The separation between the plates of a parallel plate capacitor, connected to a battery (zero resistance) of constant EMF is increased with constant (very slow) speed by external forces. During the process, w is the work done by external forces. Δ U is the change in potential energy of the capacitor, wb is work done by the battery and H is the heat loss in the circuit. Then

Question

The separation between the plates of a parallel plate capacitor, connected to a battery (zero resistance) of constant EMF is increased with constant (very slow) speed by external forces. During the process, w is the work done by external forces. Δ U is the change in potential energy of the capacitor, wb is work done by the battery and H is the heat loss in the circuit. Then (A) w+wb=Δ U (B) H≠ 0 (C) H=Δ U (D) w=0

Tardigrade Admin · Accepted Answer

Since the plates are being moved very slowly there is no heat loss. So the total work done by external agent and battery will be equal to the change in potential energy of the system

$w + w_{b} = Δ U$

$H \neq = 0$

$H = Δ U$

$w = 0$

Since the plates are being moved very slowly there is no heat loss. So the total work done by external agent and battery will be equal to the change in potential energy of the system

w+wb​=ΔU

H=0

H=ΔU

w=0

Since the plates are being moved very slowly there is no heat loss. So the total work done by external agent and battery will be equal to the change in potential energy of the system

$w + w_{b} = Δ U$

$H \neq = 0$

$H = Δ U$

$w = 0$