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(C) When a dielectric slab is inserted into a capacitor while it remains connected to a battery, the potential difference $V$ across the plates remain

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Work is done at the cost of the battery

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(C) When a dielectric slab is inserted into a capacitor while it remains connected to a battery, the potential difference $V$ across the plates remains constant.Since the capacitance $C$ increases $(C = K C_0)$, the charge $Q$ on the plates increases $(Q = CV)$.The battery must supply additional charge to the capacitor to maintain the constant potential difference.The work done by the battery is $W_{battery} = \Delta Q \cdot V = (Q_{final} - Q_{initial})V$.The change in energy stored in the capacitor is $\Delta U = \frac{1}{2} C_{final} V^2 - \frac{1}{2} C_{initial} V^2 = \frac{1}{2} \Delta C V^2$.Since $W_{battery} = \Delta C V^2$, the work done by the battery is twice the increase in stored energy.The remaining half of the energy supplied by the battery is converted into the work done by the external agent inserting the slab.Thus, the work is done at the cost of the battery.

$A$ capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process

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