question_answer

One-fourth of a sphere of radius R is removed as shown in Fig. An electric field E exists parallel to the xy plane. Find the flux through the curved part.

A) \[\pi {{R}^{2}}E\]

B) \[\sqrt{2}\pi {{R}^{2}}E\]

C) \[\pi {{R}^{2}}E/\sqrt{2}\]

D) None of these

Correct Answer: C

Solution :

[c] \[{{\phi }_{plain}}+{{\phi }_{curve}}=0\text{ or }{{\phi }_{plain}}=-{{\phi }_{curve}}\] \[{{\vec{A}}_{1}}=-\frac{\pi {{R}^{2}}}{2}\hat{i},{{\vec{A}}_{2}}=-\frac{\pi {{R}^{2}}}{2}\hat{j}\] \[\vec{E}=E\cos 45{}^\circ \hat{i}+E\sin 45{}^\circ \hat{j}\] \[=\frac{E}{\sqrt{2}}\hat{i}+\frac{E}{\sqrt{2}}\hat{j}\text{ and }\] \[\phi =\vec{E}.({{\vec{A}}_{1}}+{{\vec{A}}_{2}})\] \[=\frac{-E}{\sqrt{2}}\frac{\pi r{{R}^{2}}}{2}-\frac{E}{\sqrt{2}}\frac{\pi r{{R}^{2}}}{2}=\frac{-\pi {{R}^{2}}E}{\sqrt{2}}\] This is the flux entering. So flux is \[\frac{\pi {{R}^{2}}E}{\sqrt{2}}\]