A charge `(-q)` and another charge `(Q)` are kept at two points `A and B` respectively. Keeping the charge `(+Q)` fixed at `B`, the charge `(-q)` at `A` is moved to another point `C` such that `ABC` forms an equilateral triangle of side `l`. The net work done in moving teh charge `(-q)` is

Options

A. `(1)/(4pi epsilon_(0))(Qq)/(l)`
B. `(1)/(4pi epsilon_(0))(Qq)/(l^(2))`
C. `(1)/(4pi epsilon_(0))Qql`
D. Zero

Answer

To solve the problem step by step, we will analyze the situation involving the charges and calculate the work done in moving the charge from point A to point C.

### Step 1: Understand the Configuration
We have two charges:
- Charge `-q` at point A
- Charge `+Q` at point B

The charge `+Q` is fixed at point B, and we are moving the charge `-q` from point A to point C, where ABC forms an equilateral triangle with each side of length `l`.

### Step 2: Calculate the Electric Potential at Points A and C
The electric potential \( V \) due to a point charge is given by the formula:
\[ V = \frac{1}{4\pi \epsilon_0} \frac{Q}{r} \]
where \( r \) is the distance from the charge to the point where the potential is being calculated.

1. **Potential at Point A (VA)**:
- The distance from charge \( +Q \) at point B to point A is \( l \).
- Therefore, the potential at point A is:
\[ V_A = \frac{1}{4\pi \epsilon_0} \frac{Q}{l} \]

2. **Potential at Point C (VC)**:
- The distance from charge \( +Q \) at point B to point C is also \( l \) (since ABC is an equilateral triangle).
- Therefore, the potential at point C is:
\[ V_C = \frac{1}{4\pi \epsilon_0} \frac{Q}{l} \]

### Step 3: Calculate the Change in Electric Potential (ΔV)
The change in electric potential \( \Delta V \) when moving from point A to point C is given by:
\[ \Delta V = V_C - V_A \]

Substituting the values we calculated:
\[ \Delta V = \frac{1}{4\pi \epsilon_0} \frac{Q}{l} - \frac{1}{4\pi \epsilon_0} \frac{Q}{l} = 0 \]

### Step 4: Calculate the Work Done (W)
The work done \( W \) in moving a charge in an electric field is given by:
\[ W = -q \Delta V \]

Substituting \( \Delta V = 0 \):
\[ W = -q \cdot 0 = 0 \]

### Final Answer
The net work done in moving the charge \( -q \) from point A to point C is:
\[ \boxed{0} \]