Discussion Forum

1)

An infinitely long solid cylinder of radius. R has a uniform volume charge density $\rho$. 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 2Rfrom the axis of the cylinder, is given by the expression  $\frac{23 \rho R}{16k \epsilon_{0}}$. The value of k is 

Answer:

Option A

Explanation:

Volume of cylinder per unit length (l=1)is

  $V= \pi R^{2}l=(\pi R^{2})$

 $\therefore$ Charge per unit length

 $\lambda=$  (volume per unit length ) x ( volume charge density)

   $=(\pi R^{2} \rho)$

 Now at P

 $E_{R}=E_{T}-E_{C}$

 R= remaining portion

T= total portion and

C= cavity

 $\therefore$   

$E_{R}=\frac{\lambda}{2\pi\epsilon_{0}(2R)}-\frac{1}{4 \pi\epsilon_{0}}\frac{Q}{(2R)^{2}}$

Q= charge on sphere

      $=\frac{4}{3}\pi\left(\frac{R}{2}\right)^{3}\rho=\frac{\pi R^{3} \rho}{6}$

Substituting the values , we have

$E_{R}=\frac{(\pi R^{2} \rho)}{4 \pi\epsilon_{0}R}-\frac{1}{4\pi\epsilon_{0}}.\frac{(\pi R^{3} \rho/6)}{4R^{2}}$

=$\frac{23  \rho R}{96\epsilon_{0}}=\frac{23  \rho R}{(16)(6)\epsilon_{0}}$

$\therefore$   k=6