Electrostatics Ques 29
- Two charges, each equal to $q$, are kept at $x=-a$ and $x=a$ on the $x$-axis. A particle of mass $m$ and charge $q_{0}=\frac{q}{2}$ is placed at the origin. If charge $q_{0}$ is given a small displacement $y(y<a)$ along the $y$-axis, the net force acting on the particle is proportional to
(a) $y$
(b) $-y$
(c) $\frac{1}{y}$
(d) $-\frac{1}{y}$
(2013 Main)
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Answer:
Correct Answer: 29.(a)
Solution:
Formula:
Coulomb Force Between Two Point Charges:
$$ \begin{aligned} F_{\mathrm{net}} & =2 F \cos \theta \\ F_{\mathrm{net}} & =\frac{2 k q \frac{q}{2}}{\left(\sqrt{y^{2}+a^{2}}\right)^{2}} \cdot \frac{y}{\sqrt{y^{2}+a^{2}}} \\ F_{\text {net }} & =\frac{2 k \sin \theta}{\left(y^{2}+a^{2}\right)^{3 / 2}} \Rightarrow \frac{k q^{2} y}{a^{3}} \propto y \end{aligned} $$
BETA
