When a `2 mu C` charge is carried from point A to point B, the amount of work done by the electric field is `50 mu J`. What is the potential difference and which point is at a higher potential ?

Options

A. `25 V,B`
B. `25 V,A`
C. `20 V,B`
D. both are at same potential.

Explanation

To solve the problem, we need to find the potential difference between points A and B when a charge is moved from A to B, and determine which point is at a higher potential.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Charge (Q) = 2 µC = \(2 \times 10^{-6}\) C
- Work done (W) = 50 µJ = \(50 \times 10^{-6}\) J

2. **Use the Formula for Potential Difference:**
The potential difference (V) between two points A and B is given by the formula:
\[ V_{AB} = V_A - V_B = \frac{W_{AB}}{Q} \]
where \(W_{AB}\) is the work done in moving the charge from point A to point B.

3. **Substitute the Values into the Formula:**
\[ V_{AB} = \frac{50 \times 10^{-6} \text{ J}}{2 \times 10^{-6} \text{ C}} \]

4. **Calculate the Potential Difference:**
\[ V_{AB} = \frac{50}{2} = 25 \text{ V} \]

5. **Determine Which Point is at Higher Potential:**
Since the work done by the electric field is positive (50 µJ), it indicates that the charge is moving from a point of higher potential (A) to a point of lower potential (B). Therefore, point A is at a higher potential than point B.

### Final Answer:
- The potential difference \(V_{AB}\) is **25 V**.
- Point A is at a higher potential than point B.