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Q.

  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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a

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

b

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

c

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

d

\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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