question_answer

A capacitor of \[10\mu F\] is charged to a potential \[50\,\,V\]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) \[100\,\,V\]                       

B) \[60\,\,V\]

C) \[80\,\,V\]                                         

D) \[50\,\,V\]

Correct Answer: B

Solution :

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 \[700\mu C\] while negative plate will have previous potential. Net electric field at P is zero \[\therefore \]\[\frac{(700-q)}{2A{{\varepsilon }_{0}}}+\frac{q}{2A{{\varepsilon }_{0}}}+\left( \frac{500-q}{2A{{\varepsilon }_{0}}} \right)=\frac{q}{2A{{\varepsilon }_{0}}}\] \[\Rightarrow \]               \[700-q+q+500-q=q\] \[\Rightarrow \]               \[2q=1200\] \[\Rightarrow \]               \[q=600\mu C\] \[\therefore \]Potential difference between plates is                 \[V=\frac{q}{C}=\frac{600}{10}=60V\]