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A parallel plate capacitor has plate area $25.0 cm^2$ and a separation of $2.0\; mm$ between the plates. The capacitor is connected to a battery of $12.0\; V $ The plate separation is decreased to 1.00 mm. Find the extra charge given by the battery to the positive plate.

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Here, $A = 25.0 cm^2 = 25.0 \times 10^{–4} m^2, d = 2.0 mm = 2 × 10^{–3}, V = 12 volt.$
When plate separation is decreased to half, capacity becomes twice.
The charge $(q' = C'V),$ becomes twice.
Hence extra charge given by the battery $= q' – q = 2q – q = 1.35 \times 10^{-10} C$
Hence C is the correct answer.
answered Jun 13, 2014 by meena.p
 

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