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(B) Let the charge $q_0$ be displaced by a small distance $y$ along the $y$-axis.The distance between each charge $q$ and the charge $q_0$ is $r = \sq

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$-y$

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(B) Let the charge $q_0$ be displaced by a small distance $y$ along the $y$-axis.The distance between each charge $q$ and the charge $q_0$ is $r = \sqrt{y^2 + a^2}$.The magnitude of the electrostatic force $F$ exerted by each charge $q$ on $q_0$ is given by Coulomb's law:$F = \frac{k q q_0}{r^2} = \frac{k q (q/2)}{y^2 + a^2} = \frac{k q^2}{2(y^2 + a^2)}$.The horizontal components of the forces cancel each other out,while the vertical components add up.The net force $F_{net}$ acting on $q_0$ is directed towards the origin (restoring force):$F_{net} = -2 F \cos 	heta$,where $\cos 	heta = \frac{y}{r} = \frac{y}{\sqrt{y^2 + a^2}}$.Substituting the values:$F_{net} = -2 \left[ \frac{k q^2}{2(y^2 + a^2)} ight] \cdot \frac{y}{\sqrt{y^2 + a^2}} = -\frac{k q^2 y}{(y^2 + a^2)^{3/2}}$.Since the displacement $y$ is very small $(y Thus,$F_{net} \approx -\frac{k q^2}{a^3} y$.Since $k, q,$ and $a$ are constants,$F_{net} \propto -y$.

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

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