Verified Solution • 2025–26 Session moderate

QJEEAdv15Ph0025: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph0025 Solution: An infinitely long uniform line charge distribution of charge…

An infinitely long uniform line charge distribution of charge per unit length λ\lambda lies parallel to the y-axis in the y-z plane at z=32z = {{\sqrt 3 } \over 2} a (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 λLnε0{{\lambda L} \over {n{\varepsilon _0}}} ( ε0{{\varepsilon _0}} = permittivity of free space), then the value of n is

JEE Advanced 2015 Paper 1 Offline Physics - Electrostatics Question 27 English

Step-by-Step Verified Solution

ANBP is cross-section of a cylinder of length L. The line charge passes through the centre O and perpendicular to paper.

JEE Advanced 2015 Paper 1 Offline Physics - Electrostatics Question 27 English Explanation

AM=a2AM = {a \over 2} , MO=3a2MO = {{\sqrt 3 a} \over 2}

∴\therefore ∠AOM=tan⁡−1(AMOM)\angle AOM = {\tan ^{ - 1}}\left( {{{AM} \over {OM}}} \right)

=tan⁡−1(13)=30∘= {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) = 30^\circ

Electric flux passing from the whole cylinder

ϕ1=qinε0=λLε0{\phi _1} = {{{q_{in}}} \over {{\varepsilon _0}}} = {{\lambda L} \over {{\varepsilon _0}}}

∴\therefore Electric flux passing through ABCD plane surface (shown only AB) = Electric flux passing through cylindrical surface ANB

=(60∘360∘)(ϕ1)=λL6ε0= \left( {{{60^\circ } \over {360^\circ }}} \right)({\phi _1}) = {{\lambda L} \over {6{\varepsilon _0}}}

∴\therefore n = 6

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring