Question

A capacitor of 10 $\mu$ F is charged to a potential 50 V with a battery. The battery is n disconnected and an additional day 200 $\mu$ C is given to the positive plate of 1 capacitor. The potential difference across the capacitor will be

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

Answer 1

Answer: (B)

Charged gained by the plates of capacitor $q_0=CV=10\mu F\times 50V=500\mu C$ When an additional charge is given to the positive plate, then total charge on positive plate becomes 700 uC while negative plate will have previous potential. Net electric field at P is zero $\therefore$ $\frac{(700-q)}{2A{\epsilon }_0}+\frac{q}{2A{\epsilon }_0}+\left(\frac{500-q}{2A{\epsilon }_0}\right)=\frac{q}{2A{\epsilon }_0}$ $\Rightarrow$ $700-q+q+500-q=q$ $\Rightarrow$ $2q=1200$ $\Rightarrow$ $q=600\mu C$ $\therefore$ Potential difference between plates is $V=\frac{q}{C}=\frac{600}{10}=60V$