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