Electrostatics Question 740

Question: Flux passing through the shaded surface of a sphere when point charge q is placed 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) zero

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 ABDE is

$ {\omega _{( ABCDE )}}-{\omega _{( BCD )}}=2\pi -\pi =\pi $ Hence, flux trough ABDE is $ \phi =\frac{q}{{\varepsilon _{0}}}\frac{\pi }{4\pi }=\frac{q}{4{\varepsilon _{0}}} $



NCERT Chapter Video Solution

Dual Pane