There exists a uniform electric field E =4 `xx 10^(5) Vm^(-3)` directed along negative x-axis such that electric potential at origin is zero. A charge of - 200 `mu`C is placed at origin and a charged of + 200 `mu` C is placed at a (3 m, 0) . The electrostatic potential energy of the system is

Options

A. 120 J
B. `-120` J
C. `-240` J
D. zero

Answer

To find the electrostatic potential energy of the system, we will consider the contributions from both the interaction between the two charges and the work done by the electric field on the charges.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- Charge \( q_1 = -200 \, \mu C = -200 \times 10^{-6} \, C \) is placed at the origin (0, 0).
- Charge \( q_2 = +200 \, \mu C = +200 \times 10^{-6} \, C \) is placed at the point (3 m, 0).

2. **Calculate the Distance Between the Charges:**
- The distance \( r \) between the charges is \( 3 \, m \).

3. **Use the Formula for Electrostatic Potential Energy:**
- The formula for the potential energy \( U \) between two point charges is given by:
\[ U = k \frac{q_1 q_2}{r} \]
where \( k = 9 \times 10^9 \, Nm^2/C^2 \).

4. **Substitute the Values into the Formula:**
- Substitute \( q_1 \), \( q_2 \), and \( r \) into the equation:
\[ U = 9 \times 10^9 \frac{(-200 \times 10^{-6})(200 \times 10^{-6})}{3} \]

5. **Calculate the Product of Charges:**
- Calculate \( q_1 q_2 \):
\[ q_1 q_2 = (-200 \times 10^{-6})(200 \times 10^{-6}) = -40 \times 10^{-12} \, C^2 \]

6. **Substitute Back to Find U:**
- Now substitute back into the potential energy formula:
\[ U = 9 \times 10^9 \frac{-40 \times 10^{-12}}{3} \]

7. **Calculate the Final Value of U:**
- Simplifying gives:
\[ U = 9 \times 10^9 \times \left(-\frac{40}{3} \times 10^{-12}\right) = -120 \times 10^{-3} \, J \]
- Therefore, the electrostatic potential energy \( U \) is:
\[ U = -0.12 \, J \]

### Final Answer:
The electrostatic potential energy of the system is \( -0.12 \, J \).