Question

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) :

• A. $q/{\epsilon }_0$
• B. $q/2{\epsilon }_0$
• C. $q/4{\epsilon }_0$
• D. zero

Answer 1

The correct option is C $q/4{\epsilon }_0$

Since the charge is at the centre of the sphere, the electric field is radial.

As the electric field lines do not intersect, the flux through the shaded circle is same as the flux through the spherical cap above the region.

Consider a small surface element on the cap that subtends $d\theta$ and $d\varphi$ as shown in the figure.

The area of the element is $\overrightarrow{dA}=R^2\sin \theta d\theta d\varphi \hat{r}$

Thus, flux is $d\Phi =\overrightarrow{E}.\overrightarrow{dA}=\frac{q}{4\pi {ϵ}_0{R}^2}{R}^2\sin \theta d\theta d\varphi =\frac{q}{4\pi {ϵ}_0}\sin \theta d\theta d\varphi$

In the cap, $\theta$ varies from $0$ to ${\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}$ and $\varphi$ varies from $0$ to $2\pi$

Thus, flux $\Phi =\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi /3}{\int }}\frac{q}{4\pi {ϵ}_0}\sin \theta d\theta d\varphi =\frac{q}{4\pi {ϵ}_0}{\int }_0^{2\pi }d\varphi {\int }_0^{\pi /3}\sin \theta d\theta =\frac{q}{2{ϵ}_0}\frac{1}{2}=\frac{q}{4{ϵ}_0}$