Question

A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. The electric flux passing through the curved surface of the hemisphere is
(Angle subtended by the diameter at position of charge is $2\theta$)

• A. $\frac{Q}{2{ϵ}_0}(1-\frac{1}{\sqrt{2}})$
• B. $\frac{-Q}{2{ϵ}_0}(1-\frac{1}{\sqrt{2}})$
• C. $\frac{Q}{{ϵ}_0}(1-\frac{1}{\sqrt{2}})$
• D. $-\frac{Q}{{ϵ}_0}(1-\frac{1}{\sqrt{2}})$

Answer 1

The correct option is B $\frac{-Q}{2{ϵ}_0}(1-\frac{1}{\sqrt{2}})$


The base of hemispherical surface can be assumed to be the base of a cone having semi-vertex angle $\theta$.
Total flux through the hemisphere will be zero, since the charge is outside the hemisphere.
From geometry,

$\tan \theta =\frac{R}{R}=1\Rightarrow \theta ={45}^{\circ }$

Total flux = $0$ = Flux through curved surface + flux hrough plane surface

Now, electric flux passing through the curved surface of the hemisphere will be given as,

$\varphi =-\frac{Q}{2{ϵ}_0}(1-\cos \theta )$

We have a negative sign as the positive normal to the curved surface is outward and opposite to positive normal to the base of cone.

$\varphi =\frac{-Q}{2{ϵ}_0}(1-\cos {45}^{\circ })$

$\varphi =\frac{-Q}{2{ϵ}_0}(1-\frac{1}{\sqrt{2}})$