MHT-CET Physics · Electrostatics

Dielectrics in Capacitors

A dielectric between the plates weakens the field inside it K times, so the capacitance rises; what else changes depends on whether the battery stays connected (V fixed) or is removed (Q fixed).

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Why this matters

23 PYQs, nine HARD, five of them two dielectrics sharing the gap, where the figure decides series or parallel. The rest: a slab that fills only part of the gap, and which quantities change when a dielectric goes in with the battery on or off.

Concept 1 of 3: Battery Connected or Removed: What Changes

The dielectric's molecules line up against the field and cancel part of it, so the field inside drops K times and the capacitance rises K times. Then ask what is held fixed. With the battery removed, the charge cannot change, so the voltage and energy fall. With the battery connected, the voltage cannot change, so more charge flows on and the energy rises.

Definition

Field inside a dielectric in an external field: $E = \dfrac{E_0}{K} < E_0$ . Its job in a capacitor: to reduce the effective potential for a given charge, raising $C$ to $KC$ .
Battery removed (Q fixed): $V \to \dfrac{V}{K}$ , $E \to \dfrac{E}{K}$ , $U \to \dfrac{U}{K}$ .
Battery connected (V fixed): $Q \to KQ$ , $U \to KU$ , $E = \dfrac{V}{d}$ unchanged — inserting and removing the slab leaves the field as it was.
Two capacitors $C, C$ in parallel, one filled: $\Delta C = C(K - 1)$ . In series, one filled: $C_{\text{eq}}$ goes from $\dfrac{C}{2}$ to $\dfrac{KC}{K + 1}$ , a change of $\dfrac{C}{2}\cdot\dfrac{K - 1}{K + 1}$ .
Changing the gap and filling it: $\dfrac{C'}{C} = K\dfrac{d}{d'}$ .

Full dielectric

C = \frac{K\varepsilon_0 A}{d}; \qquad Q\text{ fixed: } U \to \frac{U}{K}; \qquad V\text{ fixed: } U \to KU

Worked example

6\,\mu

F capacitor charged to 100 V is disconnected, then filled with a dielectric of

K = 3

. New voltage and energy?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q46Moderate

Example 1 · Electrostatics · Dielectrics in Capacitors

The potential energy of charged parallel plate capacitor is

U_0

. If a slab of dielectric constant

K

is inserted between the plates, then the new potential energy will be

Not asking what is held fixed

Energy falls to

\frac{U}{K}

when the charge is fixed and rises to

KU

when the voltage is. The question tells you which only by saying 'charged and isolated' or 'battery remains connected'.

Drill 8 more on battery connected or removed: what changes

Concept 2 of 3: A Slab Filling Part of the Gap

A slab of thickness t and constant K acts like a thinner layer of air, t/K thick. So the capacitor behaves as if its gap had shrunk by t(1 − 1/K). A metal sheet is the limit K → ∞: it simply removes its own thickness from the gap.

Definition

Slab of thickness $t$ in a gap $d$ : $C = \dfrac{\varepsilon_0 A}{d - t\left(1 - \frac{1}{K}\right)}$ .
Equivalent view: air gap $d - t$ and slab $t$ in series, $\dfrac{1}{C} = \dfrac{d - t}{\varepsilon_0 A} + \dfrac{t}{K\varepsilon_0 A}$ .
Conducting sheet of thickness $t$ : $C = \dfrac{\varepsilon_0 A}{d - t}$ . A sheet $\frac{2d}{3}$ thick triples $C$ .
Keep the units apart: a numeric answer in $\varepsilon_0$ F needs $A$ in m² and $d$ in m.

Partial slab

C = \frac{\varepsilon_0 A}{d - t\left(1 - \dfrac{1}{K}\right)}

Worked example

A gap of 4 mm holds a slab 2 mm thick with

K = 4

. By what factor does the capacitance change?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q18Moderate

Example 2 · Electrostatics · Dielectrics in Capacitors

The plates of a parallel plate capacitor are separated by a distance '

d

' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by

50\%

. The thickness of the dielectric slab is

Subtracting t/K instead of t(1 − 1/K)

The slab replaces

t

of air with an air-equivalent

\frac{t}{K}

, so the gap loses

t - \frac{t}{K}

. The options also offer

t\left(1 + \frac{1}{K}\right)

and a factor of 2 in front.

Drill 4 more on a slab filling part of the gap

Concept 3 of 3: Two Dielectrics: Series or Parallel From the Figure

Look at the boundary between the two materials. If it runs parallel to the plates, the materials are stacked one after the other across the gap — capacitors in series. If it runs from one plate to the other, they sit side by side, each over part of the area — capacitors in parallel.

Definition

Stacked across the gap (boundary parallel to the plates), thicknesses $d_1, d_2$ : series, $C = \dfrac{\varepsilon_0 A}{d_1/K_1 + d_2/K_2}$ . Equal halves: $C = \dfrac{2K_1K_2}{K_1 + K_2}C_0$ .
Side by side (boundary joining the plates), areas $A_1, A_2$ : parallel, $C = \dfrac{\varepsilon_0}{d}(K_1A_1 + K_2A_2)$ . Equal halves: $C = \dfrac{K_1 + K_2}{2}C_0$ . Half filled, half air: $\dfrac{K + 1}{2}C_0$ .
Three regions: reduce the stacked pair first, then add it in parallel with the rest.

Equal halves

\text{stacked: } C = \frac{2K_1K_2}{K_1 + K_2}C_0, \qquad \text{side by side: } C = \frac{K_1 + K_2}{2}C_0

Worked example

The gap is filled with two layers of equal thickness,

K_1 = 2

and

K_2 = 6

, stacked between the plates. Then the same two fill it side by side instead. Capacitance each way, in terms of the air value

C_0

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q4Moderate

Example 3 · Electrostatics · Dielectrics in Capacitors

A parallel plate capacitor with air medium between the plates has a capacitance of

10\,\mu F

. The area of capacitor is divided into two equal halves & filled with two media having dielectric constant

K_1 = 2

K_2 = 4

. The capacitance of the system will be

The same words, two different figures

'Filled with two dielectrics as shown' has been set with both figures. The ratio

\frac{K_1 + K_2}{2K_1K_2}

belongs to the STACKED figure; side by side gives

\frac{2}{K_1 + K_2}

. Decide from the boundary line, not from memory of a past answer.

Drill 8 more on two dielectrics: series or parallel from the figure

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Battery Connected or Removed: What Changes
Full dielectric
$C = \frac{K\varepsilon_0 A}{d}; \qquad Q\text{ fixed: } U \to \frac{U}{K}; \qquad V\text{ fixed: } U \to KU$
A Slab Filling Part of the Gap
Partial slab
$C = \frac{\varepsilon_0 A}{d - t\left(1 - \dfrac{1}{K}\right)}$
Two Dielectrics: Series or Parallel From the Figure
Equal halves
$\text{stacked: } C = \frac{2K_1K_2}{K_1 + K_2}C_0, \qquad \text{side by side: } C = \frac{K_1 + K_2}{2}C_0$

Watch out for (3)

Not asking what is held fixed
→ Battery Connected or Removed: What Changes
Subtracting t/K instead of t(1 − 1/K)
→ A Slab Filling Part of the Gap
The same words, two different figures
→ Two Dielectrics: Series or Parallel From the Figure

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