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Question

$\Rightarrow \mathrm{F}_{\mathrm{net}}=2 \mathrm{F} \cos 	heta$

$F_{n e t}=\frac{2 k q\left(\frac{q}{2}ight)}{(\sqrt{y^{2}+a^{2}})^{2}} \cdot \f

Vedclass · Accepted Answer

$y$

Vedclass · Answer

$\Rightarrow \mathrm{F}_{\mathrm{net}}=2 \mathrm{F} \cos 	heta$

$F_{n e t}=\frac{2 k q\left(\frac{q}{2}ight)}{(\sqrt{y^{2}+a^{2}})^{2}} \cdot \frac{y}{\sqrt{y^{2}+a^{2}}}$

$F_{n e t}=\frac{2 k q\left(\frac{q}{2}ight) y}{\left(y^{2}+a^{2}ight)^{3 / 2}} \Rightarrow \frac{k q^{2} y}{a^{3}}$

So, $F \propto y$

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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