Electrostatics Question 740
Flux passing through the shaded surface of a sphere when a point charge q is placed at the center is (radius of the sphere is R)
Options:
A) $ q/{\varepsilon _{0}} $
B) $ q/2{\varepsilon _{0}} $
C) $ q/4{\varepsilon _{0}} $
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \alpha =60{}^\circ ,$
Solid angle subtended by BCD is $ \omega =2\pi ( 1 - \cos \alpha )=\pi $
Solid angle subtended by ABDE is
$ {\omega _{( ABCDE )}}-{\omega _{( BCD )}}=2\pi -\pi =\pi $ Hence, flux through ABDE is $ \phi =\frac{q}{{\varepsilon _{0}}}\frac{\pi }{4\pi }=\frac{q}{4{\varepsilon _{0}}} $
BETA
