A metal sheet is inserted between the plates of a parallel plate capacitor of capacitance $C$. If the sheet partly occupies the space between the plates, the capacitance: (A) remains $C$ (B) becomes greater than $C$ (C) becomes less than $C$ (D) becomes zero

> [!TLDR] > A conducting sheet has zero internal field, so it removes its own thickness $t$ from the effective gap, leaving $C' = \dfrac{\varepsilon_0 A}{d-t} > C$. **Option (B)**. Let the plate area be $A$ and separation $d$, so $C = \dfrac{\varepsilon_0 A}{d}$. Insert a metal sheet of thickness $t\,(<d)$ that partly fills the gap. 1. **Field inside a conductor is zero.** The sheet develops induced charges on its two faces and carries no field within it, so it contributes nothing to the potential drop. Only the air gaps on either side of the sheet — of total thickness $d - t$ — sustain the field. 2. **Effective separation shrinks.** The capacitor behaves as if the plate gap were reduced from $d$ to $d - t$: $$C' = \frac{\varepsilon_0 A}{d - t}.$$ 3. **Compare with $C$.** Since $t > 0$, we have $d - t \frac{\varepsilon_0 A}{d} = C.$$ The capacitance increases by the factor $\dfrac{d}{d-t}$, independent of where the sheet is placed. (Only if the sheet completely filled the gap, $t \to d$, would $C'\to\infty$; and a zero-thickness sheet, $t=0$, would leave $C$ unchanged.) > [!ANSWER] > The capacitance **(B) becomes greater than $C$.**

A metal sheet is inserted between the plates of a parallel plate capacitor of capacitance C. If the sheet part

Question

C

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Dielectric Insertion Capacitance

Dielectric Insertion Capacitance – From Intuition to Precision

Imagine you have two metal plates facing each other, separated by air. You connect them to a battery. The plates get charged — one positive, one negative — and they store energy in the electric field between them. That's a capacitor.

Now, without disconnecting the battery, you slide a slab of some insulating material (like glass, plastic, or mica) between the plates. What happens? The battery pushes more charge onto the plates. The capacitor now stores more charge for the same voltage. Its capacitance has increased.

That increase — the extra capacitance contributed by the presence of the dielectric — is what we call dielectric insertion capacitance.

Note

The word "insertion" here simply means "the capacitance that appears because you inserted a dielectric." It's not a separate device; it's the change in capacitance due to the material.

Why does the capacitance increase?

The key is polarisation. Inside a dielectric, molecules are like tiny electric dipoles — they have a positive end and a negative end. In an electric field, these dipoles rotate to align with the field. The positive ends point toward the negative plate, and the negative ends point toward the positive plate.

This alignment creates a layer of bound charge on the surfaces of the dielectric, right next to the plates. This bound charge partially cancels the electric field inside the dielectric. But here's the crucial point: if the capacitor is connected to a battery (constant voltage), the battery responds by pushing more free charge onto the plates to restore the original field. More charge for the same voltage means higher capacitance.

If the capacitor is disconnected from the battery (constant charge), the dielectric reduces the voltage across the plates. Same charge, lower voltage — again, higher capacitance.

Watch out

A common mistake: thinking the dielectric "creates" extra charge out of nothing. It doesn't. The battery supplies the extra charge in the constant-voltage case. In the constant-charge case, the voltage drops — the capacitance formula $C = Q/V$ still gives a larger $C$ because $V$ is smaller.

The precise statement

Let’s define:

$C_0$ = capacitance of the capacitor with vacuum (or air) between the plates.
$\kappa$ (or $\varepsilon_r$ ) = dielectric constant (relative permittivity) of the material. For vacuum, $\kappa = 1$ . For most solids, $\kappa > 1$ (e.g., glass ~5–10, water ~80).

When you fill the entire space between the plates with a dielectric of constant $\kappa$ , the new capacitance is:

C = \kappa \, C_0

The dielectric insertion capacitance is the additional capacitance contributed by the dielectric:

C_{\text{insertion}} = C - C_0 = (\kappa - 1) C_0

C_{\text{insertion}} = (\kappa - 1) C_0

This is the extra capacitance you get purely because you inserted the dielectric. If $\kappa = 1$ (vacuum), $C_{\text{insertion}} = 0$ — no insertion effect.

What if the dielectric only partially fills the gap?

In real problems, the slab might not fill the entire space. Then the capacitor behaves like two capacitors in series (or parallel, depending on geometry). The insertion capacitance is no longer a simple multiple — you have to compute the effective capacitance using the appropriate combination rules.

But the core idea remains: the dielectric increases capacitance because its polarisation reduces the net field (or, equivalently, allows more charge at the same voltage).

Tip

For a parallel-plate capacitor with plate area $A$ , separation $d$ , and a dielectric of thickness $t$ inserted (leaving an air gap of $d-t$ ), the effective capacitance is:

$C = \frac{\varepsilon_0 A}{d - t + \frac{t}{\kappa}}$

The insertion capacitance is then $C - C_0$ , where $C_0 = \varepsilon_0 A / d$ .

Why this matters for exams

You'll often be asked:

"A dielectric slab is inserted into a charged capacitor. Find the change in capacitance / energy / charge."
"Compare the capacitance before and after insertion."

The key is to remember:

If the battery stays connected → voltage constant, charge increases, capacitance increases.
If the battery is removed → charge constant, voltage decreases, capacitance increases.

In both cases, the capacitance increases by the factor $\kappa$ (if fully filled). The insertion capacitance $(\kappa - 1)C_0$ is the amount of that increase.

Important

Dielectric insertion capacitance is not a new kind of capacitance — it's the difference $C - C_0$ . Always start from the base capacitance $C_0$ and multiply by $\kappa$ for a full fill. For partial fills, use the series/parallel model.

Final takeaway

Think of the dielectric as a "capacitance booster." It doesn't create energy — it just allows the same geometry to store more charge (or maintain the same charge at a lower voltage). The insertion capacitance quantifies exactly how much boost the material provides.

How inserting a dielectric changes a capacitor's capacitance is a heavily tested numerical topic in the NCERT Class 12 Physics chapter on electrostatic potential and capacitance, appearing often in CBSE boards and JEE Main. Anyone searching "effect of dielectric on capacitance class 12 physics numericals" will find this constant-voltage-versus-constant-charge distinction is the key reasoning examiners look for.

A conducting sheet has zero internal field, so it removes its own thickness $t$ from the effective gap, leaving $C' = \dfrac{\varepsilon_0 A}{d-t} > C$ . Option (B).

Let the plate area be $A$ and separation $d$ , so $C = \dfrac{\varepsilon_0 A}{d}$ . Insert a metal sheet of thickness $t\,(<d)$ that partly fills the gap.

Field inside a conductor is zero. The sheet develops induced charges on its two faces and carries no field within it, so it contributes nothing to the potential drop. Only the air gaps on either side of the sheet — of total thickness $d - t$ — sustain the field.
Effective separation shrinks. The capacitor behaves as if the plate gap were reduced from $d$ to $d - t$ :

C' = \frac{\varepsilon_0 A}{d - t}.

Compare with $C$ . Since $t > 0$ , we have $d - t < d$ , hence

C' = \frac{\varepsilon_0 A}{d-t} > \frac{\varepsilon_0 A}{d} = C.

The capacitance increases by the factor $\dfrac{d}{d-t}$ , independent of where the sheet is placed. (Only if the sheet completely filled the gap, $t \to d$ , would $C'\to\infty$ ; and a zero-thickness sheet, $t=0$ , would leave $C$ unchanged.)

✓Final answer

C

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$Q. The capacitance of a parallel plate capacitor ________ (increases/decreases) when a dielectric is inserted between the plates.$
›Reveal solutionSolution

Inserting a dielectric between the plates of a parallel plate capacitor increases its capacitance by a factor equal to the dielectric constant K.

For a parallel plate capacitor of plate area A and separation d, in vacuum $C_0 = \dfrac{\varepsilon_0 A}{d}$ . When a dielectric slab of dielectric constant K completely fills the gap, the dielectric gets polarised by the field between the plates, and the induced bound charge on its surfaces sets up a field that partly cancels the field of the free charges on the plates. For the same free charge Q, the field (and hence the potential difference V = Ed) between the plates is reduced. Since $C = Q/V$ , a smaller V for the same Q means a larger capacitance:

$C = KC_0 = \dfrac{K\varepsilon_0 A}{d}$

Because K > 1 for every dielectric, C always goes up when a dielectric is inserted.

✓Final answer
$C = KC_0$
CBSE 2025Set 55/5/11 markMCQ
$C$
›Reveal solutionSolution
A conducting sheet has zero internal field, so it removes its own thickness $t$ from the effective gap, leaving $C' = \dfrac{\varepsilon_0 A}{d-t} > C$ . Option (B).

Let the plate area be $A$ and separation $d$ , so $C = \dfrac{\varepsilon_0 A}{d}$ . Insert a metal sheet of thickness $t\,(<d)$ that partly fills the gap.

Field inside a conductor is zero. The sheet develops induced charges on its two faces and carries no field within it, so it contributes nothing to the potential drop. Only the air gaps on either side of the sheet — of total thickness $d - t$ — sustain the field.

Effective separation shrinks. The capacitor behaves as if the plate gap were reduced from $d$ to $d - t$ :

$C' = \frac{\varepsilon_0 A}{d - t}.$

Compare with $C$ . Since $t > 0$ , we have $d - t < d$ , hence

$C' = \frac{\varepsilon_0 A}{d-t} > \frac{\varepsilon_0 A}{d} = C.$
The capacitance increases by the factor $\dfrac{d}{d-t}$ , independent of where the sheet is placed. (Only if the sheet completely filled the gap, $t \to d$ , would $C'\to\infty$ ; and a zero-thickness sheet, $t=0$ , would leave $C$ unchanged.)

✓Final answer
$C$
CBSE 2025Set ANNUAL1 markMCQ
$Q. If a dielectric is kept between two plates of capacitor, its capacitance :$
(a) increases
(b) decreases
(c) First increases then decreases
(d) none of these
›Reveal solutionSolution

Inserting a dielectric between the plates of a capacitor increases its capacitance by a factor equal to the dielectric constant $K$ (K > 1).

For a parallel plate capacitor, capacitance without dielectric is $C_0 = \varepsilon_0 A/d$ . When a dielectric slab of dielectric constant $K$ fills the gap, the dielectric partially cancels the field between the plates (due to induced polarization charges), so more charge can be stored for the same potential difference. The new capacitance becomes

$C = K C_0$

Since $K > 1$ for any real dielectric, the capacitance increases.

✓Final answer
$Capacitance increases — option (a).$
CBSE 2025Set ANNUAL1 markMCQ
$C$
(a) $C$
(b) $2C$
(c) $C/2$
(d) $C/4$
›Reveal solutionSolution

Inserting a dielectric of dielectric constant $k$ between the plates of a parallel-plate capacitor multiplies its capacitance by $k$ ; here $k=2$ gives $C'=2C$ .

Setup

For a parallel-plate capacitor with air (or vacuum) between the plates, plate area $A$ and separation $d$ :

$C = \frac{\varepsilon_0 A}{d}$

When the space between the plates is completely filled with a dielectric medium of dielectric constant $k$ , the permittivity increases from $\varepsilon_0$ to $k\varepsilon_0$ , so the new capacitance is

$C' = \frac{k\varepsilon_0 A}{d} = kC$

Substituting the given value

Here the oil has dielectric constant $k = 2$ , so

$C' = 2C$

✓Final answer
$2C$
CBSE 2022Set ANNUAL1 mark
$Q. What happens to the capacitance of a capacitor when a dielectric slab is placed between its plates?$
›Reveal solutionSolution

Inserting a dielectric always increases capacitance.

When a dielectric slab is placed between the plates of a capacitor, the dielectric gets polarised by the field between the plates, producing an induced (bound) charge on its surfaces that opposes the original field. This reduces the net electric field between the plates for the same charge $Q$ , so the potential difference $V = Ed$ decreases. Since $C = Q/V$ , a smaller $V$ for the same $Q$ means the capacitance increases. If the dielectric fills the whole gap, the new capacitance is $C' = KC_0$ , where $K > 1$ is the dielectric constant of the slab and $C_0$ is the capacitance with vacuum/air between the plates.

✓Final answer
$K$
CBSE 2020Set OC1 markMCQ
$k=6$
(a) remains unaffected
(b) reduces to 1/6 times
(c) becomes 6 times
(d) None of the above
›Reveal solutionSolution

A dielectric always increases a capacitor's capacitance by exactly the factor of its dielectric constant.

Formula

$C_{medium} = kC_{vacuum}$

With $k=6$ : $C' = 6C$ — the capacitance becomes 6 times the original.

✓Final answer
$Capacitance becomes 6 times — option (c).$
CBSE 2018Set ANNUAL1 markMCQ
$Q. A parallel-plate capacitor, with air between the plates, has capacitance 3 uF. If the capacitor is immersed in a liquid of dielectric constant 4.0, its capacitance will be$
(a) 0.75 uF
(b) 1.5 uF
(c) 6 uF
(d) 12 uF
›Reveal solutionSolution

A dielectric of constant K increases capacitance K-fold: C' = 4.0 x 3 uF = 12 uF, option (d).

For a parallel-plate capacitor with air, $C_0 = \dfrac{\varepsilon_0 A}{d}$ . Introducing a dielectric of constant $K$ that fills the gap gives

$C' = K\,C_0 = \dfrac{K\varepsilon_0 A}{d}$

Step 1 — Here $C_0 = 3\ \mu\text{F}$ and $K = 4.0$ .

Step 2 — $C' = 4.0 \times 3\ \mu\text{F} = 12\ \mu\text{F}$ .

The dielectric molecules polarise and set up an opposing field, lowering the net field and the potential difference for a given charge, which raises the capacitance.

✓Final answer
$(d) 12 uF.$

Q.A metal sheet is inserted between the plates of a parallel plate capacitor of capacitance CCC. If the sheet partly occupies the space between the plates, the capacitance: (A) remains CCC (B) becomes greater than CCC (C) becomes less than CCC (D) becomes zero

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Q.A metal sheet is inserted between the plates of a parallel plate capacitor of capacitance $C$ . If the sheet partly occupies the space between the plates, the capacitance: (A) remains $C$ (B) becomes greater than $C$ (C) becomes less than $C$ (D) becomes zero

Unlock everything free for 14 days