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. $V$
• B. $V+\frac{Q}{C}$
• C. $V+\frac{Q}{2C}$
• D. $V-\frac{Q}{C},ifQ<CV$

Answer 1

Answer: (C)

$\text{Step 1: Electric field inside capacitor}$$\text{[Ref. Fig]}$

Initial charge $q=CV$

Final charge on plate $1=CV+Q$

Final charge on plate $2=-CV$

Electric field due to plate $\left(1\right)$

$|E_1|=\frac{{\sigma }_1}{2{\in }_0}=\frac{CV+Q}{A\left(2{\in }_0\right)}$ $({\sigma }_1=\frac{CV+Q}{A})$

Electric field due to plate $\left(2\right)$

$|E_2|=\frac{{\sigma }_2}{2{\in }_0}=\frac{CV}{A\left(2{\in }_0\right)}$ $({\sigma }_2=\frac{CV}{A}(Negative\left)\right)$

Electric field direction is away from positive charge and towards negative charge.

Therefore, inside capacitor both $E_1$ and $E_2$ will be in right direction.

So, $E_{net}=E_1+E_2$ $=\frac{CV+Q}{2A{\in }_0}+\frac{CV}{2A{\in }_0}=\frac{CV}{A{\in }_0}+\frac{Q}{2A{\in }_0}$

$\text{Step 2: Potential difference across capacitor}$

$V^{\prime }=E_{net}\times d$

$\Rightarrow V^{\prime }=(\frac{CV}{A{\in }_0}+\frac{Q}{2A{\in }_0})d$

$\Rightarrow V^{\prime }=CV\left(\frac{d}{A{\in }_0}\right)+\frac{Q}{2}\left(\frac{d}{A{\in }_0}\right)$

Since capacitance of parallel plate capacitor: $C=\frac{A{\in }_0}{d}$

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

Hence, Option ${}^{\prime }{C}^{\prime }$ is correct.