Question

A sphere of radius R has total charge Q. The volume charge density (C/m^{3}) within the sphere is      \(\rho=\rho_{0}(1-(r^{2}/R^{2}))\)     

This charge desity decreases quadratically from      \(\rho_{0}\)     

b) Show that the electric field inside the sphere points radially outward with magnitude      $ \(E=(Qr/8\pi\epsilon_{0}R^{3})(5-3(r^{2}/R^{2}))\) $      

c) Show that your results of part (b) has the expected value at r=R.

Answer 1

b)

integrating charge density

int rho_0 (1-r^2/R^2) *dV = Q

we get rho_0 = Q*15/(8R^3)--1

by guass law

E*4 pi r^2 = charge enclosed/epsilon_0---2

charge enclosed = int rho * dV

= rho_0 *4 pi (r^3/3 - r^5/(5R^2))--3

substituting 2 and 1 and 3 in 2

we get

E = Qr /(8 pi epsilon_0*R^3)(5-3(r/R)^2)--4

c)

at r=R

charge enclosed = Q

E*4 pi R^2 = Q/epsilon_0

==> E = Q/[4pi R^2*epsilon_0]

we get the same by substituting r =R in 4