In the given figure two tiny conducting balls | vedclass

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Question

(a) In equilibrium $F_e = T sin	heta $ ....... $(i)$
$mg = T cos	heta $ ....... $(ii)$
$	an 	heta = \frac{{{F_e}}}{{mg}} = \frac{{{q^2}}}{{4\pi

Vedclass · Accepted Answer

${\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$

Vedclass · Answer

(a) In equilibrium $F_e = T sin	heta $ ....... $(i)$
$mg = T cos	heta $ ....... $(ii)$
$	an 	heta = \frac{{{F_e}}}{{mg}} = \frac{{{q^2}}}{{4\pi {\varepsilon _o}{x^2} 	imes mg}}$ also $	an 	heta \approx \sin 	heta = \frac{{x/2}}{L}$
Hence $\frac{x}{{2L}} = \frac{{{q^2}}}{{4\pi {\varepsilon _o}{x^2} 	imes mg}}$
$==>$ ${x^3} = \frac{{2{q^2}L}}{{4\pi {\varepsilon _o}mg}}$ $==>$ $x = {\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _o}mg}}} ight)^{1/3}}$

In the given figure two tiny conducting balls of identical mass $m$ and identical charge $q$ hang from non-conducting threads of equal length $L$. Assume that $\theta$ is so small that $\tan \theta \approx \sin \theta $, then for equilibrium $x$ is equal to

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