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March 28, 2015

March 28, 2015

Posted by **Jessica** on Thursday, February 28, 2008 at 9:19pm.

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

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**Damon**, Thursday, February 28, 2008 at 9:34pmInside 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 -
**Jessica**, Thursday, February 28, 2008 at 9:45pmso the electic field outside the sphere is 9*10^9 p (r/a)^2/a^2?

- physics -
**Damon**, Thursday, February 28, 2008 at 9:47pmWhoops, 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

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**rory**, Thursday, February 28, 2008 at 9:49pmcould i get some help damon

- physics - typo -
**Damon**, Thursday, February 28, 2008 at 9:50pm(4/3) p * pi r^5 /a^2

- physics -
**Damon**, Thursday, February 28, 2008 at 9:53pmThe 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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