Spherical conducting shell enclosing a non-conducting core

[hmparticle9]

Homework Statement

A solid non-conducting sphere of uniform charge density and total charge $-Q$ and radius $a$ is surrounded by a concentric conducting spherical shell of inner radius $b$ and outer radius $c$ with $a < b < c$. The outer shell has charge$2Q$. I am interested in the case $r \in [b,c]$

Relevant Equations

$$\int_S \mathbf{E} \cdot d \mathbf{S} = \frac{q}{\epsilon_0}$$

The case I am interested in is $r \in [b,c]$. Because the outer shell is conducting and the outer shell encloses a charge $-Q$ would it be correct to say that for the case $r \in [b,c)$ the charge inside the outer shell arranges itself in a way to "cancel" as much of the contained charge as it can.

Because the outer shell has $2Q$ to play with we are okay. $Q$ of the $2Q$ charge arranges itself on the shell boundary $r = b$, hence $E(r) = 0, r \in [b,c)$.

Am I correct in saying that the remaining charge $Q$ arranges on the outer shell $r=c$? When we use Gauss's law we obtain
$$E(c) = \frac{Q}{4 \pi \epsilon_0 c^2}$$

If we had less than $Q$ charge in the outer shell I am trying to figure out what $E(r)$ would look like in the outer shell...

 

[kuruman]

Start by considering what the electric field is at $b<r<c$, i.e. inside the conducting material. What does Gauss's law have to say about that?

 

[hmparticle9]

Surely the charge $\hat{Q} < Q$ would accumulate at $r = b$. Gauss's law would say

$$\int_S \mathbf{E} \cdot d \mathbf{S} = \frac{Q - \hat{Q}}{\epsilon_0}$$ within the shell?

 

[kuruman]

Surely the charge $\hat{Q} < Q$ would accumulate at $r = b$. Gauss's law would say

$$\int_S \mathbf{E} \cdot d \mathbf{S} = \frac{Q - \hat{Q}}{\epsilon_0}$$ within the shell?

Sure, but you didn't answer my question. What is the value of $\mathbf E$ on the left-hand side of the equation? Hint: Under static conditions, a conductor is an equipotential.

 

[hmparticle9]

Sorry ! :)

$$\int_S \mathbf{E} \cdot \text{d} \mathbf{S} = E(r) 4 \pi r^2 = \frac{Q - \hat{Q}}{\epsilon_0} \implies E(r) = \frac{Q - \hat{Q}}{4\pi r^2\epsilon_0}$$

 

[kuruman]

Sorry ! :)

$$\int_S \mathbf{E} \cdot \text{d} \mathbf{S} = E(r) 4 \pi r^2 = \frac{Q - \hat{Q}}{\epsilon_0} \implies E(r) = \frac{Q - \hat{Q}}{4\pi r^2\epsilon_0}$$

That is not correct.

What is the direction of the electric field inside the conductor? Remember that electric field lines point from regions of high electric potential to regions of low electric potential and that conductors are equipotentials. Put it together and come up with a numerical value for the normal component of the electric field on the surface of the Gaussian surface inside the conductor.