Q.A parallel plate capacitor $C$ is connected to a cell of emf $E$ through a key $K$ in a simple single loop. Consider two situations: (A) the key $K$ is kept closed and the plates are pulled apart using an insulating handle; (B) the key $K$ is opened first and then the plates are pulled apart using an insulating handle. Choose the correct option(s).

Concept

For a parallel-plate capacitor $C = \dfrac{\varepsilon_0 A}{d}$, with $Q = CV$. A connected battery clamps $V$; a disconnected (isolated) capacitor conserves $Q$.

Steps

1. Increasing $d$ decreases $C$ in both situations. …

Method: Deciding What Stays Fixed When Capacitor Geometry Changes

Use this method whenever a capacitor's geometry (plate separation, area, or dielectric) is changed either with the battery still connected or after it has been disconnected.

Steps

Step 1: Start from the geometry formula

For a parallel-plate capacitor, $C=\dfrac{\varepsilon_0 A}{d}$ (or $\dfrac{k\varepsilon_0 A}{d}$ with a dielectric present). Note exactly how $C$ responds to the geometric change described — e.g. increasing $d$ always decreases $C$.

Step 2: Check whether the switch/key is open or closed at the moment of the change

This is the single most important fact in the whole problem — it decides which of $Q$ or $V$ is held fixed.

Step 3: If the key is closed (battery connected), hold $V$ fixed

The battery clamps the plates at $V=E$ regardless of geometry. As $C$ changes, use $Q=CV$ with $V$ fixed to find how $Q$ must change. …