Two point charges of $1 × 10^{-6} C$ and $4 × 10^{-6} C$ are placed at a certain distance away from each other. The resultant electric field intensity is zero at a distance of 15 cm from the charge $1 × 10^{-6} C$. The distance between the two point charges will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 45 cm
Let the distance between the two charges be $d$.
Let $E_1$ be the field due to $q_1 = 1 \times 10^{-6} \text{ C}$
Let $E_2$ be the field due to $q_2 = 4 \times 10^{-6} \text{ C}$
Given that net electric field is zero at a point 15 cm from $q_1$:
So, $E_1 = E_2$
$\frac{1}{4\pi\epsilon_0} \cdot \frac{q_1}{(15)^2} = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_2}{(d - 15)^2}$
$\frac{1}{(15)^2} = \frac{4}{(d - 15)^2}$
Taking square roots on both sides:
$\frac{1}{15} = \frac{2}{d - 15}$
$d - 15 = 30$
$d = 45 \text{ cm}$