\(\tan \theta  = \frac{{{F_e}}}{{mg}} = \frac{{{q^2}}}{{4\pi {\varepsilon _o}{x^2} \times mg}}\) તથા \(\tan \theta  \approx \sin \theta  = \frac{{x/2}}{L}\)

માટે \(\frac{x}{{2L}} = \frac{{{q^2}}}{{4\pi {\varepsilon _o}{x^2} \times mg}}\,\, \Rightarrow \,\,{x^3} = \frac{{2{q^2}L}}{{4\pi {\varepsilon _o}mg}}\,\,\, \Rightarrow \,\,x = {\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _o}mg}}} \right)^{1/3}}\)