Question 732: Electrostats & Capacitance

Question

Two equal positive point charges are separated by a distance $$2 a$$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $$\mathrm{q}_{0}$$ becomes maximum is $$\frac{a}{\sqrt{x}}$$. The value of $$x$$ is __________.

Accepted Answer

$F_{P}=q_{0} E_{p}=q_{0} \frac{k q z}{\left(a^{2}+z^{2}ight)^{3 / 2}}$
$$
	ext { or } F_{P}=\frac{k q q_{0} z}{\left(a^{2}+z^{2}ight)^{3 / 2}}
$$
To maximize $\frac{d F_{P}}{d z}=0$
or $k q q_{0} \frac{\left(a^{2}+z^{2}ight)^{3 / 2}-z \frac{3}{2} 	imes 2 z\left(a^{2}+z^{2}ight)^{\frac{1}{2}}}{\left(a^{2}+z^{2}ight)^{3}}=0$
$\Rightarrow z=\frac{a}{\sqrt{2}}$

Electrostats & Capacitance

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