Electric field discontinuity in a spherical shell?

[twist.1995]

Homework Statement

Lets say, there is a non-uniform charge distribution, given as

in a spherical shell that has a cavity with radius a and the radius b to the outer surface. I am wondering if the field is discontinuous just on the surface of this sphere.

Homework Equations

http://blob:https://www.physicsforums.com/94bddccf-abed-4c26-bd7c-fdd42ade2eca [/B]
Φ=∫E.da=Q/ε₀

The Attempt at a Solution

I know that the infinitely charged plane with a surface charge density σ is discontinuous.

I am wondering if this is true for any surfaces regardless of shape and size.

Using Gauss' law, I obtained

, which does not say anything about the discontinuity at b.
Also, if I slice a tiny portion of the outer surface, will I be able to get a surface charge density since the charge is symmetrically distributed all over the shell?

 

[Dr Transport]

no electric field inside the shell, no charge enclosed...

 

[twist.1995]

no electric field inside the shell, no charge enclosed...

The shell is an insulator, since there is a non-uniform volume charge density. If it was a conductor, there will be no charge inside the shell, and it will all be distributed on the inner and outer surfaces. Therefore, the integral gives the electric field inside the shell, as I have mentioned previously.
I am interested in what the electric field is on the surface just outside the sphere. I took the infinitely charged plane as an example and found that the electric field emanating outwards in the normal direction to the plane is σ/ε0. I need to know what will happen on the surface of the sphere (spherical shell),- can I consider it as an infinite plane with a constant electric field on the surface patch on it will take the value of what the Gauss' law gave me. In other words, I found that the electric field is continuous everywhere (not inside the cavity), but is it continuous just on the surface of the sphere or not?

 

[Dr Transport]

you need to check Gauss's law inside the cavity of sphere, you enclose no charge, so there is no electric field...

 

[rude man]

Why would it be discontinuous? Just use Gauss's law at r = just inside b and just outside b. Same E field!

 

[kuruman]

Why would it be discontinuous?

If the shell is an insulator (assumed to be a linear, isotropic and homogeneous dielectric), is it not true that the normal component of $\vec D$ (not $\vec E$) is continuous while the tangential component of $\vec E$ is continuous? In that case
$$ \epsilon \vec{E}_{inside} \cdot \hat n= \epsilon_0 \vec{E}_{outside} \cdot \hat n $$
If there is spherical symmetry, there is no tangential component, but there is still discontinuity in the normal component of the electric field. The surface polarization causes the discontinuity.

 

[rude man]

If the shell is an insulator (assumed to be a linear, isotropic and homogeneous dielectric), is it not true that the normal component of $\vec D$ (not $\vec E$) is continuous while the tangential component of $\vec E$ is continuous? In that case
$$ \epsilon \vec{E}_{inside} \cdot \hat n= \epsilon_0 \vec{E}_{outside} \cdot \hat n $$
If there is spherical symmetry, there is no tangential component, but there is still discontinuity in the normal component of the electric field. The surface polarization causes the discontinuity.

Only if there is polarization to begin with. There was no mention in the problem statement that the relative dielectric coefficient was > 1 so the E field is the same inside and outside the sphere at r=b.

If there is polarization your statement is correct.