Two uniformly charged large plane sheets `S_1` and `S_2` having charge densities `sigma_1` and `sigma_2` (`sigma_1gtsigma_2`) are placed at a distance d parallel to each other. A charge `q_0` is moved along a line of length a(altd) at an angle `45^@` with the normal to `S_1`. Calculate the work done by the electric field

Explanation

To solve the problem, we need to calculate the work done by the electric field when a charge \( q_0 \) is moved along a line at an angle of \( 45^\circ \) with the normal to the charged plane sheet \( S_1 \).

### Step-by-Step Solution:

1. **Identify the Electric Fields**:
The electric field due to a uniformly charged infinite plane sheet is given by:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

For the two sheets \( S_1 \) and \( S_2 \):
- The electric field \( E_1 \) due to sheet \( S_1 \) is:
\[ E_1 = \frac{\sigma_1}{2\epsilon_0} \]
- The electric field \( E_2 \) due to sheet \( S_2 \) is:
\[ E_2 = \frac{\sigma_2}{2\epsilon_0} \]

2. **Determine the Net Electric Field**:
Since the charge densities are given as \( \sigma_1 > \sigma_2 \), the net electric field \( E \) between the two sheets is:
\[ E = E_1 - E_2 = \frac{\sigma_1}{2\epsilon_0} - \frac{\sigma_2}{2\epsilon_0} = \frac{\sigma_1 - \sigma_2}{2\epsilon_0} \]

3. **Calculate the Force on the Charge**:
The force \( F \) acting on the charge \( q_0 \) in the electric field is given by:
\[ F = q_0 E = q_0 \left( \frac{\sigma_1 - \sigma_2}{2\epsilon_0} \right) \]

4. **Determine the Displacement**:
The charge \( q_0 \) is moved along a line of length \( a \) at an angle of \( 45^\circ \) with the normal to \( S_1 \). The displacement \( s \) can be expressed as:
\[ s = d \]

5. **Calculate the Work Done**:
The work done \( W \) by the electric field when moving the charge is given by:
\[ W = F \cdot s \cdot \cos(\theta) \]
where \( \theta = 45^\circ \). Thus, substituting the values:
\[ W = \left( q_0 \frac{\sigma_1 - \sigma_2}{2\epsilon_0} \right) d \cos(45^\circ) \]
Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ W = q_0 \frac{\sigma_1 - \sigma_2}{2\epsilon_0} d \cdot \frac{1}{\sqrt{2}} \]
Therefore, the final expression for the work done is:
\[ W = \frac{q_0 d (\sigma_1 - \sigma_2)}{2\epsilon_0 \sqrt{2}} \]

### Final Answer:
\[ W = \frac{q_0 d (\sigma_1 - \sigma_2)}{2\epsilon_0 \sqrt{2}} \]