Two balls of same mass and carrying equal cha | vedclass

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Question

$	ext { In equilibrium, } \mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin 	heta$

$\mathrm{mg}=\mathrm{T} \cos 	heta$

$	an 	heta  = \frac{{{F_e

Vedclass · Accepted Answer

${l^{1/3}}$

Vedclass · Answer

$	ext { In equilibrium, } \mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin 	heta$

$\mathrm{mg}=\mathrm{T} \cos 	heta$

$	an 	heta  = \frac{{{F_e}}}{{mg}} = \frac{{{q^2}}}{{4\pi {_0}\,{x^2}}} 	imes mg$

$	ext { also } 	an 	heta \approx \sin =\frac{x / 2}{\ell}$

Hence, $\frac{x}{2 \ell}=\frac{q^{2}}{4 \pi \epsilon_{0} x^{2} 	imes m g}$

$\Rightarrow x^{3}=\frac{2 q^{2} \ell}{4 \pi \epsilon_{0} m g}$

$	herefore x=\left(\frac{q^{2} \ell}{2 \pi \epsilon_{0} m g}ight)^{1 / 3}$

Therefore $x \propto \ell^{1 / 3}$

Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, $x$ between the balls is proportional to

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