Question

A solid sphere of radius R has a charge Q distributed in its volume with a charge density rho=kr^a, where k and a are constants and r is the distance from its centre. If the electric field at r=(R)/(2) is 1/8 times that r=R, find the value of a.

Answer 1

Value of a = 2

Explanation:

Applying Gauss' law, we get:

∮.E.dA = 1/ϵo. ∫(pdv)  

    = 1/ϵo. ∫(kr^a) range of 0 to R * 4πr²dr

​E * 4πR² = (4πk/ϵo)( R^(a+3) / (a+3) )

Given r = R/2.

E2 = k (R/2)^(a+1) / ϵo(a+3)

Given E2 = E1/8

So k (R/2)^(a+1) / ϵo(a+3) = kR^(a+1) / 8ϵo(a+3)

  1/ 2^a+1 = 1/8

So 2^a+1 = 8

  2^3 = 8

So a + 1 = 3

Therefore a = 3-2 = 2