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 electric 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 .
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Answer

6

Explanation

Volume of cylinder per unit length `( l= 1)` is `V = pi R^(2) 1 = (pi R)^(2)`
`therefore ` Charge per unit length ,
`lambda` = (volume per unit length) `xx` ( volume charge density ) ` = ( pi R^(2) rho)`
Now at P `E_(r) = E_(T) - R_(C)` .
Remaining portion, Total proton and, Cavity
`therefore " " E_(R) = (lambda)/(2 pi epsi_(0) (2R)) - (1)/( 4 pi epsi_(0)R), ` Q = charge on sphere `= (4)/(3) pi ((R)/(2))^(3) rho (pi R^(3) rho)/(6)`
Substituing the vlaue , where value we have `E_(R) = ((pi R^(2)rho))/(4 pi epsi_(0) R) - (1)/(4 pi epsi_(0)R) ((pi R^(3) rho //6))/(4 R^(2)) = (23 rho p)/(96 epsi_(0)) = ( 23 rho R)/((16)(6) epsi_(0))`. `therefore k = 6`