An initially uncharged capacitor C is being charged by a battery of emf E through a resistance R upto the instant when the capacitor is charged to the potential E/2, the ratio of the work done by the battery to the heat dissipated by the resistor is given by :-

Options

A. `2:1`
B. `3 :1`
C. `4 :3`
D. `4 :1`

Explanation

To solve the problem, we need to find the ratio of the work done by the battery to the heat dissipated by the resistor when a capacitor is charged to a potential of \( \frac{E}{2} \).

### Step-by-Step Solution:

1. **Understand the Charging of the Capacitor**:
The capacitor \( C \) is charged by a battery of emf \( E \) through a resistor \( R \). The voltage across the capacitor when charged is \( V = \frac{E}{2} \).

2. **Calculate Charge on the Capacitor**:
The charge \( Q \) on the capacitor when it is charged to \( \frac{E}{2} \) is given by:
\[ Q = C \cdot V = C \cdot \frac{E}{2} = \frac{CE}{2} \]

3. **Determine the Total Charge**:
The maximum charge \( Q_0 \) that the capacitor can hold when fully charged to \( E \) is:
\[ Q_0 = C \cdot E \]

4. **Use the Charging Equation**:
The charge \( Q \) at time \( t \) during charging is given by:
\[ Q = Q_0 \left(1 - e^{-\frac{t}{RC}}\right) \]
Setting \( Q = \frac{CE}{2} \) and \( Q_0 = CE \), we have:
\[ \frac{CE}{2} = CE \left(1 - e^{-\frac{t}{RC}}\right) \]
Simplifying gives:
\[ \frac{1}{2} = 1 - e^{-\frac{t}{RC}} \]
Rearranging leads to:
\[ e^{-\frac{t}{RC}} = \frac{1}{2} \]
Taking the natural logarithm:
\[ -\frac{t}{RC} = \ln\left(\frac{1}{2}\right) \implies t = -RC \ln\left(\frac{1}{2}\right) = RC \ln(2) \]

5. **Calculate Work Done by the Battery**:
The work done \( W \) by the battery in charging the capacitor is given by:
\[ W = Q \cdot V = \left(\frac{CE}{2}\right) \cdot E = \frac{CE^2}{2} \]

6. **Calculate Heat Dissipated by the Resistor**:
The heat \( H \) dissipated in the resistor can be calculated using the formula:
\[ H = \int_0^t I^2 R \, dt \]
The current \( I \) can be expressed as:
\[ I = \frac{dQ}{dt} = \frac{CE}{R} e^{-\frac{t}{RC}} \]
Therefore,
\[ H = \int_0^{RC \ln(2)} \left(\frac{CE}{R} e^{-\frac{t}{RC}}\right)^2 R \, dt \]
Simplifying this integral leads to:
\[ H = \frac{C^2 E^2}{R} \int_0^{RC \ln(2)} e^{-\frac{2t}{RC}} \, dt \]
Evaluating the integral yields:
\[ H = \frac{C^2 E^2}{R} \left[-\frac{RC}{2} e^{-\frac{2t}{RC}}\right]_0^{RC \ln(2)} = \frac{C^2 E^2}{R} \cdot \frac{RC}{2} \left(1 - \frac{1}{4}\right) = \frac{3C E^2}{8} \]

7. **Calculate the Ratio**:
Finally, the ratio of work done by the battery to the heat dissipated is:
\[ \text{Ratio} = \frac{W}{H} = \frac{\frac{CE^2}{2}}{\frac{3CE^2}{8}} = \frac{4}{3} \]

### Final Answer:
The ratio of the work done by the battery to the heat dissipated by the resistor is \( \frac{4}{3} \).