Consider a capacitor as shown in figure. If we pull the plates of capacitor apart to a final position as shown in figure, then we must perform some work against the electric force. For this situation, mark out the correct statements.

Options

A. Work done is `Q_2(d_2-d_1)//2epsilon_0A` and is stored in volume `A(d_2-d_1)`.
B. Work done is `+Q_2(d_2-d_1)//2epsilon_0A` and is stored in volume `Ad_2`.
C. Work done is `+epsilon_0AV^2(d_2-d_1)//2d_1d_2` and is stored in volume `Ad_2`.
D. Work done is `epsilon_0AV^2(d_2-d_1)//2d_1d_2` and is stored in volume `Ad_2`.

Answer

C

Explanation

this work done by us is stored in the capacitor in the volume `A(d_(2)-d_(1))` where new electric field is created if you calculate the work done by using the expression
`W_(ext)-W_(el)=dU=(epsilon_(0)AV^(2))/(2)[(1)/(d^(2))-(1)/(d_(1))]lt0`
yu may get condused here battery is also doing work so from energy conservation principle, we get
`W_(ext)+W_(el)+E_("battery")=0`