An infinitely long uniform line charge distribution of charge per unit length λ lies parallel to the y -axis in the y - z plane at z = 3 2 α (see figure). If the magnitude of the flux of the electric field through the

Correct answer: 0 dϕ = λ 2π ϵ _ 0 d ^ 2 + x ^ 2 . d d ^ 2 + x ^ 2 Ldx ϕ = 2π ϵ _ 0 ∫ _ - (a)/(2) ^ (a)/(2) dx d ^ 2 + x ^ 2 = λ(d)(L) 2π ϵ _ 0 ((1)/(d))2 [ ^ - 1 (x)/(d)] _ 0 ^ (a)/(2) = π ϵ _ 0 ( ^ - 1 a 2 . a 3 2 ) = π ϵ _ 0 ((π)/(6)) ∴ n = 6

JEE Advanced 2015Paper 1

Exam: JEE Advanced
Year: 2015
Paper: Paper 1
Subject: Physics
Chapter: Electrostatics

JEE Advanced 2015 · Paper 1 · Physics · Electrostatics

An infinitely long uniform line charge distribution of charge per...

An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the $y$ -axis in the $y$ - $z$ plane at $z = \dfrac{\sqrt{3}}{2}\alpha$ (see figure). If the magnitude of the flux of the electric field through the rectangular surface $ABCD$ lying in the $x$ - $y$ plane with its centre at the origin is $\dfrac{\lambdaL}{n{ε}_{0}}$ ( ${ε}_{0} =$ permittivity of free space), then the value of $n$ is

An infinitely long uniform line charge distribution of charge per unit length λ lies parallel to the y -axis in the y -

Correct answer

Step-by-step solution

An infinitely long uniform line charge distribution of charge per unit length λ lies parallel to the y -axis in the y -

dϕ = \dfrac{\lambda}{2\pi{ϵ}_{0}\sqrt{{d}^{2} + {x}^{2}}} . \dfrac{d}{\sqrt{{d}^{2} + {x}^{2}}}Ldx

ϕ = \dfrac{\lambdadL}{2\pi{ϵ}_{0}}\mathop{∫}_{- \dfrac{a}{2}}^{\dfrac{a}{2}}\dfrac{dx}{{d}^{2} + {x}^{2}}

= \dfrac{\lambda\left(d\right)\left(L\right)}{2\pi{ϵ}_{0}}\left(\dfrac{1}{d}\right)2{\left[\tan^{- 1} \dfrac{x}{d}\right]}

_{0}^{\dfrac{a}{2}} = \dfrac{\lambdaL}{\pi{ϵ}_{0}}\left(\tan^{- 1} \dfrac{a}{2 . \dfrac{a\sqrt{3}}{2}}\right) = \dfrac{\lambdaL}{\pi{ϵ}_{0}}\left(\dfrac{\pi}{6}\right)

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∴ n = 6

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