Exams › JEE Main › Physics

A solid sphere of radius R carries charge q spread throughout its volume with volume charge density given by ρ = k r^a, where k and a are constants and r is measured from the centre. If the electric field at the point r = R/2 is one-eighth of the field at the surface r = R, determine the value of a.

1. 2
2. 3
3. 5
4. 6

Correct answer: 2

Solution

Q_enc = integral of k r^a *4 pi r^2 dr gives E(r) = k r^(a+1)/(eps0 (a+3)), so E ~ r^(a+1). E(R/2)/E(R) = (1/2)^(a+1) = 1/8 = (1/2)^3, so a+1=3, a=2.

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 →