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

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

Q 3 :
A capacitor has air as dielectric medium and two conducting plates of area 12 ${cm}^{2}$ and they are 0.6 cm apart. When a slab of dielectric having area 12 ${cm}^{2}$ and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm to keep the capacitance same as in previous case. The dielectric constant of the slab is

(Given $ε_{0} = 8.834 \times 10^{- 12}$ F/m) [2024]

1
0.66
1.33
1.50

1

2

3

4

(4)

Parallel plate capacitance, $C = \frac{ϵ_{o} A}{d_{1}}$

For partially filled capacity, $C = \frac{ϵ_{o} A}{(d_{2} - t + \frac{t}{k})}$

For case (i) $C_{1} = \frac{ϵ_{o} A}{0.6}$ , For case (ii) $C_{2} = \frac{ϵ_{o} A}{0.8 - 0.6 + \frac{0.6}{K}}$

Now $C_{1} = C_{2}, \frac{ϵ_{o} A}{0.6} = \frac{ϵ_{o} A}{0.2 + \frac{0.6}{K}}$ ,

$0.6 = 0.2 + \frac{0.6}{K} \Rightarrow K = \frac{3}{2}$

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