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Question: The energy stored in capacitor is – ![](https://cdn.pureessence.tech/canvas_446.png?top_left_x=1061...

The energy stored in capacitor is –

A

U = 12\frac { 1 } { 2 } ε0 V2 d\frac { \varepsilon _ { 0 } \mathrm {~V} ^ { 2 } } { \mathrm {~d} } [l + x (K + 1)]

B

U = ε02dV2\frac { \varepsilon _ { 0 } } { 2 \mathrm { dV } ^ { 2 } } [l + x (K + 1)]

C

U = [l + x (K – 1)]

D

U = ε0bV2ℓK2 d\frac { \varepsilon _ { 0 } \mathrm { bV } ^ { 2 } \ell \mathrm { K } } { 2 \mathrm {~d} }

Answer

U = l+x(K–1)l + x (K – 1)

Explanation

Solution

U = 12\frac { 1 } { 2 } CV2

U=12\frac { 1 } { 2 } [ε0 b(ℓ−x)d+ε0bxdK]v2\left[ \frac { \varepsilon _ { 0 } \mathrm {~b} ( \ell - \mathrm { x } ) } { \mathrm { d } } + \frac { \varepsilon _ { 0 } \mathrm { bx } } { \mathrm { d } } \mathrm { K } \right] \mathrm { v } ^ { 2 } = (l–x+xK)

U = [l + k (K – 1)]