There exists a uniform electric field `E=4xx10^(5) Vm^(-1)` directed along negative x-axis such that electric potential at origin is zero. A charge of `-200muC` is palced at origin, and a charge of `+200muC` is placed at (3m,0). The electrostatic potential energy of the system is

Options

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

Explanation

To solve the problem of finding the electrostatic potential energy of the system with the given charges and electric field, we can follow these steps:

### Step 1: Understand the Configuration
We have two charges:
- Charge \( Q_1 = -200 \, \mu C \) placed at the origin (0, 0).
- Charge \( Q_2 = +200 \, \mu C \) placed at (3 m, 0).

There is a uniform electric field \( E = 4 \times 10^5 \, V/m \) directed along the negative x-axis. The electric potential at the origin is given to be zero.

### Step 2: Calculate the Electric Potential at (3, 0)
The potential \( V \) at a point in an electric field can be calculated using the formula:
\[ V = -E \cdot d \]
where \( d \) is the distance moved in the direction of the field. Here, we move from the origin to (3, 0), which is against the electric field.

Thus, the potential at (3, 0) is:
\[ V(3, 0) = -E \cdot (3) = -4 \times 10^5 \, V/m \times 3 \, m = -12 \times 10^5 \, V \]

### Step 3: Calculate the Electrostatic Potential Energy
The electrostatic potential energy \( U \) of the system can be calculated using the formula:
\[ U = k \frac{Q_1 Q_2}{r} + Q_1 V(3, 0) + Q_2 V(0, 0) \]
where:
- \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, N m^2/C^2) \)
- \( r \) is the distance between the charges, which is 3 m.

#### Calculate \( U_{12} \) (Potential Energy due to Interaction of Charges)
\[ U_{12} = k \frac{Q_1 Q_2}{r} = (8.99 \times 10^9) \frac{(-200 \times 10^{-6})(200 \times 10^{-6})}{3} \]
Calculating this gives:
\[ U_{12} = (8.99 \times 10^9) \frac{-40 \times 10^{-12}}{3} = -119.87 \, J \]

#### Calculate \( U_1 \) (Potential Energy due to Electric Field)
Since the potential at the origin is zero:
\[ U_1 = Q_1 \cdot V(0, 0) = -200 \times 10^{-6} \cdot 0 = 0 \]

#### Calculate \( U_2 \) (Potential Energy due to Electric Field)
\[ U_2 = Q_2 \cdot V(3, 0) = 200 \times 10^{-6} \cdot (-12 \times 10^5) = -240 \, J \]

### Step 4: Total Electrostatic Potential Energy
Now, we can sum all contributions to find the total potential energy:
\[ U_{total} = U_{12} + U_1 + U_2 = -119.87 + 0 - 240 = -359.87 \, J \]

### Final Answer
The electrostatic potential energy of the system is approximately:
\[ U_{total} \approx -360 \, J \]