Question

20.A mass m carrying a charge q is suspended from a string and placed in a uniform horizontal electric field of intensity E.The angle made by the string with the vertical in the equilibrium position is ---------

Answer 1

$$\begin{gathered} \mathrm{Along}\mathrm{vertical}\mathrm{direction}: \\ \mathrm{Vertical}\mathrm{component}\mathrm{of}\mathrm{tension}=\mathrm{Weight} \\ \mathrm{Tcos\theta }=\mathrm{mg} \\ \mathrm{Along}\mathrm{horizontal}\mathrm{direction}: \\ \mathrm{Horizontal}\mathrm{component}\mathrm{of}\mathrm{tension}=\mathrm{Force}\mathrm{due}\mathrm{to}\mathrm{electric}\mathrm{field} \\ \mathrm{Tsin\theta }=\mathrm{Eq} \\ \mathrm{Hence}, \\ \frac{\mathrm{Tsin\theta }}{\mathrm{Tcos\theta }}=\frac{\mathrm{Eq}}{\mathrm{mg}} \\ \mathrm{tan\theta }=\frac{\mathrm{Eq}}{\mathrm{mg}} \\ \mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{Eq}}{\mathrm{mg}}\right) \end{gathered}$$