JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Capacitance

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A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key 'K' is closed, the total energy stored across the combination is

E_{1}

. Now key 'K' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now

E_{2}

. The ratio

E_{1} / E_{2}

will be :

(A) $\frac{1}{10}$
(B) $\frac{2}{5}$
(C) $\frac{5}{13}$ ✓ Correct
(D) $\frac{5}{26}$

Answer: (C)

Solution

When the switch is closed, the capacitors are in parallel:

C_{eq} = 2 C

, so

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

. When the switch is opened, the right capacitor retains charge CV while the left capacitor remains at potential V. After inserting dielectric (K=5) in the left capacitor, its capacitance becomes 5C. The energy is

E_{2} = \frac{1}{2} (5 C) V^{2} + \frac{( C V ) ^{2}}{2 ( 5 C )} = \frac{5 C V ^{2}}{2} + \frac{C V ^{2}}{10} = \frac{13 C V ^{2}}{5}

. Therefore,

\frac{E _{1}}{E _{2}} = \frac{C V ^{2}}{13 C V ^{2} /5} = \frac{5}{13}

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Capacitance

Solution

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