At a certain distance from a point charge, the field intensity is `500 V/m` and the potential is `-3000 V`. The distance to the charge and the magnitude of the charge respectively are

Options

A. `6m` and `6muC`
B. `4m` and `2muC`
C. `6m` and `4muC`
D. `6m` and `2muC`

Explanation

To solve the problem, we need to find the distance to the charge (r) and the magnitude of the charge (q) given the electric field intensity (E) and the electric potential (V) at a certain distance from the charge.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Electric Field Intensity (E) = 500 V/m
- Electric Potential (V) = -3000 V

2. **Use the Formula for Electric Field**:
The electric field (E) due to a point charge (q) at a distance (r) is given by the formula:
\[ E = \frac{k \cdot q}{r^2} \]
where \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

3. **Use the Formula for Electric Potential**:
The electric potential (V) due to a point charge (q) at a distance (r) is given by:
\[ V = \frac{k \cdot q}{r} \]

4. **Set Up the Equations**:
From the electric field equation:
\[ \frac{k \cdot q}{r^2} = 500 \quad \text{(Equation 1)} \]
From the electric potential equation:
\[ \frac{k \cdot q}{r} = -3000 \quad \text{(Equation 2)} \]

5. **Divide Equation 2 by Equation 1**:
\[ \frac{\frac{k \cdot q}{r}}{\frac{k \cdot q}{r^2}} = \frac{-3000}{500} \]
This simplifies to:
\[ \frac{r}{1} = -6 \]
Therefore, we find:
\[ r = 6 \, \text{m} \]

6. **Substitute r back into Equation 2 to find q**:
Substitute \( r = 6 \) m into Equation 2:
\[ \frac{k \cdot q}{6} = -3000 \]
Rearranging gives:
\[ k \cdot q = -3000 \cdot 6 \]
\[ k \cdot q = -18000 \]

7. **Solve for q**:
Substitute \( k = 9 \times 10^9 \):
\[ 9 \times 10^9 \cdot q = -18000 \]
\[ q = \frac{-18000}{9 \times 10^9} \]
\[ q = -2 \times 10^{-6} \, \text{C} \]
The magnitude of the charge is:
\[ |q| = 2 \, \mu\text{C} \]

### Final Answers:
- Distance to the charge (r) = 6 m
- Magnitude of the charge (|q|) = 2 µC