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 \[V=\frac{q}{C}=\frac{600}{10}=60\,V\]You need to login to perform this action.
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