡ No. Position-independent. C = ε₀A/(d−t) only.
Why: For a conductor, the inside has E = 0. The slab acts as if it replaces a portion of the gap with zero effective length, regardless of where it sits.

❓ Two capacitors of equal capacitance C are charged to the SAME voltage V and then connected in parallel. How much energy is lost?

⚡ Zero. Energy is lost only when initial voltages differ. Same V → no charge redistribution → no loss.
Why: Loss formula = ½·C₁C₂/(C₁+C₂)·(V₁−V₂)². If V₁ = V₂, the factor (V₁−V₂)² is zero.

❓ A point charge q is placed at the centre of a HOLLOW spherical conductor. What is the potential at any point INSIDE the conducting wall?

⚡ Equal to the potential of the conductor surface = kQ_total/R_outer. The whole conductor is equipotential.
Why: Inside a conductor, E = 0, so V is constant. Match the value with the surface.

❓ For a parallel-plate capacitor with battery ON, a dielectric slab fills HALF the gap (d/2 of thickness). The free-charge density on the plates becomes…

⚡ Increases by factor \(2K/(K+1)\). Because C/C₀ = 2K/(K+1) and Q = CV.
Why: σ = Q/A and Q scales with C when V is fixed.

❓ A dielectric slab is being slowly inserted into a DISCONNECTED capacitor. As the slab enters, the voltage across the capacitor…

⚡ Decreases continuously. Q fixed, C increases, so V = Q/C drops.
Why: Disconnect freezes Q. Any insertion that raises C must drop V.

❓ In a network of capacitors, two nodes are at the SAME potential (by symmetry or by inspection). Can you join them with a wire?

⚡ Yes, freely. A wire between two nodes at equal potential carries no charge and doesn't change anything.
Why: Equipotential nodes have no driving voltage to push charge. The wire is a 'do nothing' element.

❓ A capacitor with a dielectric slab inserted is in equilibrium (battery off). If you let the slab go, in which direction does it move and why?

⚡ Slab moves INTO the capacitor. The system seeks lower energy. \(U \propto 1/C\). Inserting raises C, lowers U.
Why: Mechanical systems move toward lower potential energy. Slab insertion lowers U for fixed Q.

❓ In a parallel-plate capacitor, by what factor does ENERGY DENSITY change when a dielectric K is inserted with battery ON?

⚡ Energy density u = ½ε₀KE². E stays fixed (battery ON), so u multiplies by K.
Why: u = ½ε₀KE² is the dielectric-modified energy density. With E fixed by battery, u scales linearly with K.

❓ Two capacitors are charged to opposite polarities. When connected, what is the formula for energy lost?

⚡ Replace (V₁−V₂)² with (V₁+V₂)² in the standard formula. ΔU_lost = ½ · C₁C₂/(C₁+C₂) · (V₁+V₂)².
Why: Opposite-polarity connection means equivalent voltage difference is V₁+V₂. Loss formula structure stays the same.

❓ Force on a dielectric slab in a parallel-plate capacitor: does it depend on how far the slab has been inserted?

⚡ No. Force is constant during insertion (linear geometry, so dC/dx is constant).
Why: F = ½V²·dC/dx and dC/dx = ε₀b(K−1)/d is constant. Force stays the same all the way in until the slab is fully inside.

❓ If you double both plate area A and plate gap d simultaneously, what happens to capacitance?

⚡ Unchanged. C = ε₀A/d — factors of 2 cancel.
Why: Both numerator and denominator scale by 2.

❓ In a capacitor with battery ON, you slip in a slab of dielectric constant 5. The battery does work W on the capacitor. What fraction of W goes into stored energy?

⚡ Half. Always ½. With battery on, stored energy = ½·battery work; the other half is lost (as heat in wires, EM radiation during transient).
Why: ΔU = ½(K−1)CV². W_bat = (K−1)CV² = 2·ΔU. So stored fraction = 1/2.

3 What students miss in this topic

High-yield extras the NCERT chapter mentions briefly but NEET tests heavily — and that most quick-revision summaries skip.

Bound vs free charges

Free charges = electrons that move on the conductor plates. Bound charges = induced on the dielectric's surface; they cannot leave the dielectric. NEET trap: σ_total = σ_free − σ_bound. The 'effective' surface density seen by the field inside the dielectric is reduced.

\[ \sigma_{\text{bound}} = \sigma_{\text{free}}(1 - 1/K) \]

Susceptibility χₑ and permittivity ε

χₑ = K − 1. ε = Kε₀. For vacuum, χₑ = 0, K = 1, ε = ε₀.

\[ K = 1 + \chi_e,\quad \varepsilon = K\,\varepsilon_0,\quad \mathbf{P} = \varepsilon_0 \chi_e \mathbf{E} \]

Why E vector E ≠ D vector D inside a dielectric

E (electric field) describes the force per unit charge. D (electric displacement) is associated with FREE charges only. Inside a dielectric: E reduces (gets shielded), but D stays the same as without the dielectric (it 'sees through' the dielectric).

\[ \mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} = \varepsilon_0 K \mathbf{E} \] \[ \oint \mathbf{D}\cdot d\mathbf{A} = Q_{\text{free}} \]

Energy density formula with dielectric

u = ½ε₀E² in vacuum becomes u = ½ε₀KE² = ½εE² with dielectric. This is THE formula most students forget. Energy per cubic metre of dielectric-filled space is K times higher for the same E.

\[ u = \tfrac{1}{2}\varepsilon_0 K E^2 = \tfrac{1}{2}\varepsilon E^2 \]

Capacitance of a sphere INSIDE a dielectric

Isolated sphere capacitance becomes 4πε₀KR. For concentric spherical capacitor with K-filled gap: C = 4πε₀K·ab/(b−a).

\[ C_{\text{isolated, dielectric}} = 4\pi\varepsilon_0 K R \]

What's the dielectric constant of a CONDUCTOR?

Effectively K → ∞. Every electric field inside vanishes, so dividing E by K reproduces this. Useful trick: a metal slab is just the K → ∞ limit of the partial-fill formula.

\[ \text{Metal slab: } C = \dfrac{\varepsilon_0 A}{d - t}\;\;\Leftarrow\;\; \lim_{K \to \infty} \dfrac{\varepsilon_0 A}{d - t + t/K} \]

Memorise the dielectric constants

Air ≈ 1.0006 (effectively 1). Water ≈ 80 (huge — polar molecule). Glass ≈ 4–7. Mica ≈ 4–8. Paper ≈ 2–4. Teflon ≈ 2.1. Vacuum = 1 (exact). NEET sometimes asks 'which is best for high-voltage applications' — answer: high K AND high dielectric strength (mica).

Why polar dielectrics' K drops with temperature

Polar molecules rely on alignment to produce P. Thermal motion (kT) disturbs alignment. Higher T = less alignment = lower P = lower K. Non-polar dielectrics don't have this effect.

Capacitor inside a fluid

Submerging a capacitor in oil (K ≈ 2.2) or distilled water (K ≈ 80) acts the same as inserting a slab of that K filling the gap fully. NEET twist: 'capacitor placed in oil' is just 'C → KC'.

Force per unit area on a charged plate

Electrostatic pressure = σ²/(2ε₀). With dielectric K: σ²/(2ε₀K). This is the formula used in plate-attraction and soap-bubble problems.

\[ P_{\text{electric}} = \dfrac{\sigma^2}{2\varepsilon_0 K} \]

Series & parallel decision tree (quick recall)

Question: are the dielectrics SIDE BY SIDE in the gap? → parallel. Are they STACKED on top of each other? → series. Parallel ⇒ equivalent K = average (weighted by area). Series ⇒ equivalent K = harmonic mean (weighted by thickness).

Dielectric breakdown is irreversible

Once a dielectric breaks down, the molecules are damaged. It conducts even after the voltage is removed. NEET fact: this is why mica and ceramic are used for high-voltage capacitors — they have high dielectric strength, not just high K.