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(C) When a capacitor remains connected to a battery,the potential difference $V$ across its plates remains constant.When a dielectric slab of constant

Vedclass · Accepted Answer

Energy from the battery is consumed in this process.

Vedclass · Answer

(C) When a capacitor remains connected to a battery,the potential difference $V$ across its plates remains constant.When a dielectric slab of constant $K$ is inserted,the capacitance increases from $C$ to $C' = KC$.The charge on the capacitor increases from $Q = CV$ to $Q' = KCV$.The energy stored in the capacitor changes from $U = \frac{1}{2}CV^2$ to $U' = \frac{1}{2}KCV^2$.The change in energy is $\Delta U = U' - U = \frac{1}{2}(K-1)CV^2$,which is positive,meaning the energy stored increases.The work done by the battery is $W_{battery} = \Delta Q \cdot V = (Q' - Q)V = (K-1)CV^2$.Since $W_{battery} = 2\Delta U$,the battery supplies energy equal to the increase in stored energy plus an equal amount of work done by the external agent (or the force of attraction pulling the slab in).Thus,energy from the battery is consumed to increase the stored energy and perform work.

When a dielectric slab is inserted between the plates of a capacitor while it remains connected to a battery,which of the following occurs during this process?

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