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. 6 m
B. 12 m
C. 36 m
D. 144 m

Answer

To solve the problem, we will use the relationships between electric field intensity (E), electric potential (V), distance (d), and the charge (q).

### Step-by-Step Solution:

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

2. **Use the relationship between electric potential and electric field:**
The electric field intensity \( E \) is related to the potential \( V \) and distance \( d \) by the formula:
\[ E = \frac{V}{d} \]
Rearranging this gives us:
\[ d = \frac{V}{E} \]

3. **Substitute the known values to find distance \( d \):**
\[ d = \frac{-3000 \, \text{V}}{500 \, \text{V/m}} = -6 \, \text{m} \]
Since distance cannot be negative, we take the absolute value:
\[ d = 6 \, \text{m} \]

4. **Use the formula for electric potential to find the charge \( q \):**
The electric potential \( V \) due to a point charge \( q \) at a distance \( d \) is given by:
\[ V = \frac{kq}{d} \]
Where \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

5. **Rearranging the formula to solve for \( q \):**
\[ q = \frac{V \cdot d}{k} \]

6. **Substituting the known values to find charge \( q \):**
\[ q = \frac{-3000 \, \text{V} \cdot 6 \, \text{m}}{9 \times 10^9 \, \text{N m}^2/\text{C}^2} \]
\[ q = \frac{-18000}{9 \times 10^9} \]
\[ q = -2 \times 10^{-6} \, \text{C} \]
The magnitude of the charge is:
\[ |q| = 2 \, \mu\text{C} \]

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