Electric field intensity at a point B due to a point charge Q kept at point A is `24 NC^(-1)`, and electric potential at B due to the same charge is `12 JC^(-1)`. Calculate the distance `AB` and magnitude of charge.

Options

A. `10^(-6)C`
B. `10^(-7)C`
C. `10^(-10)C`
D. `10^(-9)C`

Explanation

To solve the problem, we need to find the distance \( AB \) and the magnitude of the charge \( Q \) using the given electric field intensity \( E \) and electric potential \( V \).

### Step 1: Write down the known values
- Electric field intensity at point B, \( E = 24 \, \text{N/C} \)
- Electric potential at point B, \( V = 12 \, \text{J/C} \)

### Step 2: Use the relationship between electric potential and electric field
The relationship between electric potential \( V \), electric field \( E \), and distance \( r \) is given by:
\[ V = E \cdot r \]
From this, we can rearrange to find the distance \( r \):
\[ r = \frac{V}{E} \]

### Step 3: Substitute the known values to find \( r \)
Substituting the values of \( V \) and \( E \):
\[ r = \frac{12 \, \text{J/C}}{24 \, \text{N/C}} = \frac{12}{24} = 0.5 \, \text{m} \]

### Step 4: Use the formula for electric potential to find the charge \( Q \)
The formula for electric potential due to a point charge \( Q \) at a distance \( r \) is:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r} \]
Rearranging this to solve for \( Q \):
\[ Q = V \cdot \frac{4 \pi \epsilon_0}{r} \]

### Step 5: Substitute the known values to find \( Q \)
Using \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \) and \( r = 0.5 \, \text{m} \):
\[ Q = 12 \cdot \frac{4 \pi (8.85 \times 10^{-12})}{0.5} \]
Calculating \( 4 \pi \epsilon_0 \):
\[ 4 \pi \epsilon_0 \approx 4 \cdot 3.14 \cdot 8.85 \times 10^{-12} \approx 1.11 \times 10^{-10} \, \text{C}^2/\text{N m}^2 \]
Now substituting this value back:
\[ Q = 12 \cdot \frac{1.11 \times 10^{-10}}{0.5} = 12 \cdot 2.22 \times 10^{-10} = 2.664 \times 10^{-9} \, \text{C} \]

### Step 6: Final answer for the charge
Thus, the magnitude of the charge \( Q \) is approximately:
\[ Q \approx 0.667 \times 10^{-9} \, \text{C} \quad \text{or} \quad 6.67 \times 10^{-10} \, \text{C} \]

### Summary of Results
- Distance \( AB = 0.5 \, \text{m} \)
- Magnitude of charge \( Q \approx 0.667 \, \text{nC} \)