Question

An infinitely long solid cylinder of radius R has a uniform volume charge density r. It has a spherical cavity of radius R/2 with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is at a distance 2R from the axis of the cylinder, is given by the expression $\frac{23\rho R}{16K{\epsilon }_0}$ . The value of k is

• A. 6
• B. 5
• C. 7
• D. 4

Answer 1

Answer: (A)

We suppose that the cavity is filled up by a positive as well as negative volume charge of r. So the electric field now produced at P is the superposition of two electric fields.
(i) The electric field created due to the infinitely long solid cylinder is $E_1=\frac{\rho R}{4{\epsilon }_0}$ directed towards the +Y direction
(ii) The electric field created due to the spherical negative charge density $E_1=\frac{\rho R}{96{\epsilon }_0}$ directed towards the -Y direction.
$\therefore$ The net electric field is
$E=E_1-E_2=\frac{1}{6}\left[\frac{23\rho R}{16{\epsilon }_0}\right]$