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 If the dielectric slab is now withdrawn from the capacitor, how will the following be affected?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.

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 If the dielectric slab is now withdrawn from the capacitor, how will the following be affected? (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.

**Key fact:** The battery is disconnected, so the charge Q on the plates stays **constant**. Let the dielectric constant be K. **(a) Capacitance — decreases** With slab, $C = K\varepsilon_0 A/d$; on removing it, $C' = \varepsilon_0 A/d =$ **$C/K$**. Capacitance falls by the factor K. **(b) Energy stored — increases** $U = Q^2/2C$. Q is fixed and C decreases to $C/K$, so $U' =$ **$KU$** — energy increases K-fold. The extra energy equals the work done by us in pulling the slab out against the attractive pull. **(c) Potential difference — increases** $V = Q/C$. With Q fixed and capacitance $C/K$, $V' =$ **$KV$** — the potential difference rises by the factor K.

Electrostatic Potential and Capacitance

Capacitor with dielectric withdrawn (isolated)

Official Answer

Key fact: The battery is disconnected, so the charge Q on the plates stays constant. Let the dielectric constant be K.

(a) Capacitance — decreases

With slab, $C = K\varepsilon_0 A/d$ ; on removing it, $C' = \varepsilon_0 A/d =$ $C/K$ . Capacitance falls by the factor K.

(b) Energy stored — increases

$U = Q^2/2C$ . Q is fixed and C decreases to $C/K$ , so $U' =$ $KU$ — energy increases K-fold. The extra energy equals the work done by us in pulling the slab out against the attractive pull.

(c) Potential difference — increases

$V = Q/C$ . With Q fixed and capacitance $C/K$ , $V' =$ $KV$ — the potential difference rises by the factor K.

battery disconnectedcharge constantdielectric constant Kcapacitance decreasesenergy increasespotential difference increasesU = Q^2/2CV = Q/C

Marking Scheme

11 mark: (a) capacitance decreases to $C/K$ , justified by $C = K\varepsilon_0 A/d$ .
21 mark: (b) energy increases to $KU$ , justified by $U = Q^2/2C$ with Q constant.
31 mark: (c) potential difference increases to $KV$ , justified by $V = Q/C$ with Q constant.

Hint

Battery disconnected ⇒ charge Q is constant; use $C = K\varepsilon_0 A/d$ , $V = Q/C$ and $U = Q^2/2C$ with $K \to 1$ on removal.

Quick Oral Answer

Since the battery is disconnected the charge is fixed; pulling out the dielectric lowers the capacitance to C over K, so both the voltage and the stored energy rise K-fold — the energy increase being the work we do against the plates' pull.

Analysis & Explanation

Concept

The single most important cue is 'battery disconnected' — this pins the charge Q constant. (Had the battery stayed connected, V would be fixed instead, and every conclusion would flip.) Always identify which quantity is held constant before judging the changes.

Chain of reasoning

Removing the dielectric lowers capacitance: $C \to C/K$ .
With Q fixed, $V = Q/C$ must rise $\to KV$ .
Energy $U = Q^2/2C$ rises $\to KU$ .

Exam trap

Students who assume V is constant (battery-connected case) will wrongly say energy decreases. Read whether the battery is connected or removed.
Justification is compulsory here ('Justify in each case') — stating the trend without the formula loses half the marks.

Real-world link

The increase in stored energy comes from real mechanical work: the charged plates attract the dielectric, so pulling it out requires effort, and that effort is stored in the field — the principle behind variable capacitors and some electret microphones.

Common Mistakes

1Assuming potential difference stays constant (that is the battery-connected case) and hence wrongly concluding the energy decreases.
2Saying the charge changes when the slab is removed — with the battery disconnected, Q is fixed.
3Giving the trends without justification, whereas the question explicitly demands a reason (formula) in each case.

Interesting Facts

The extra stored energy is exactly the mechanical work you do pulling the slab out — the charged plates literally grab the dielectric and try to suck it back in.

This isolated-capacitor behaviour is the basis of the classic 'dielectric being pulled into a capacitor' problem and of MEMS variable capacitors used in tuning circuits and sensors.

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Frequently Asked Questions

Why does removing the dielectric increase the stored energy here?

With the battery disconnected, the charge Q is fixed and energy is $U = Q^2/2C$ . Withdrawing the slab reduces $C$ to $C/K$ , so $U$ rises to $KU$ . The added energy is the mechanical work you do pulling the slab out against the plates' attraction.

How would the answers change if the battery stayed connected?

Then V would be constant instead of Q. Removing the dielectric would still lower $C$ to $C/K$ , but now $Q = CV$ would fall to $Q/K$ and energy $U = \frac{1}{2}CV^2$ would decrease to $U/K$ — the opposite trend for charge and energy compared with the disconnected case.