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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2018 · Paper 2 · Physics · Electrostatics

An infinitely long thin non-conducting wire is parallel to the z -...

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

Options

  1. AThe electric flux through the shell is 3Rλ/ϵ0\sqrt{3}R\lambda / {ϵ}_{0}
  2. BThe z - component of the electric field is zero at all the points on the surface of the shell
  3. CThe electric flux through the shell is 2Rλ/ϵ0\sqrt{2}R\lambda / {ϵ}_{0}
  4. DThe electric field is normal to the surface of the shell at all points

Correct answer

A. The electric flux through the shell is 3Rλ/ϵ0\sqrt{3}R\lambda / {ϵ}_{0}

Step-by-step solution

An infinitely long thin non-conducting wire is parallel to the z - axis and carries a uniform line charge density λ .
Field due to straight wire is perpendicular to the wire and radially outward. Hence Ez=0{E}_{z} = 0
Length, PQ=2Rsin60=3RPQ = 2Rsin 60 = \sqrt{3}R According to Gauss's law
Total flux =∮E→.ds→=qin∈0=λ3R∈0= ∮ \overset{\to}{E} . \overset{\to}{ds} = \dfrac{{q}_{in}}{{\in}_{0}} = \dfrac{\lambda\sqrt{3}R}{{\in}_{0}}
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