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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

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

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

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

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

answer is C.

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Detailed Solution

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 Q₂)

Potential difference between the plates $\frac{{ch\arg e}}{{capaci\tan ce}} = = \frac{{{Q_2}}}{C} = \frac{{2{Q_2}}}{{2C}} = \frac{{{Q_2} - ( - {Q_2})}}{{2C}} = \frac{{{Q_2} - {Q_3}}}{{2C}}.$

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