Electrostatics Question 748

Question: 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.

Options:

A) $ \pi R^{2}E $

B) $ \sqrt{2}\pi R^{2}E $

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

D) None of these

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Answer:

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 rR^{2}}{2}-\frac{E}{\sqrt{2}}\frac{\pi rR^{2}}{2}=\frac{-\pi R^{2}E}{\sqrt{2}} $ This is the flux entering. So flux is $ \frac{\pi R^{2}E}{\sqrt{2}} $



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