Inside the sphere, any imaginary volume you think of will have no charge inside it. Therefore no Electric Field is coming out of it. Thus there is no electric field within the sphere.
The instant you get outside the sphere, the charge looks like it is all at the center, and it is the usual 9*10^9 q/r^2 kind of field.
so the electic field outside the sphere is 9*10^9 p (r/a)^2/a^2?
Whoops, sorry, was thinking charge on surface of sphere.
If uniformly distributed throughout the volume, then my answer is the same outside the sphere
However inside the sphere you will surround chagres the will look as if they are at the center
The amount you surround will be the volume at radius r times the charge density.
V = (4/3) pi r^3
q =V (charge density)= (4/3) pi r^3 (p) (r^2/a^2)
(4/2) p *pi ^5/a^2
that times is q
E = 9*10^9 q/r^2
= 9 * 10^9 (4/3)p* pi r^3/a^2
when you reach r = a
E = 9*10^9 (4/3) p* pi r^3 /a^2
which is of course
9*10^9 ( q total)/r^2 from then on
could i get some help damon
The answers Dan H gave you looked reasonable to me Rory.
yep i noticed that =P
(4/3) p * pi r^5 /a^2
The idea is that inside the sphere, the only charge that counts is what is inside the radius you are at. That looks like it is at the center and is the charge density times the volume inside the radius you are at.
Outside the sphere, the entire charge, the density times the entire volume of the sphere (4/3) pr a^3 * density is inside your radius and no longer increases/