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Q.

The energy and capacity of a charged parallel plate capacitor are U and C respectively. Now a dielectric slab of $ε_{r} = 6$ is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant):

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a

$\frac{U}{6}, 6 C$

b

U, 6C

c

U, C

d

6U, 6C

answer is C.

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Detailed Solution

$V' = \frac{V}{6} C' = 6 C H e n c e e n e r g y = \frac{1}{2} (6 C) \frac{v^{2}}{36} = \frac{U}{6}$

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The energy and capacity of a charged parallel plate capacitor are U and C respectively. Now a dielectric slab of εr=6 is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant):