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Question

A charge of +2.0×108 C is placed on the positive plate and a charge of 1.0×108 C on the negative plate of a parallel-plate capacitor of capacitance 1.2×103 μF.

Calculate the potential difference developed between the plates.


Solution

q1=2.0×108 C,

q2=1.0×108 C

Net charge =(q1q2)2

=1.5×108 C

q=VC

V=1.5×1081.2×109=12.5 V

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