From a point charge, there is a fixed point A. At a, there is an electric field of 500 V/m and potential difference of 3000 V. Distance between point charge and A will be

Options

A. 6m
B. 12m
C. 16 m
D. 24m

Explanation

To find the distance between the point charge and point A, we can use the relationship between electric field (E), potential difference (V), and distance (D). The formula we will use is:

\[ E = \frac{V}{D} \]

Where:
- \( E \) is the electric field (in V/m),
- \( V \) is the potential difference (in V),
- \( D \) is the distance (in m).

### Step-by-Step Solution:

1. **Identify the given values:**
- Electric field, \( E = 500 \, \text{V/m} \)
- Potential difference, \( V = 3000 \, \text{V} \)

2. **Rearrange the formula to solve for distance (D):**
\[ D = \frac{V}{E} \]

3. **Substitute the known values into the equation:**
\[ D = \frac{3000 \, \text{V}}{500 \, \text{V/m}} \]

4. **Calculate the distance:**
\[ D = \frac{3000}{500} = 6 \, \text{m} \]

5. **Conclusion:**
The distance between the point charge and point A is \( 6 \, \text{m} \).