A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery

NCERT Exemplar · Q20

Q. A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

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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:

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

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When the dielectric is removed after disconnecting the battery, the charge stays constant, capacitance decreases, voltage increases, electric field increases, and stored energy increases — because the system is isolated and work must be done to remove the polarised material.

The key to this problem is understanding when the battery is disconnected. If the battery stays connected, the voltage is fixed. But here, the battery is disconnected before the dielectric is removed. That changes everything.

Let’s walk through it step by step.

1. The initial situation — dielectric present, battery connected

A capacitor with a dielectric of constant $K$ is connected to a DC source. The battery forces a fixed voltage $V_0$ across the plates. The capacitance with the dielectric is:

C = K C_0

where $C_0$ is the capacitance with vacuum (or air) between the plates.

The charge stored is:

Q = C V_0 = K C_0 V_0

The electric field inside the dielectric is $E = V_0 / d$ , where $d$ is the plate separation. The energy stored is:

U = \frac{1}{2} C V_0^2 = \frac{1}{2} K C_0 V_0^2

All these values are set by the battery.

2. Battery is disconnected — charge is now trapped

When the battery is disconnected, the charge on the plates has no path to leave. So the charge $Q$ becomes fixed — it cannot change, no matter what we do next.

Watch out

A very common mistake is to think that removing the dielectric somehow lets charge leak away. It does not — the plates are isolated. The charge stays exactly the same.

So after disconnection:

Charge stored: remains constant (same $Q$ ).

3. Dielectric is removed — what changes?

Now we pull out the dielectric slab. The material that was polarised and reducing the internal field is gone. The capacitance drops back to its vacuum value:

C_{\text{new}} = C_0

Since $K > 1$ , we have $C_{\text{new}} < C$ — capacitance decreases.

4. Voltage must adjust — because $Q$ is fixed

The fundamental relation $Q = CV$ still holds. Since $Q$ is constant and $C$ has decreased, the voltage must increase:

V_{\text{new}} = \frac{Q}{C_0} = \frac{K C_0 V_0}{C_0} = K V_0

So voltage increases by a factor of $K$ .

Tip

Think of it this way: with the dielectric, the polarised molecules partially cancel the field, so a smaller voltage is enough to hold the same charge. Remove the dielectric, and you need a larger voltage to maintain that same charge — hence $V$ goes up.

5. Electric field — directly proportional to voltage

The electric field between parallel plates is $E = V / d$ . Since $d$ is fixed and $V$ increases by factor $K$ , the field also increases by factor $K$ :

E_{\text{new}} = \frac{V_{\text{new}}}{d} = \frac{K V_0}{d} = K E_0

So electric field increases.

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Concept understanding — Dielectric Insertion Capacitance

Dielectric Insertion Capacitance – From Intuition to Precision

Why does the capacitance increase?

The precise statement

What if the dielectric only partially fills the gap?

1. The initial situation — dielectric present, battery connected

2. Battery is disconnected — charge is now trapped

3. Dielectric is removed — what changes?

4. Voltage must adjust — because $Q$ is fixed

5. Electric field — directly proportional to voltage

Method: Battery-Connected vs Battery-Disconnected Capacitor Problems

Steps

1. The initial situation — dielectric present, battery connected

2. Battery is disconnected — charge is now trapped

3. Dielectric is removed — what changes?

4. Voltage must adjust — because QQQ is fixed

5. Electric field — directly proportional to voltage

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4. Voltage must adjust — because $Q$ is fixed

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