Question

A capacitor of capacitance $10\mu F$ is charged a potential $50V$ with a battery. The battery is now disconnected and an additional charge $200\mu C$ is given to the positive plate of the capacitor. The potential difference across the capacitor will be :

• A. $50V$
• B. $100V$
• C. $60V$
• D. $80V$

Answer 1

Answer: (C)

Charge acquired by the plates of the capacitor $q_0=CV=\left(10\mu F\right)\times \left(50V\right)=500\mu C$

Now, let the charge distribution is as follows.

Total charge on positive plate has now become $700\mu C$ while that in negative plate is still $-500\mu C$.

Here, charges are in $\mu C$.

Net electric field at point $P$ is zero.

$\therefore \frac{(700-q)}{2A{\epsilon }_0}+\frac{q}{2A{\epsilon }_0}+\left(\frac{500-q}{2A{\epsilon }_0}\right)=\frac{q}{2A{\epsilon }_0}$

$\Rightarrow q=600\mu C$

$\therefore$ Potential difference between the plates is

$\Delta V=\frac{q}{C}=\frac{600\mu C}{10\mu F}=60V$