Question

In the R−C circuit shown in the figure, the total energy of $3.6\times {10}^{-3}J$ is dissipated in the $10\Omega$ resistor when the switch $S$ is closed. The initial charge on the capacitor is:

• A. $60\mu C$
• B. $120\mu C$
• C. $60\sqrt{2}\mu C$
• D. $\frac{60}{\sqrt{2}}\mu C$

Answer 1

The correct option is B

$120\mu C$

Step 1: Given data and Diagram

Total energy $=3.6\times {10}^{-3}J$

Resistance $=10\Omega$

Capacitance $C=2\mu F$

Step 2: Explanation Energy Stored in a Charged Capacitor

1. In order to charge up a capacitor, work is to be done against the electrostatic forces present.
2. This work is stored in the capacitor as electrostatic potential energy.
3. Let us start with an uncharged capacitor.
4. To charge it up we must remove electrons from one conductor and carry them to the other conductor.
5. Suppose at an intermediate stage during charging the charge on the positive conductor is $q$ and the potential difference across the capacitor is $\varphi =\frac{q}{C}$, where $C$ is the capacitance.

Work done to increase the charge by an amount $dq$ is

$dW=\varphi dq=\frac{q}{C}dq$

Therefore, the total work done in charging the capacitor from an uncharged condition to a state with charge $Q$ would be

$W={\int }_0^Q\frac{1}{C}qdq=\frac{Q^2}{2C}$.

Step 3: Calculate the initial charge

Given that $C=2\mu F=2\times {10}^{-6}F$

The energy stored in the capacitor will be dissipated through a resistor when $S$ is closed.

Thus, $\frac{Q^2}{2C}=3.6\times {10}^{-3}$

$\Rightarrow Q^2=3.6\times {10}^{-3}\times 2\times 2\times {10}^{-6}$

$=144\times {10}^{-10}$

$\Rightarrow Q=12\times {10}^{-5}C$

$=120\mu C$

Hence, the correct answer is option (B).