In an isolated parallel plate capacitor of capacitance C , four surface have charges {Q₁} , {Q₂}

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2. Electric Potential and Capacitance

hard

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

$\frac{{{Q_1} + {Q_2} + {Q_3} + {Q_4}}}{{2C}}$

$\frac{{{Q_2} + {Q_3}}}{{2C}}$

$\frac{{{Q_2} - {Q_3}}}{{2C}}$

$\frac{{{Q_1} + {Q_4}}}{{2C}}$

(IIT-1999)

Solution

(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}}.$

Standard 12

Physics

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medium

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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

Solution

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