Exams › JEE Main › Physics

At a distance R from a point charge q the electric field is E. If the same charge q is now placed on a copper sphere of radius R, what is the field magnitude at the surface of the sphere?

1. E/4
2. E/2
3. E
4. 2E

Correct answer: E

Solution

For a uniformly charged conducting sphere, the field at and outside the surface equals that of a point charge q at the centre. At the surface (r = R), E_surface = kq/R², which is exactly the field E of the point charge at distance R. So the field is unchanged, E.

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 →