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A parallel plate capacitor of plate area A, plate seperation d is charged potential difference V. The battery is then removed and a slab of dielectric constant k is inserted between plates of capacitor. If $Q,\in $ and $w$ denotes charge , electric field (after slab is inserted ) and work done on the system in process of inserting the slab, the incorrect relation is :

$(A)\;Q= \frac{\in _0 AV}{d} \\ (B)\;E= \frac{V}{kd} \\ (C)\; w= \frac{\in_0 AV^2}{2 kd} \\ (D)\;w= \frac{\in _0 AV^2}{2d} (1-\frac{1}{k}) $

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1 Answer

After inserting slab
New capacitance $C'=KC=\large\frac{K \in _0 A}{d}$
New potential difference, $V'=\large\frac{V}{K}$
New electric field $E'= \large\frac{V'}{d}=\frac{V}{kd}$
New charge $Q=C'V'= \large\frac{\in _0 AV}{d}$
Work= Final energy - Initial energy
$\qquad= \large\frac{1}{2}$$ C'V'^2 -\large\frac{1}{2} CV^2$
$\qquad=\large\frac{1}{2} $$ (KC ) \bigg ( 1-\large\frac{1}{K}\bigg)$
=> $ |w| =\large\frac{\in _0 AV^2}{2d} \bigg(1 -\large\frac{1}{k}\bigg)$
Hence C is the correct answer.


answered Jan 6, 2014 by meena.p
edited Jan 6, 2014 by meena.p
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