In an isolated parallel plate capacitor of capacitance $C$, the four surface have charges ${Q_1}$, ${Q_2}$, ${Q_3}$ and ${Q_4}$ as shown. The potential difference between the plates is
IIT 1999, Diffcult
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(c) Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal, ${Q_2} = - {Q_3}$.
The charge on a capacitor means the charge on the inner surface of the positive plate (here it is ${Q_2}$)
Potential difference between the plates $ = \frac{{charge}}{{capacitance}}$$ = \frac{{{Q_2}}}{C} = \frac{{2{Q_2}}}{{2C}}$$ = \frac{{{Q_2} - ( - {Q_2})}}{{2C}} = \frac{{{Q_2} - {Q_3}}}{{2C}}.$
The charge on a capacitor means the charge on the inner surface of the positive plate (here it is ${Q_2}$)
Potential difference between the plates $ = \frac{{charge}}{{capacitance}}$$ = \frac{{{Q_2}}}{C} = \frac{{2{Q_2}}}{{2C}}$$ = \frac{{{Q_2} - ( - {Q_2})}}{{2C}} = \frac{{{Q_2} - {Q_3}}}{{2C}}.$
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