Separation between the plates of a parallel plate capacitor, connected to a battery (zero resistance) of constant e.m.f., is increased with constant (very slow) speed by external forces. During the process, W is the work done by external forces. If U is the change in potential energy of the capacitor, $W_{b}$ is the work done by the battery and H is the heat loss in the circuit. Then

W + W_{b} = Δ U

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H \neq 0

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H = Δ U

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W = 0

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Solution

The correct option is A $W + W_{b} = Δ U$
$C = \frac{ϵ_{0} A}{X} a n d q = \frac{ϵ_{0} A}{X} E$
$i = \frac{d q}{d t} = - \frac{ϵ_{0} A E}{x^{2}} \frac{d x}{d t}$

Since current is finite in the circuit and resistance is zero, therefore heat loss in the circuit is zero.

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The correct option is A W+Wb=ΔUC=ϵ0AX andq=ϵ0AXE i=dqdt=−ϵ0AEx2dxdt Since current is finite in the circuit and resistance is zero, therefore heat loss in the circuit is zero.

The correct option is A $W + W_{b} = Δ U$
$C = \frac{ϵ_{0} A}{X} a n d q = \frac{ϵ_{0} A}{X} E$
$i = \frac{d q}{d t} = - \frac{ϵ_{0} A E}{x^{2}} \frac{d x}{d t}$

Since current is finite in the circuit and resistance is zero, therefore heat loss in the circuit is zero.