Question

A capacitor of capacitance $C$ is charged to a potential difference $V$ from a cell and then disconnected from it. A charge $+Q$ is now given to its positive plate. The potential difference across the capacitor is now

• A. $V+\frac{Q}{2C}$
• B. $V+\frac{Q}{C}$
• C. $V-\frac{Q}{C}$
• D. $V$

Answer 1

The correct option is A $V+\frac{Q}{2C}$

Let the area of the plate be $A$ and distance between the plates be $d$.

Initial charge on the plates,

$Q_0=CV$

As, $C=\frac{{\epsilon }_{0}A}{d}$

$\Rightarrow Q_0=\frac{A{\epsilon }_0}{d}V.....\left(1\right)$

Now, when charge $Q$ is added to the positive plate, electric field between the plates will be,

$E=\frac{Q+Q_0}{2A{\epsilon }_0}+\frac{Q_0}{2A{\epsilon }_0}$



Now new voltage across the plates will be,

$V^{\prime }=Ed=\frac{Qd+Q_{0}d}{2A{\epsilon }_0}+\frac{Q_{0}d}{2A{\epsilon }_0}$

$\Rightarrow V^{\prime }=\frac{Qd}{2A{\epsilon }_0}+\frac{Q_{0}d}{2A{\epsilon }_0}+\frac{Q_{0}d}{2A{\epsilon }_0}$

$\Rightarrow V^{\prime }=\frac{Qd}{2A{\epsilon }_0}+\frac{Q_{0}d}{A{\epsilon }_0}$

From equation (1), $V=\frac{Q_{0}d}{A{\epsilon }_0}$ and $C=\frac{{\epsilon }_{0}A}{d}$, we have

$V^{\prime }=\frac{Q}{2C}+V$

Hence, option (c) is the +correct answer.

Key concept : Effect on potential difference on PPC (Parallel Plate Capacitor) with battery removed after charging and when some more charge is given to its positive plate after removal of battery.