JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Capacitance

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A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

(A)increase by $50%$ ✓ Correct
(B)decrease by $15%$
(C)increase by $25%$
(D)increase by $33%$

Answer: (A)

Solution

When a capacitor is connected to a battery, the voltage remains constant. The energy stored is:

E = \frac{1}{2} (K C) V^{2}

where

K

is the dielectric constant.

Percentage change in energy:

% change = \frac{E _{2} - E _{1}}{E _{1}} \times 100 = \frac{\frac{1}{2} K _{2} C V ^{2} - \frac{1}{2} K _{1} C V ^{2}}{\frac{1}{2} K _{1} C V ^{2}} \times 100

= \frac{K _{2} - K _{1}}{K _{1}} \times 100 = \frac{15 - 10}{10} \times 100 = 50%

increase

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Capacitance

Solution

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