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11 Jun 2012 19:03:16 IST

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Capacitors

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A charge of +2.0 x 10^-8 C is placed on the positive plate and a charge of -1.0 x 10^-8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10^-3 uF. Calculate the potential difference developed between the plates.

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Sagar Saxena

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Joined: 8 Oct 2008

Posts: 8064

17 Jun 2012 14:07:13 IST

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here a capacitor of charge (q1+q2)/2=5*10^-9

V=q/c

V=4.16*10^-6V

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