Question

A capacitor of capacitance $C_1$ is charged by connecting it to a battery. The battery is now removed and this capacitor is connected to a second uncharged capacitor of capacitance $C_2$ . If the charge gets distributed equally on the two capacitors after connection, the ratio of the total energy stored in the capacitors after connection to the total energy stored in them before connection is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{2}$

If Q is the initial charge on capacitor with capacitance $C_1$, the initial energy stored is

$U_i=\frac{Q^2}{2C_1}$

When the two capacitors are connected together, the potential difference across the two capacitors will be the same because they are connected in parallel.

Given that, charge on capacitors distribute equally after connection.

From charge conservation,

$Q_1+Q_2=Q$

$Q_1=Q_2$

$Q_1$ and $Q_2$ are the final charges on capacitors after connection.

Thus, $Q_1=Q_2=\frac{Q}{2}$

Using $(C=\frac{Q}{V}$), provided that $V_1=V_2$

It can be inferred that their capacitances are equal.

Thus,$C_1=C_2=C\left(\text{say}\right)$

$\Rightarrow U_i=\frac{Q^2}{2C}$

Therefore, final energy stored in the two capacitors is

$U_f=\frac{{Q_1}^2}{2C}+\frac{{Q_2}^2}{2C}$

$\Rightarrow U_f=\frac{Q^2}{8C}+\frac{Q^2}{8C}=\frac{Q^2}{4C}$

$\Rightarrow \frac{U_f}{U_i}=\frac{1}{2}$

$\therefore$Option $\left(b\right)$ is the correct answer.