A parallel plate capacitor of capacitance 2 F is charged to a potential V. The energy stored in the capacitor is $\mathrm{E}_{1}$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_{2}$. The ratio $E_{2} / E_{1}$ is :

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Capacitance

A parallel plate capacitor of capacitance 2 F is charged to a potential V. The energy stored in the capacitor is

E_{1}

. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is

E_{2}

. The ratio

E_{2} / E_{1}

is :

Answer: (B)

With charge

Q = C V

the initial energy is

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

. Joining the charged capacitor to an identical uncharged one spreads the same charge

Q

over a total capacitance

2 C

, so the common potential drops to

\frac{V}{2}

. The stored energy becomes

E_{2} = \frac{1}{2} (2 C) (\frac{V}{2})^{2} = \frac{C V ^{2}}{4}

. Therefore

\frac{E _{2}}{E _{1}} = \frac{C V ^{2} /4}{C V ^{2} /2} = \frac{1}{2}

, that is

E_{2} : E_{1} = 1 : 2

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option