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

2

Explanation

Due to symmetry of charge distribution the electric field at a point outside the cylinder is `_|_r` to the axis of cyclinder , outwards. Let `vec(E_(1))` be the electric field intensity at `P` due to charge on solid cylinder.
Draw a cylinderical gaussain surface of radius `2R` and height `R` as shown in Fig, so that point `P` lies on the curved surface of gaussain cylindrical.
Electric flux over the gaussain cylindrical surface `= oint vec(E_(1)). vec(ds) = (1)/(in_(0)) xx` charge enclosed by the surface
`:. E_(1) (2pi xx 2R) R = (1)/(in_(0)) (pi R^(2)) R rho`
`= (1)/(in_(0)) pi R^(3) rho`

or `E_(1) = (1)/(in_(0)) (pi R^(3) rho)/(4pi R^(2)) = (R rho)/(4in_(0))` ..(i)
Charge on cavity of radius `R//2`,
`q = (4)/(3) pi (R//2)^(3) rho = pi R^(3) rho//6`
Electric fiedl intensity at `P` due to charge on spherical cavity is
`E_(2) = (1)/(4pi in_(0)) = (q)/((2R)^(2)) = (1)/(4pi in_(0)) xx (pi R^(3) rho//6)/(4 R^(2))`
`= (R rho)/(96in_(0))` ...(i)
Hence net electric field, `E = E_(1) - E_(2)`
`= (R rho)/(4in_(0)) - (R rho)/(96in_(0)) = (23rho R)/(96 in_(0))`
As per question, `= (23rho R)/(96in_(0)) = (23rho R)/(16 k in_(0))`
`:. 16k = 96` or `k = 6`