question_answer

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

Correct Answer: B

Solution :

                Charged gained by the plates of capacitor \[{{q}_{0}}=CV=10\mu F\times 50\,V=500\mu C\] 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 \[\therefore \] \[\frac{(700-q)}{2A{{\varepsilon }_{0}}}+\frac{q}{2A{{\varepsilon }_{0}}}+\left( \frac{500-q}{2A{{\varepsilon }_{0}}} \right)=\frac{q}{2A{{\varepsilon }_{0}}}\] \[\Rightarrow \]               \[700-q+q+500-q=q\] \[\Rightarrow \]                               \[2q=1200\] \[\Rightarrow \]                               \[q=600\mu C\] \[\therefore \] Potential difference between plates is \[V=\frac{q}{C}=\frac{600}{10}=60\,V\]