A line charge of length ' a ' 2 is kept at the center of an edge B C of a cube A B C D E F G H having edge length ' a ' as shown in the figure. If the density of line charge is λ C per unit length, then the total

Correct answer: D. λ a 8 _0 aligned & Charge of the line charge =(a λ)/(2) & Portion of wire inside cube =(1)/(4) & q_ e n =(1)/(4)((a λ)/(2))=(a λ)/(8) & φ= q_ e n _0 =(a λ)/(8 _0) aligned

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Physics
Chapter: Electrostatics

JEE Main 2025 · 22 January · Morning Shift · Physics · Electrostatics

A line charge of length ' a ' 2 is kept at the center of an edge B...

A line charge of length

\frac{\text{ ' } a \text{ ' }}{2}

is kept at the center of an edge

B C

of a cube A B C D E F G H

having edge length '

' as shown in the figure. If the density of line charge is

\lambda

C per unit length, then the total electric flux through all the faces of the cube will be _______. (Take,

\epsilon_0

as the free space permittivity)

Options

A $\frac{\lambda \mathrm{a}}{2 \epsilon_0}$
B $\frac{\lambda \mathrm{a}}{4 \epsilon_0}$
C $\frac{\lambda \mathrm{a}}{16 \epsilon_0}$
D $\frac{\lambda \mathrm{a}}{8 \epsilon_0}$

Correct answer

D. $\frac{\lambda \mathrm{a}}{8 \epsilon_0}$

Step-by-step solution

\begin{aligned} & \text{ Charge\text{ of } the line charge }=\frac{a \lambda}{2} \\ & \text{ Portion of wire inside cube }=\frac{1}{4} \\ & \therefore \quad q_{e n}=\frac{1}{4}\left(\frac{a \lambda}{2}\right)=\frac{a \lambda}{8} \\ & \qquad \phi=\frac{q_{e n}}{\varepsilon_0}=\frac{a \lambda}{8 \varepsilon_0}\end{aligned}

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