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 cylinder, as shown in the figure. The magnitude of the electic field at the point `P`, which is at a distance `2 R` form the axis of the cylinder, is given by the expression `( 23 r R)/( 16 k e_0)` . The value of `k` is .
.

Answer

A,C,D

Explanation

We have `l = r p R^2`
`vec E_P = overlineE_(clinder (+ r)) + overline_(sphere(-r))`
`vec E_P = overlineE_(clinder (+ r)) + overline_(sphere(-r))`
For cylinder ` …(1)
` E _(cyjlinder (+ r)) = ( 2 Kl)/r =2' 1/( 4 pe_0) ' 1/((2R)) ' rpR^@ ` ...(2)
`E_(sphere) = (KQ)/r^2 = 1/(4 pe_0) ' 1/(4 pe_0) ' 1/((2R)^@) ' r'`
. ...(3)
From (1) , (2) and (3)
` E= ( 23 rR)/( 4' 24 e_0) = ( 23 r R)/( 6' 16 e_0) , k=6`
`E_1 2 pi (2R) l= (rhopi R^2l)/( varepsilon_0) :. E_1 ( rhoR)/(4pi varepsilon_0)`
`E_2 4 pi (2 r)^2 = (rho 4/3 pi (R/2)^2)/(pi varepsilon_0)`
` :. E_2 = (rhoR)/( 24 xx 4 pi varepsilon_0)`
` E_1 -E_2 = (rhoR)/(4 pi varepsilon_0) ( 1- 1/( 24)) = (23 rho R)/( 4 xx 24 pi varepsilon_0)`
` :. K = 6`.