The work done in carrying a charge of `5 mu C` form a point `A` to a point `B` in an electric field is `10mJ`. The potential difference `(V_(B) - V_(A))` is then

Options

A. `+2kV`
B. `-2kV`
C. `+200 kV`
D. `-200 kV`

Answer

To find the potential difference \( V_B - V_A \) when carrying a charge from point A to point B in an electric field, we can use the relationship between work done, charge, and potential difference.

### Step-by-Step Solution:

1. **Identify the given values:**
- Charge \( Q = 5 \, \mu C = 5 \times 10^{-6} \, C \)
- Work done \( W = 10 \, mJ = 10 \times 10^{-3} \, J \)

2. **Use the formula for work done in terms of potential difference:**
The work done \( W \) in moving a charge \( Q \) through a potential difference \( V_B - V_A \) is given by:
\[ W = Q \cdot (V_B - V_A) \]

3. **Rearrange the formula to find the potential difference:**
\[ V_B - V_A = \frac{W}{Q} \]

4. **Substitute the known values into the equation:**
\[ V_B - V_A = \frac{10 \times 10^{-3} \, J}{5 \times 10^{-6} \, C} \]

5. **Calculate the potential difference:**
\[ V_B - V_A = \frac{10 \times 10^{-3}}{5 \times 10^{-6}} = \frac{10}{5} \times 10^{3} = 2 \times 10^{3} \, V \]
\[ V_B - V_A = 2000 \, V = 2 \, kV \]

### Final Answer:
The potential difference \( V_B - V_A \) is \( 2 \, kV \).

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