Q. 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
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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 $
