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
$(x$ and $a$ are in metres, $\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N-m^2{C}^{-2})$

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

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 $F_{\text{net }}=F\cos \theta +F\cos \theta$
$=2F\cdot \cos \theta$
From Coulomb's law,
$F=k\cdot \frac{q_1{q}_2}{r^2}$
$k=\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2/C^2$
$F=9\times 10^9\times \frac{10\times 10^{-6}\times 20\times 10^{-6}}{{\left(\sqrt{a^2+x^2}\right)}^2}$
$=9\times 10^9\times \frac{200\times 10^{-12}}{\left(a^2+x^2\right)}$
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 $x^2$
$F_{\text{net }}=3.6\frac{x}{{\left(a^2\right)}^{3/2}}\approx \frac{3.6x}{a^3}N$