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
${\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
${\left( {\frac{{q{L^2}}}{{2\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
${\left( {\frac{{{q^2}{L^2}}}{{4\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
${\left( {\frac{{{q^2}L}}{{4\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$
Two positive point charges of unequal magnitude are placed at a certain distance apart. A small positive test charge is placed at null point, then
Three charges $+Q, q, +Q$ are placed respectively, at distance, $0, \frac d2$ and $d$ from the origin, on the $x-$ axis. If the net force experienced by $+Q$, placed at $x = 0$, is zero, then value of $q$ is
Two free point charges $+q$ and $+4q$ are a distance $R$ apart. $A$ third charge is so placed that the entire system is in equilibrium. Then the third charge is :-
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 $q_0$ becomes maximum is $\frac{a}{\sqrt{x}}$. The value of $x$ is $................$
Write general equation of Coulombian force on ${q_1}$ by system of charges ${q_1},{q_2},.......,{q_n}$.