At a certain distance from a point charge, the field intensity is `500 V//m` and the potentil 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 point.

### Step-by-Step Solution:

1. **Given Values**:
- Electric field intensity, \( E = 500 \, \text{V/m} \)
- Electric potential, \( V = -3000 \, \text{V} \)

2. **Formulas**:
- The electric field due to a point charge is given by:
\[ E = \frac{k \cdot |Q|}{R^2} \]
- The electric potential due to a point charge is given by:
\[ V = \frac{k \cdot Q}{R} \]
where \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

3. **Setting Up Equations**:
- From the electric field equation:
\[ 500 = \frac{9 \times 10^9 \cdot |Q|}{R^2} \quad \text{(Equation 1)} \]
- From the electric potential equation:
\[ -3000 = \frac{9 \times 10^9 \cdot Q}{R} \quad \text{(Equation 2)} \]

4. **Solving for R**:
- Rearranging Equation 2 gives:
\[ Q = -\frac{3000R}{9 \times 10^9} \]
- Substitute \( Q \) from Equation 2 into Equation 1:
\[ 500 = \frac{9 \times 10^9 \cdot \left(-\frac{3000R}{9 \times 10^9}\right)}{R^2} \]
- Simplifying this:
\[ 500 = -\frac{3000}{R} \]
- Rearranging gives:
\[ R = -\frac{3000}{500} = 6 \, \text{m} \]

5. **Finding the Magnitude of Charge (Q)**:
- Now substitute \( R = 6 \, \text{m} \) back into Equation 2:
\[ Q = -\frac{3000 \cdot 6}{9 \times 10^9} = -\frac{18000}{9 \times 10^9} = -2 \times 10^{-6} \, \text{C} \]
- The magnitude of the charge is:
\[ |Q| = 2 \, \mu\text{C} \]

### Final Answers:
- Distance to the charge, \( R = 6 \, \text{m} \)
- Magnitude of the charge, \( |Q| = 2 \, \mu\text{C} \)