Question

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 $(y<<a)$ along the $y$-axis, the net force acting on the particle is proportional to:

• A. $-y$
• B. $\frac{1}{y^2}$
• C. $-\frac{1}{y}$
• D. $y$

Answer 1

The correct option is D $y$

$F_{net}=2F\cos \theta$$=2\frac{k\cdot q\cdot q/2}{(\sqrt{a^2+y^2}{)}^2}\cdot \frac{y}{\sqrt{a^2+y^2}}$$=\frac{k{q}^{2}y}{{(a^2+y^2)}^{3/2}}$

As $y<<a$ we write
$F_{net}=\frac{k{q}^{2}y}{\left(a^3\right)}$
i.e. $F\propto y$