JEE Main 2024 · 9 April · evening · Single correct · +4 / −1

Five charges +q,+5 q,-2 q,+3 q and -4 q are situated as shown in the figure. The electric flux due to this configuration through the surface S is :

JEE Main PYQ figure
  1. (a) q/ε₀
  2. (b) (3 q)/ε₀
  3. (c) (5 q)/ε₀
  4. (d) (4 q)/ε₀
Answer and solution

Answer (d) (4 q)/ε₀

Solution

Physics sir · 3 steps

Gauss's law only cares about charges inside S, so first sort the five charges into inside and outside.

  1. Gauss's law. The total flux through a closed surface depends only on the net charge enclosed by it. Charges outside do send field lines through S, but every line that enters also leaves, so their net flux is zero.

    Φ = q_(enclosed)/ε₀

    ChargePositionCounts?
    +5qinside Syes
    +qinside Syes
    -2qinside Syes
    +3qoutside Sno
    -4qoutside Sno
  2. Add the inside ones. q_(enclosed) = 5q + q - 2q = 4q.
  3. So. The flux through S is 4q/ε₀, which is option (d).

The idea to keep. For flux through a closed surface, add only the enclosed charges, with their signs. The shape of the surface and the charges outside it do not matter.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Electrostatics I — Charges, Field & Gauss's Law

Asked about 150 times in JEE Main since 2020. This question is from 2024.