Question

A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in figure. Which of the following statements is/are correct?

(A) The electric flux passing through the curved surface of the hemisphere is $-{\displaystyle \frac{Q}{2{\epsilon }_0}}(1-{\displaystyle \frac{1}{\sqrt{2}}})$
(B) The component of the electric field normal to the flat surface is constant over the surface
(C) Total flux through the curved and the flat surfaces is ${\displaystyle \frac{Q}{{\epsilon }_0}}$
(D) The circumference of the flat surface is an equipotential

Answer 1

Hint:To find electric flux we use Gauss’ Law. Electric flux is the measure of the electric field through a given area. The electric flux ${\varphi }_E$ through any closed surface is equal to ${\displaystyle \frac{1}{{\epsilon }_0}}$ times the net charge $Q$ closed by the surface.

Complete step by step solution:
Given: radius of the hemisphere is $R$
The solid angle subtended by the flat surface at $\mathrm{\Omega }=2\pi (1-\cos \theta )$ $...(\text{i})$

From the above figure, we can say that $\theta ={45}^{\circ }$
Flux passing through curved surface is given by,
$\varphi =-{\displaystyle \frac{Q}{{\epsilon }_0}}\times {\displaystyle \frac{\mathrm{\Omega }}{4\pi }}$ $...(\text{ii})$
Put the value of equation $(\text{i})$ in equation $(\text{ii})$, we get
$\varphi =-{\displaystyle \frac{Q}{{\epsilon }_0}}\times {\displaystyle \frac{2\pi (1-{\displaystyle \frac{1}{\sqrt{2}}})}{4\pi }}$
$\varphi =-{\displaystyle \frac{Q}{2{\epsilon }_0}}(1-{\displaystyle \frac{1}{\sqrt{2}}})$
Hence, the option (A) is correct.
The component of electric field normal to the flat surface is $E\cos \theta$.

Here $E$ and $\theta$ changes for different points on the flat surface. Therefore option (B) is incorrect.
The total flux due to charge $Q$ is $\varphi$ and $\varphi$ through the curved and flat surface will be less than ${\displaystyle \frac{Q}{{\epsilon }_0}}$. Hence option (C) is incorrect.
Potential at any point on the circumference of the flat surface is given by,
$V={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{\sqrt{R^2+R^2}}}$
$V={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{\sqrt{2}R}}$
Since, the circumference is equidistant from charge $Q$ it will be an equipotential surface.

Therefore option (D) is correct.

Note:An equipotential surface is that at every point electrical potential is the same. The electric field $\overrightarrow{E}$ is directed perpendicular to the equipotential surface. Work done in moving charge over an equipotential surface is zero. Two equipotential surfaces can never intersect.