Exams › JEE Main › Physics

For a given charge configuration enclosed (partly) by a spherical Gaussian surface, when the FLUX of the electric field through that spherical surface is being calculated, the electric field appearing in the flux integral is due to:

1. q2 (the enclosed charge only)
2. Only the positive charges
3. +q1 and -q1 only
4. All the charges

Correct answer: All the charges

Solution

In Gauss's law, the net flux equals q_enclosed/epsilon0 and depends only on the charge inside. However, the electric field vector E at any point on the Gaussian surface is the total field due to ALL charges - both inside and outside the surface. Contributions of external charges cancel out in the net flux integral, but they still contribute to E itself. Hence the field in the flux integrand is due to all the charges.

Related JEE Main Physics questions

• What is the SI unit of the permittivity of free space, ε₀?
• What is the SI unit used for electric flux?
• A constant electric field E is directed along the positive x-axis. If a charge of 0.5 C is displaced by 2 m along a line inclined at 60° to the x-axis, and the work done in this process is 10 J, what is the magnitude of the electric field?
• An electric dipole is located at the origin O along the x-axis. A point P lies 20 cm from O, and the line OP makes an angle π/3 with the x-axis. If the electric field at P is inclined at an angle θ to the x-axis, what is θ?
• A uniformly charged semicircular wire has radius a and linear charge density λ. The magnitude of the electric field at the centre of the semicircle is
• A point Q is located on the perpendicular bisector of an electric dipole whose dipole moment is p. If Q is at a distance r from the dipole and r is much greater than the dipole size, the electric field at Q varies as

⚔️ Practice JEE Main Physics free + battle 1v1 →