JEE Advanced 2017 · Paper 2 · Physics Q10

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PhysicsMultiple correct+4 / -2Question 10 of 54

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?

An imaginary hemisphere of radius R with its flat circular face at the bottom (dashed ellipse); the charge +Q sits just outside the top of the curved surface, on the axis.

The electric flux passing through the curved surface of the hemisphere is

-\frac{Q}{2\varepsilon_0}\left(1 - \frac{1}{\sqrt{2}}\right)

Total flux through the curved and the flat surfaces is

\frac{Q}{\varepsilon_0}

The component of the electric field normal to the flat surface is constant over the surface

The circumference of the flat surface is an equipotential

Answer and worked solution

Official answer: A, D

Total flux. The charge is outside the closed hemisphere, so the total flux is 0. (B) is false.

Flat face. It subtends a cone of half-angle $45^\circ$ at the charge (radius $R$ at distance $R$ ):
$\Omega=2\pi(1-\cos45^\circ),\qquad\Phi_{flat}=\frac Q{4\pi\epsilon_0}\Omega=\frac Q{2\epsilon_0}\left(1-\frac1{\sqrt2}\right)\ \text{(outward)}.$

Curved face. It carries the equal inward flux, $-\dfrac Q{2\epsilon_0}\left(1-\dfrac1{\sqrt2}\right)$ . (A) is true.

(C) The normal field on the flat face varies with distance from the centre. False.
(D) Every point of the rim is $\sqrt2R$ from the charge, so the rim is an equipotential. True.

(A), (D)

Topics:Electric flux and Gauss's law Electric potential

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JEE Advanced 2017 · Paper 2, Physics question 10