The potential at a point A is – 500 V and that at another point B is + 500 V. What is the work done by external agent to take 2 units (S.I.) of negative charge from B to A.

Explanation

To find the work done by an external agent to move a negative charge from point B to point A, we can follow these steps:

### Step 1: Understand the given information
We have:
- Potential at point A, \( V_A = -500 \, \text{V} \)
- Potential at point B, \( V_B = +500 \, \text{V} \)
- Charge to be moved, \( Q = -2 \, \text{C} \) (negative charge)

### Step 2: Calculate the potential difference (\( \Delta V \))
The potential difference between points A and B is given by:
\[ \Delta V = V_A - V_B \]
Substituting the values:
\[ \Delta V = -500 \, \text{V} - 500 \, \text{V} = -1000 \, \text{V} \]

### Step 3: Use the work done formula
The work done \( W \) by an external agent to move a charge in an electric field is given by:
\[ W = Q \cdot \Delta V \]
Substituting the values of \( Q \) and \( \Delta V \):
\[ W = (-2 \, \text{C}) \cdot (-1000 \, \text{V}) = 2000 \, \text{J} \]

### Step 4: Conclusion
The work done by the external agent to move the charge from point B to point A is:
\[ W = 2000 \, \text{J} \]