A parallel plate capacitor of capacitance $C$ has a dielectric slab between its plates. It is charged to a potential difference $V$ by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected? (a) Capacitance of the capacitor, (b) Energy stored in the capacitor, and (c) The potential difference between the plates of the capacitor. Justify your answer in each case.

A parallel plate capacitor of capacitance C has a dielectric slab between its plates. It is charged to a poten

Question

C

(a) Capacitance of the capacitor,

(b) Energy stored in the capacitor, and

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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:

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

Start your 14-day free trial to unlock the full solution →

When a dielectric slab is withdrawn from an isolated charged capacitor, the capacitance decreases by a factor of $K$ , the charge remains constant, the potential difference increases by a factor of $K$ , and the stored energy increases by a factor of $K$ (work is done against the attractive force on the dielectric).

Understanding Dielectric Withdrawal from an Isolated Capacitor

The key to this problem lies in recognizing what remains constant when the battery is disconnected. Once isolated, the capacitor cannot gain or lose charge—the charge $Q$ on the plates is frozen. Everything else adjusts in response to the changing capacitance.

A dielectric material increases capacitance because its molecules polarize in the electric field, creating an internal field that partially opposes the applied field. This reduces the net field between the plates, allowing more charge to be stored at the same voltage. When we remove the dielectric, we lose this benefit.

Important

The Isolation Constraint: After disconnecting the battery, the charge $Q$ on the capacitor plates remains constant. This is the anchor for all subsequent reasoning.

Step-by-Step Analysis

1. Initial conditions with dielectric present

The capacitor with dielectric constant $K$ has capacitance:

C = K C_0

where $C_0$ is the capacitance without the dielectric (air-filled). Charged to potential $V$ , it stores charge:

Q = CV = KC_0 V

2. Effect on Capacitance (a)

When the dielectric slab is withdrawn, the capacitance reverts to its air-filled value:

C_{\text{final}} = C_0 = \frac{C}{K}

The capacitance decreases by a factor of $K$ . For most dielectrics, $K > 1$ (typically 2–10 for common materials), so this is a substantial drop.

3. Effect on Potential Difference (c)

Since $Q$ is constant and $C$ has decreased, we use $V = \frac{Q}{C}$ :

V_{\text{final}} = \frac{Q}{C_0} = \frac{KC_0 V}{C_0} = KV

The potential difference increases by a factor of $K$ . This makes physical sense: with the dielectric gone, the same charge creates a stronger net electric field between the plates, hence a larger voltage.

Watch out

A common mistake is to think the voltage stays constant because it was "set" by the battery. Once disconnected, the battery no longer enforces any voltage—only the charge is fixed.

4. Effect on Stored Energy (b)

The energy stored in a capacitor is $U = \frac{1}{2}QV$ . Initially:

U_{\text{initial}} = \frac{1}{2}Q \cdot V = \frac{1}{2}(KC_0 V) \cdot V = \frac{1}{2}KC_0 V^2

After withdrawal:

U_{\text{final}} = \frac{1}{2}Q \cdot V_{\text{final}} = \frac{1}{2}(KC_0 V) \cdot (KV) = \frac{1}{2}K^2 C_0 V^2

Therefore:

U_{\text{final}} = K \cdot U_{\text{initial}}

The energy increases by a factor of $K$ . …

Start your 14-day free trial to unlock the full solution →

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 sm …

Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 $d$ :

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

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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

$C_{medium} = kC_{vacuum}$

…

Start your 14-day free trial to see the full solution →
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 — $C' = 4.0 \times 3\ \mu\text{F} = 12\ \mu\text{F}$ .

…

Start your 14-day free trial to see the full solution →

Q.A parallel plate capacitor of capacitance CCC has a dielectric slab between its plates. It is charged to a potential difference VVV by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected?

Understanding Dielectric Withdrawal from an Isolated Capacitor

Step-by-Step Analysis

🎓Unlock everything free for 14 days

Q.A parallel plate capacitor of capacitance $C$ has a dielectric slab between its plates. It is charged to a potential difference $V$ by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected?

Unlock everything free for 14 days