Question

Two charges each of charge $+10\, \mu \,C$ are kept on $Y$ - axis at $y = -a$ and $y = +a$, respectively. Another point charge $-2\,\mu\, C$ is placed at the origin and given a small displacement $x(x < < a)$ along $X$ - axis. The force acting on the point charge is
$\left(x\right.$ and $a$ are in metres, $\frac{1}{4 \pi \varepsilon_{0}}\left.=9 \times 10^{9} \,N - m ^{2}\, C ^{-2}\right)$

• A. $\frac{3.6 x}{a^{2}} N$
• B. $\frac{2.4 x^{2}}{a} N$
• C. $\frac{3.6 x}{a^{3}} N$
• D. $\frac{4.8 x}{a^{2}} N$

Answer 1

Answer: (C)

Given,
$ q_{1}=+10 \mu C \text { at } y=\pm a $
$q_{2}=+10 \mu C$
$+10 \mu C $

Let the force acting on each charge $+10 \mu C$ be $F, x$ be
the displaced position the component of each force will $F \cos \theta$ in $x$ -direction.
So, the net force $F_{\text {net }}=F \cos \theta+F \cos \theta$
$=2 F \cdot \cos \theta$
From Coulomb's law,
$F =k \cdot \frac{q_{1} q_{2}}{r^{2}}$
$k =\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} / C ^{2} $
$F =9 \times 10^{9} \times \frac{10 \times 10^{-6} \times 20 \times 10^{-6}}{\left(\sqrt{a^{2}+x^{2}}\right)^{2}}$
$=9 \times 10^{9} \times \frac{200 \times 10^{-12}}{\left(a^{2}+x^{2}\right)}$
Net force,
$F_{\text {net }}=2 \times 9 \times 10^{9} \times \frac{200 \times 10^{-12}}{\left(a^{2}+x^{2}\right)} \cdot \cos \theta$
$\because \cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{x}{\sqrt{a^{2}+x^{2}}}$
$\therefore F_{\text {net }}=2 \times 9 \times \frac{2 \times 10^{-1}}{\left(a^{2}+x^{2}\right)}\left(\frac{x}{\sqrt{a^{2}+x^{2}}}\right)$
$=36 \times 10^{-1} \frac{x}{\left(a^{2}+x^{2}\right)^{3 / 2}}$
Given, $x << a$
Neglecting $x^{2}$
$ F_{\text {net }}=3.6 \frac{x}{\left(a^{2}\right)^{3 / 2}} \approx \frac{3.6 x}{a^{3}} N$