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Q.

A parallel plate capacitor filled with a medium of dielectric constant 10 , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

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a

decrease by 15 %

b

increase by 50 %

c

increase by 33 %

d

increase by 25 %

answer is A.

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Detailed Solution

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$E \Rightarrow \frac{1}{2} (KC) v^{2}$
$∴ %$ change
$\begin{array}{l} \Rightarrow \frac{\frac{1}{2} K_{2} {CV}^{2} - \frac{1}{2} K_{1} {CV}^{2}}{\frac{1}{2} K_{1} {CV}^{2}} = \frac{K_{2} - K_{1}}{K_{1}} \times 100 \\ \Rightarrow \frac{15 - 10}{10} \times 100 = 50 % \end{array}$

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