Separation between plates of a parallel plate capacitor of capacitance -o A-d connected to a battery of EMF E is doubled very slowly- Choose the correct options-

Work done by battery is W = E q It is clear that q is negative here. nbsp; q = CE/2 CE = CE/2 = ϵ_o E A/2d W = ϵ_o E^2 A/2d. As the process is done ...

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Standard XIII

Physics

Capacitance of Parallel Plate Capacitor

Separation be...

Question

Separation between plates of a parallel plate capacitor of capacitance $\frac{ϵ_{o} A}{d}$ connected to a battery of EMF E is doubled very slowly. Choose the correct option(s).

Heat generated in circuit is

\frac{E^{2} ϵ_{o} A}{2 d}

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Work done by battery when the separation is increasing is

- \frac{E^{2} ϵ_{o} A}{2 d}

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Work done by electrostatic force is positive during process.

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Heat generated in circuit is zero.

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Solution

The correct option is D Heat generated in circuit is zero.
Work done by battery is $W = E △ q$
It is clear that $△ q$ is negative here.

$△ q = \frac{C E}{2} - C E = - \frac{C E}{2} = - \frac{ϵ_{o} E A}{2 d}$
$W = - \frac{ϵ_{o} E^{2} A}{2 d} .$

As the process is done very slowly, heat generated in process is zero.

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Separation between plates of a parallel plate capacitor of capacitance ϵoAd connected to a battery of EMF E is doubled very slowly. Choose the correct option(s).

The correct option is D Heat generated in circuit is zero. Work done by battery is W=E △q It is clear that △q is negative here. △q=CE2−CE=−CE2=−ϵoEA2d W=−ϵo E2A2d. As the process is done very slowly, heat generated in process is zero.

Separation between plates of a parallel plate capacitor of capacitance $\frac{ϵ_{o} A}{d}$ connected to a battery of EMF E is doubled very slowly. Choose the correct option(s).