JEE Advanced 2018 · Paper 2 · Physics Q3

← JEE Advanced 2018 · Paper 2

PhysicsMultiple correct+4 / -2Question 3 of 54

An infinitely long thin non-conducting wire is parallel to the $z$ -axis and carries a uniform line charge density $\lambda$ . It pierces a thin non-conducting spherical shell of radius $R$ in such a way that the arc $PQ$ subtends an angle $120^\circ$ at the centre $O$ of the spherical shell, as shown in the figure. The permittivity of free space is $\epsilon_0$ . Which of the following statements is (are) true?

Spherical shell of radius R centred at O, with the z-axis (dashed) through O. A vertical line charge λ, parallel to z and to the left of O, pierces the shell at P (top) and Q (bottom); radii OP and OQ (length R) make an angle of 120° at O.

The electric flux through the shell is

\sqrt{3}R\lambda/\epsilon_0

The

z

-component of the electric field is zero at all the points on the surface of the shell

The electric flux through the shell is

\sqrt{2}R\lambda/\epsilon_0

The electric field is normal to the surface of the shell at all points

Answer and worked solution

Official answer: A, B

Charge enclosed. Only the chord $PQ$ of the wire is inside the shell. It subtends $120^\circ$ at the centre, so
$PQ=2R\sin60^\circ=\sqrt3R,\qquad q_{enc}=\sqrt3R\lambda.$

Flux. By Gauss's law, $\Phi=\dfrac{\sqrt3R\lambda}{\epsilon_0}$ . (A) is true and (C) is false.

Direction. An infinite line charge parallel to $z$ produces a field perpendicular to itself, i.e. in the $xy$ -plane. So $E_z=0$ everywhere, including on the shell. (B) is true.

Normal? The field is radial from the wire, not from the centre $O$ , so it is not normal to the sphere in general. (D) is false.

(A), (B)

Topics:Electric flux and Gauss's law Electric field

← Previous question Next question →

JEE Advanced 2018 · Paper 2, Physics question 3