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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
${\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
${\left( {\frac{{q{L^2}}}{{2\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
${\left( {\frac{{{q^2}{L^2}}}{{4\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
${\left( {\frac{{{q^2}L}}{{4\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
Solution
(a) In equilibrium $F_e = T sin\theta $ ……. $(i)$
$mg = T cos\theta $ ……. $(ii)$
$\tan \theta = \frac{{{F_e}}}{{mg}} = \frac{{{q^2}}}{{4\pi {\varepsilon _o}{x^2} \times mg}}$ also $\tan \theta \approx \sin \theta = \frac{{x/2}}{L}$
Hence $\frac{x}{{2L}} = \frac{{{q^2}}}{{4\pi {\varepsilon _o}{x^2} \times mg}}$
$==>$ ${x^3} = \frac{{2{q^2}L}}{{4\pi {\varepsilon _o}mg}}$ $==>$ $x = {\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _o}mg}}} \right)^{1/3}}$
