Electric Potential and Capacitance | Physics JEE previous year questions with Solutions

Q 1 :
A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab $(ε_{r} = 6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _______ $\times 10^{- 12} J .$ [2024]

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(750) Before inserting dielectric, capacitance is given as $C_{0} = 12.5$ pF and charge on the capacitor is $Q = C_{0} V$ . After inserting dielectric capacitance will become $C_{f} = ε_{r} C_{0} .$

Change in potential energy of the capacitor

$∆ U = U_{0} - U_{f} = \frac{Q^{2}}{2 C_{0}} - \frac{Q^{2}}{2 C_{f}} = \frac{Q^{2}}{2 C_{0}} [1 - \frac{1}{ε_{r}}]$

$∆ U = \frac{{(C_{0} V)}^{2}}{2 C_{0}} [1 - \frac{1}{ε_{r}}] = \frac{1}{2} C_{0} V^{2} [1 - \frac{1}{ε_{r}}]$

Using $C_{0} = 12.5 p F,$ $V = 12$ $V,$ $ε_{r} = 6$

$∆ U = \frac{1}{2} (12.5) \times 12^{2} [1 - \frac{1}{6}] = \frac{1}{2} (12.5) \times 12^{2} \times \frac{5}{6}$

$= 750 p J = 750 \times 10^{- 12} J$

Q 2 :
A parallel plate capacitor with plate separation 5mm is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 mm, while keeping the battery connections intact, the capacitor draws 25% more charge from the battery than before. The dielectric constant of the sheet is __________. [2024]

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(2) Without dielectric $Q = \frac{A \in_{0}}{d} V$

with dielectric

$Q = \frac{A \in_{0} V}{d - t + \frac{t}{K}}$

Given $\frac{A \in_{0} V}{d - t + \frac{t}{K}} = (1.25) \frac{A \in_{0} V}{d}$

$\Rightarrow 1.25 (3 + \frac{2}{K}) = 5 \Rightarrow K = 2$

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