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The plates of parallel plate capacitor of capacity $ 100 \mu F$ are charged to potential $100 v $and then seperated from each other so that distance between them is doubled. How much energy is spent in doing so ?

$(A)\;0.5 \;J \\ (B)\;0.25 \;J \\ (C)\; 2.5 \;J \\ (D)\;5\;J $

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1 Answer

$w= \large\frac{1}{2} \bigg( \large\frac{C}{2} \bigg) (2V^2) ^2 - \large\frac{1}{2}$$CV^2$
$\qquad=\large\frac{1}{2} $$CV^2$
$\qquad= \large\frac{1}{2} (100 \times 10^{-6} ) \times (100)^2$
$\qquad= 0. 5 J$
Hence A is the correct answer.

 

answered Jan 6, 2014 by meena.p
edited Jul 31, 2014 by thagee.vedartham
 

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