Electric Field inside a Hollow non conducting sphere.

[sl33py.]

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

 

[Doc Al]

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?

 

[kerimek]

.

I would approach this problem by first understanding the concept of electric field and how it is affected by the distribution of charge. In this case, we have a hollow non-conducting sphere with a positive charge distributed throughout its shell.

Using Gauss's Law, we can relate the electric field to the enclosed charge within a closed surface. In this case, the closed surface would be a spherical shell with a radius of 2R, since the charge is distributed throughout the entire shell.

The electric field at the outer surface of the sphere can be calculated using the equation E = q/(4πε0R^2), where q is the total charge distributed throughout the shell and ε0 is the permittivity of free space. We can use this equation to find the electric field at the outer surface of the sphere.

To find the electric field at a radial depth x from the outer surface, we can use the same equation but with a different radius, 2R-x. This is because at this depth, the enclosed charge would be the charge distributed within the spherical shell with a radius of 2R-x.

Now, the question asks at what radial depth the electric field strength is half of the electric field at the surface. This means we need to find the value of x that satisfies the equation E(x) = 1/2 E(2R).

Substituting the values of E(x) and E(2R) into the equation, we get:

q/(4πε0(2R-x)^2) = 1/2 * q/(4πε0R^2)

Simplifying the equation, we get:

(2R-x)^2 = 2R^2

Expanding and rearranging, we get:

x = R/3

Therefore, the electric field strength is half of the electric field at the surface at a radial depth of R/3 from the outer surface. This means that at this depth, the electric field strength will be 1/2 * E(2R) = 1/2 * q/(4πε0R^2) = q/(8πε0R^2).

In conclusion, the electric field strength at a radial depth x from the outer surface of a hollow non-conducting sphere is given by the equation E(x) = q/(4πε0(2R-x)^2) and at a radial depth of R/