Question

A charge $Q$ is placed at the centre of the open end of a cylindrical vessel of radius $R$ and height $2R$ as shown in figure. The flux of the electric field through the surface (curved surface $+$ base) of the vessel is

• A. $\frac{Q}{{ϵ}_0}$
• B. $\frac{Q}{2E_{\circ }}\left(1+\frac{1}{\sqrt{2}}\right)$
• C. $\frac{Q}{4{ϵ}_0}$
• D. $\frac{Q}{\sqrt{5}{ϵ}_0}$

Answer 1

The correct option is B $\frac{Q}{2E_{\circ }}\left(1+\frac{1}{\sqrt{2}}\right)$

$\begin{array}{c}\cos \theta =\frac{1}{\sqrt{2}}\\ \Omega =2\pi \left(1-\cos \theta \right)\\ =2\pi \left(1-\frac{1}{\sqrt{2}}\right)\\ \therefore {\varphi }^{\prime }=\frac{Q}{4\pi E_{\circ }}.2\pi \left(1-\frac{1}{\sqrt{2}}\right)\\ =\frac{Q}{2E_{\circ }}\left(1-\frac{1}{\sqrt{2}}\right)\\ \therefore totalflux=\frac{Q}{E_{\circ }}-\frac{Q}{2E_{\circ }}\left(1-\frac{1}{\sqrt{2}}\right)\\ =\frac{Q}{2E_{\circ }}+\frac{Q}{2\sqrt{2}{E}_{\circ }}\\ Hence,\\ =\frac{Q}{2E_{\circ }}\left(1+\frac{1}{\sqrt{2}}\right)\\ SooptionBiscorrectanswer.\end{array}$