ExamHabit
Practise free

JEE Main 2023 · 29 Jan · Shift 1

JEE Main 2023 Physics question on Electrostatics

In a cuboid of dimension 2L×2L×L2L \times 2L \times L, a charge qq is placed at the center of the surface SS having area of 4L24L^2. The flux through the opposite surface to SS is given by
  1. A.
    q6ε0\dfrac{q}{6\varepsilon_0}
  2. B.
    q12ε0\dfrac{q}{12\varepsilon_0}
  3. C.
    q3ε0\dfrac{q}{3\varepsilon_0}
  4. D.
    q2ε0\dfrac{q}{2\varepsilon_0}
Show answer and solution

Correct option: A

Solution

Place the charge at the center of face S (area 4L²). Use the image trick: imagine a mirror cuboid on the other side of S so the charge sits at the center of a 2L×2L×2L cube. By symmetry, the total flux through all 6 faces of this larger cube is q/ε₀, and each face gets q/(6ε₀). The face opposite to S in the original cuboid corresponds exactly to one face of this symmetric cube, so the flux through it is q/(6ε₀).

[JEE Main 2023 · 29 Jan · Shift 1]

Practise Electrostatics the way it is asked

A timed set of real JEE Main Electrostatics questions, graded as the exam grades. No sign-up.

Start a Electrostatics test

More: all JEE Main Electrostatics questions · JEE Main 2023 · Physics chapter list