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

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

Answer 1

Answer: (A)

$y$

Answer 2

\hspace15mm F_{net} =2F cos\, \theta
\hspace15mm F_{net} =\frac{2kq\bigg(\frac{q}{2}\bigg)}{(\sqrt{y^2+a^2})^2}.\frac{y}{\sqrt{y^2+a^2}}
\hspace15mm F_{net} =\frac{2kq\bigg(\frac{q}{2}\bigg)y}{(\sqrt{y^2+a^2})^{3/2}}
\Rightarrow\hspace10mm \frac{kq^2y}{a^3} \propto y