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The capacitance of capacitors of plate area $A_2 and A_1 (A_2 < A_1) $ at a distance l is

$(A)\;\frac{\in _0 A_2}{l} \\ (B)\;\frac{\in _0A_1}{l} \\ (C)\; \frac{\in_0 \sqrt {A_1A_2}}{l} \\ (D)\;None $

1 Answer

Electric field between the two plates is perpendicular to surface of conductor as shown, therefore effective area is $A_2$
=> $C=\large\frac{\in _0 A_2}{l}$
Hence A is the correct answer.
answered Jan 8, 2014 by meena.p
 

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