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Physics · Electrostatics

JEE Advanced 2018 — Paper 2 — Question 3

An infinitely long thin non-conducting wire is parallel to the z -axis and carries a uniform line charge density λ\lambda. It pierces a thin nonconducting spherical shell of radius RR in such a way that the arc PQP Q subtends an angle 120∘120^{\circ} at the centre OO of the spherical shell, as shown in the figure. The permittivity of free space is ε0\varepsilon_{0}. Which of the following statements is (are) true?

Question figure
  1. Option A:

    The electric flux through the shell is 3Rλ/ε0\sqrt{3} R \lambda / \varepsilon_{0}

    Correct
  2. Option B:

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

    Correct
  3. Option C:

    The electric flux through the shell is 2Rλ/ε0\sqrt{2} \mathrm{R} \lambda / \varepsilon_{0}

  4. Option D:

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

Answer: A, B

Step-by-step solution

Charge inside the spherical shell =λ⋅2Rcos⁡30∘=λR3=\lambda \cdot 2 \mathrm{R} \cos 30^{\circ}=\lambda \mathrm{R} \sqrt{3}

So, electric flux =3Rλε0=\frac{\sqrt{3} R \lambda}{\varepsilon_{0}}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics
Topic
Electric flux and Gauss's Law

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