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
Show Answer
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}} $
BETA
