Electric field intensity at a point B due to a point charge Q kept at 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.

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 formulas for electric field and electric potential.
The electric field \( E \) due to a point charge \( Q \) at a distance \( R \) is given by:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} \]
The electric potential \( V \) due to the same point charge is given by:
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