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Electricity

New kid on the Block

 Joined: 28 Dec 2011 Post: 1
11 Jun 2012 19:03:16 IST
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Capacitors
Engineering Entrance , Medical Entrance , AIPMT , JEE Main , AIIMS , JEE Advanced , Physics , Electricity

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.

Blazing goIITian

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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