Zero electric field in empty region of hollow conductor

Question

A hollow conducting shell having a net charge of +Q has a point charge -Q placed at its centre.

From the diagram, it looks like

Suppose the radius of the inner shell from the origin is a and the radius of the outer shell from the origin is b.

When we work the electric field for the region $r < a$ algebraically, the electric field works to be $\vec{E}=\vec{0}$.

Similar conceptual questions to this question popped up in my first year Physics and I had to let it go since I couldn't understand qualitatively what is going on.

Now that I have the mathematical ability to verify the electric field is indeed 0, the qualitative concept doesn't match to what the math suggests.

What does net charge means in this context? I would suppose that the overall charge of the hollow conducting sphere is +Q after taking into account the -Q charge. However, I have doubts. Or does net charge refers to charges residing on the surface of a conductor?

In many texts, the statement "The electric field inside a conductor in electrostatic equilibrium is zero". What does this means? Are they referring to the conductor itself or inside a conductor? I.e, the empty region inside a hollow conducting sphere?

Could someone provide a good verbose explanation to my question?

Answer 1

I have found your question and the diagram a little difficult to interpret.

I have redrawn you diagram to show a charge of $+Q$ on the outer shell and a charge of $-Q$ at the centre together with two conducting shells shaded grey.

What else the electric field inside the conductors is zero.
If there was an electric field then the mobile charge carrier would feel a force on them and move so it would cease to be static electricity.
It is the $+Q$ and $-Q$ induced charges on the surface of the inner conducting shell which ensure that the electric field inside the conductor is zero.
The net charge on that inner shell is $+Q + (-Q) = 0$.
Let $r$ be the distance from the centre.

In region $A$ the electric field is $\dfrac {1}{4 \pi \epsilon_o} \dfrac {-Q}{r^2}$.

In region $B$ inside the conducting inner shell, the electric field is zero.

In region $C$ the electric field is again $\dfrac {1}{4 \pi \epsilon_o} \dfrac {-Q}{r^2}$.

In region $D$ inside the conducting outer shell, the electric field is zero.

Application of Gauss' law requites that the net charge $+Q$ on the outer shell must reside on the inside as shown in the diagram.
As that is the case then the net field in region $E$ is zero. The net charge of this whole arrangement is zero $-Q + Q + (-Q) +Q = 0$

Suppose that the net charge of the whole arrangement was $+Q$ which would mean that the outer shell would have a charge of $+2Q$ on it.

Then the diagram would be exactly the same except that there would be a $+Q$ charge on the outside of the outer shell and the electric field in region $E$ would be $\dfrac {1}{4 \pi \epsilon_o} \dfrac {+Q}{r^2}$.
The net charge on the outer shell is $+Q+Q = +2Q$

Suppose the net charge on the outer shell was $-Q$ then again all the charges and electric field would be the same except that on the outside of the outer shell there would be a charge of $-2Q$, the electric field in region $E$ would be $\dfrac {1}{4 \pi \epsilon_o} \dfrac {-2Q}{r^2}$ and the net charge on the outer sphere would be $+Q+(-2Q)= -Q$.

Answer 2

In the situation you depicted, the electric field is different from $0$ in the hollow region enclosed by the conductor and equal to $0$ inside the volume of the conductor (at equilibrium).

• Why is $\vec E = 0$ inside the volume of the conductor at equilibrium?

We will proceed by reductio ad absurdum. Since we are at equilibrium by hypothesis, there can be no net flow of charge (charge cannot be moving). Suppose that $\vec E$ was different from $0$ inside the conductor: then charges would feel the electric field, and since they are free to move inside a conductor they would start to move until a situation in which $\vec E = 0$ is reached. So our hypothesis (that there was equilibrium) was false, and we must conclude that $\vec E = 0$ inside the volume of the conductor.

• Why is $\vec E \neq 0$ inside the hollow region?

This is easier, although we're gonna use some math. We will use Gauss'law, which states that the net flux of the electric field $\Phi$ through a closed surface $S$ enclosing a volume $V$ is $Q/\epsilon$, where $Q$ is the total charge inside $V$:

$$\Phi = \int_S \vec E \cdot d\vec S= \frac{q}{\epsilon}$$

Let's take an imaginary spherical surface $S$ of radius $R$ containing the charge $-Q$ at its center. If we apply the above equation, we will obtain, since the field lines are orthogonal to the surface,

$$\Phi = 4 \pi R^2 E = \frac{-Q}{\epsilon} \to E =-\frac{Q}{4 \pi \epsilon R^2}$$

So the field inside is just the field of a point charge.

• What about the field outside the conductor?

We just apply Gauss' law again and take an imaginary surface enclosing the whole sphere. Since this time the net charge inside the imaginary sphere is $0$ ($Q-Q=0$) and the field lines are orthogonal to the surface of the sphere, Gauss' law tells us that this time

$$E=0$$

(Why are the field orthogonal to the surface? Again, reductio ad absurdum: if they weren't, there would be a parallel component of the electric field acting on the charge inside the conductor and they would start to move, so we wouldn't be at equilibrium)

Hope this answers your question.