## Electric Field inside a Hollow non conducting sphere.

1. The problem statement, all variables and given/known data

Positive charge is distributed throughout a non-conducting spherical shell of inner radius R and outer radius 2R at what radial depth beneath the outer surface the electric field strength is on half to the elextirc field at the surface

2. Relevant equations

Gauss's Law:
integral ( E.dA ) = q/e0

3. The attempt at a solution

suppose that a charge 'q' is distributed in the sphere then
The electric field at the outer surface will be :
int ( E.dA ) = q/e0
=> E. 4.pi.(2R)^2 = q/e0
=> 16E.pi.r^2 = q/e0
The volume of tht sphere will be :
4/3.pi. (2R)^3- 4/3.pi.R^3
=> 28/3.piR^3
So the charge density will be:
density= charge/volume
p=3q/(28.pi.R^3)
Now imagine a gaussian surface having a radial depth from the outer surface "x", the radial distance for that surface will be 2R-x
i am stuck after that, help would be appreciated
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 Mentor Blog Entries: 1 So what's the field at the surface? (In terms of q and R.) Use the same thinking to find the field at r = X, where R

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