Electrostatics 3 Question 14
14. A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct? (2017 Adv.)
(a) The electric flux passing through the curved surface of the hemisphere is $-\frac{Q}{2 \varepsilon_{0}} 1-\frac{1}{\sqrt{2}}$.
(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 $\frac{Q}{\varepsilon_{0}}$
(d) The circumference of the flat surface is an equipotential
Show Answer
Answer:
Correct Answer: 14. (a,d)
Solution:
- (a) $\Omega=2 \pi(1-\cos \theta) ; \theta=45^{\circ}$
$ \begin{aligned} \varphi & =-\frac{\Omega}{4 \pi} \times \frac{Q}{\varepsilon_{0}}=-\frac{2 \pi(1-\cos \theta)}{4 \pi} \frac{Q}{\varepsilon_{0}} \\ & =-\frac{Q}{2 \varepsilon_{0}} 1-\frac{1}{\sqrt{2}} \end{aligned} $
(b) The component of the electric field perpendicular to the flat surface will decrease so we move away from the centre as the distance increases (magnitude of electric field decreases) as well as the angle between the normal and electric field will increase. Hence, the component of the electric field normal to the flat surface is not constant.
Alternate solution
$ \begin{aligned} x & =\frac{R}{\cos \theta} \\ E & =\frac{K Q}{x^{2}}=\frac{K Q \cos ^{2} \theta}{R^{2}} \Rightarrow E_{\perp}=\frac{K Q \cos ^{3} \theta}{R^{2}} \end{aligned} $
As we move away from centre $\theta \uparrow \cos \theta \downarrow$ so $E_{\perp} \downarrow$
(c) Total flux $\varphi$ due to charge $Q$ is $\frac{Q}{\varepsilon_{0}}$.
So, $\varphi$ through the curved and flat surface will be less than $\frac{Q}{\varepsilon_{0}}$.
(d) Since, the circumference is equidistant from $Q$ it will be equipotential $V=\frac{K Q}{\sqrt{2} R}$.
