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

Explanation

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. The formula we will use is:

\[ W = Q \cdot (V_B - V_A) \]

Where:
- \( W \) is the work done,
- \( Q \) is the charge,
- \( V_B - V_A \) is the potential difference.

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

### Step 2: Rearrange the formula to find the potential difference
We can rearrange the formula to solve for \( V_B - V_A \):

\[ V_B - V_A = \frac{W}{Q} \]

### Step 3: Substitute the known values into the equation
Now we can substitute the values of \( W \) and \( Q \):

\[ V_B - V_A = \frac{10 \times 10^{-3} \, \text{J}}{5 \times 10^{-6} \, \text{C}} \]

### Step 4: Calculate the potential difference
Now, perform the calculation:

\[ V_B - V_A = \frac{10 \times 10^{-3}}{5 \times 10^{-6}} = \frac{10}{5} \times \frac{10^{-3}}{10^{-6}} = 2 \times 10^{3} = 2000 \, \text{V} \]

### Final Answer
Thus, the potential difference \( V_B - V_A \) is \( 2000 \, \text{V} \).

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