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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