Question

An infinitely long solid cylinder of radius R has a uniform volume charge density p. 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 $\frac{23\rho R}{16k{ϵ}_0}$. The value of k is:

• A. 5
• B. 12
• C. 6
• D. 24

Answer 1

The correct option is C 6

The given system of cylinder with cavity that can be expressed as super position of infinite cylinder with charge $+p$ and a sphere with charge density $-p$.

Now, field due to infinite cylinder is given by $E_{cyl}=\frac{\lambda }{2\pi {dE}_0}$. Here $\lambda$ is the charge per unit charge $=\lambda =p\times A=\pi R^{2}p$ and $d=2R$

Thus, $E_{cyl}=\frac{\pi R^{2}p}{2\pi \left(2R\right)E_0}=\frac{pR}{{4E}_0}$

Field due to sphere is given by $E_{sph}=\frac{1}{4\pi E_0}\frac{Q}{d^2}$

here, now we have to solute $Q=$ The total charge in the sphere,

$Q=-pV$

$=-\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\times p$

and $d=2R$

Thus, $E_{sph}=\frac{4\pi {\left(\frac{R}{2}\right)}^{3}p}{3\times 4\pi E_0{\left(2R\right)}^2}=-\frac{pR}{96E_0}$

Thus, the net electric field is $E$ $=\frac{pR}{E_0}(\frac{1}{4}-\frac{1}{96})$

Hence $=\frac{23pR}{96E_0}$

Thus, $16k=96$

or, $k=6$