Question

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

Answer 1

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

Just after disconnecting from cell, charge on positive plate, $Q_1=Q_0$ and on negative plate, $Q_2=-Q_0$
Where $Q_0=CV$

After giving charge +Q to positive plate, charge on positive plate,$Q_1^{{}^{\prime }}=Q_0+Q$
Here, $\overrightarrow{E_1}$is due to $Q_1^{{}^{\prime }}$ on left plate and $\overrightarrow{E_2}$ due to $Q_2$ on right plate.

So,$V^{{}^{\prime }}=(E_1+E_2)d=(\frac{Q_0+Q}{2{\epsilon }_{0}A}+\frac{Q_0}{2{\epsilon }_{0}A})d$

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


$\Rightarrow V^{{}^{\prime }}=\frac{Q_0}{C}+\frac{Q}{2C}$ $[\because C=\frac{{\epsilon }_{0}A}{d}]$

$\Rightarrow V^{{}^{\prime }}=V+\frac{Q}{2C}$$[\because C=\frac{Q_0}{V}]$