Electric Field inside a Hollow non conducting sphere.

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Homework Statement



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



Homework Equations



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


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
 

Answers and Replies

  • #2
Doc Al
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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<X<2R. What's the total charge contained in the gaussian surface at r = X?
 

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