Question

Two charges of 10 $\mu$ C and -90 $\mu$ C are separated by a distance of 24 cm. Electrostatic field strength from the smaller charge is zero at a distance of

• A. 12 cm
• B. 24 cm
• C. 36 cm
• D. 48 cm

Answer 1

Answer: (A)

Let $x$ be the distance where $E$ is zero.
$\Rightarrow E=\frac{k\left(10\times 10^{-6}\right)}{x^2}(-\hat{x})+\frac{k\left(90\times 10^{-6}\right)}{(x+24{)}^2}(+\hat{x})$
$\Rightarrow 0=-\frac{10}{x^2}+\frac{90}{(x+24{)}^2}$
$\Rightarrow \frac{1}{x^2}=\frac{9}{(x+24{)}^2}$
$\Rightarrow \frac{1}{x}=\frac{3}{x+24}$
$\Rightarrow x+24=3x$
$\Rightarrow x=12$