An infinitely long thin non-conducting wire is parallel to the -axis and carries a uniform line charge density . It pierces a thin non-conducting spherical shell of radius in such a way that the arc subtends an angle at the centre of the spherical she

Question

An infinitely long thin non-conducting wire is parallel to the $z$ -axis and carries a uniform line charge density $λ$ . It pierces a thin non-conducting spherical shell of radius $R$ in such a way that the arc $P Q$ subtends an angle $120 °$ at the centre $O$ of the spherical shell, as shown in the figure. The permittivity of free space is $ε_{0}$ . Which of the following statements is (are) true [2018]

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(1, 2)

According to Gauss's law, electric flux

$ϕ = \frac{1}{ε_{0}} q_{in} = \frac{1}{ε_{0}} [λ \times 2 R \sin 60 °]$

$= \frac{\sqrt{3} λ R}{ε_{0}}$

$A B = R \sin 60 °$ or $A C = 2 R \sin 60 °$

Also, the electric field is perpendicular to the wire therefore its z-component is zero.

Solution

1 The electric flux through the shell is $\frac{\sqrt{3} R λ}{ε_{0}}$

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

3 The electric flux through the shell is $\frac{\sqrt{2} R λ}{ε_{0}}$

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

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Solution

1 The electric flux through the shell is 3Rλε0

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

3 The electric flux through the shell is 2Rλε0

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

Ans. (1, 2) According to Gauss's law, electric flux ϕ=1ε0qin=1ε0[λ×2Rsin60°] =3λRε0 AB=Rsin60° or AC=2Rsin60° Also, the electric field is perpendicular to the wire therefore its z-component is zero.

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1 The electric flux through the shell is $\frac{\sqrt{3} R λ}{ε_{0}}$

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

3 The electric flux through the shell is $\frac{\sqrt{2} R λ}{ε_{0}}$