JEE Advance - Physics (2018 - Paper 2 Offline - No. 3)

Explanation
Let Gaussian surface be same as the non-conducting spherical shell. The charge enclosed by this Gaussian surface is the charge on the line segment PQ i.e.,
$${q_{enc}} = PQ\lambda = \sqrt 3 R\lambda $$,
where we used law of cosines in triangle OPQ to get $$PQ = \sqrt {{R^2} + {R^2} - 2(R)(R)\cos 120^\circ } = \sqrt 3 R$$. The electric flux through the shell (Gaussian surface) is given by Gauss's law
$$\phi = {q_{enc}}/{ \in _0} = \sqrt 3 R\lambda /{ \in _0}$$.
The electric field at a point due to the infinitely long line charge is radial i.e., perpendicular to the line PQ. The non-conducting spherical shell does not affect or alter the electric field. Thus, electric field at a point lying on the surface of the shell is perpendicular to the z-axis i.e., its z component is zero. The figure shows electric field on the surface of the shell for a positive line charge. Note that the magnitude of electric field is inversely proportional to the distance of the point from the line charge.
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