# physics

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A sphere of radius a carries a volume charge density p=p (r/a)^2. Find the electric field inside and outside the sphere.

I don't know what equation to use. Do I use electirc field due to point charge?

• physics -

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.

• physics -

so the electic field outside the sphere is 9*10^9 p (r/a)^2/a^2?

• physics -

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)
which is
(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
you get
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

• physics -

could i get some help damon

• physics -

The answers Dan H gave you looked reasonable to me Rory.

• physics -

yep i noticed that =P

• physics - typo -

(4/3) p * pi r^5 /a^2

• physics -

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/

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