# What does it mean for charge to drain off of a conductor when it is grounded?

I am self-studying Griffiths' Electrodynamics. I am on a section about basic properties of conductors. My question is about electric fields and electric potential for a conductor with a cavity inside.

My question is based on a problem in Chapter 2 of the book, which is as follows

2.38 A metal sphere of radius $$R$$, carrying charge $$q$$, is surrounded by a thick concentric metal shell (inner radius $$a$$, outer radius $$b$$, as in Fig. 2.48). The shell carries no net charge.

(a) Find the surface charge density $$\sigma$$ at $$R$$, at $$a$$, and at $$b$$.

(b) Find the potential at the center, using infinity as the reference point.

(c) Now the outer surface is touched to a grounding wire, which drains off charge and lowers its potential to zero (same as infinity). How do your answers to (a) and (b) change?

My question below will be about item (c) and what it means to ground the outer surface.

A question about this problem has been asked before. That question gave me insights about what it means to ground an object, but did not elucidate my question on what exactly it means in physical terms (not just mathematical terms regarding electric potential) when grounding occurs.

Here is the cited Fig. 2.48

and here is my depiction of the situation

In words, the metallic sphere in the center is a charged conductor. Therefore, all the charge is on the surface and there is no electric field inside. This surface charge induces a surface charge on the interior of the metallic shell, $$q_{induced}$$.

Using Gauss' law with a Gaussian surface that is entirely inside the metallic shell, we find that

$$\int_{G.S} \vec{E}\cdot\hat{n}dS=0=\frac{q_{enclosed}}{\epsilon_0}$$

$$q_{enclosed}=q_{induced}+q=0$$

$$q_{induced}=-q$$

There is an electric field in between the interior sphere and the outer shell, represented by the blue field lines in the depiction above.

The outer surface of the metallic shell has an induced positive charge that equals q because as a whole the charge on the shell is zero.

Because the shell is a conductor, there is no electric field inside. There is, however, an electric field emanating radially outward from the outer shell because of the positive charge on the outer surface.

What does it mean to ground the outer surface? Specifically, what does it mean to drain off charge?

It seems to mean that the potential of the "grounded object", in this case the metallic shell, is brought to the potential of the reference point (in our case infinity). Therefore, the metallic shell has potential zero after being grounded.

If the difference in potential between infinity and the metallic shell is zero, then the electric field between them is zero. Ie

$$\int_{\infty}^{\vec{r}_b} \vec{E}\cdot d\vec{r}=0 \implies \vec{E}=0$$

(I am not sure exactly why $$\vec{E}$$ couldn't be something nonzero that happens to have this line integral be zero)

If the electric field is zero outside the metallic shell then Gauss' law with a Gaussian surface being a sphere concentric with our inner sphere and surrounding the metallic shell tells us that the charge enclosed (that of inner sphere plus metallic shell) must be zero.

Apparently this means necessarily that the charge on the outer surface of the metallic shell is zero.

So my question is: how does this happen, and why? Ie, What does it mean that "charge is drained" from the metallic shell?

It seems to mean that positive charge is replaced by no charge at all. Does this happen because electrons flow in to the outer surface? Does the charge on the inner surface mean there are more electrons there than on the outer surface? What determines the amount of electrons on a surface that gives neutral charge?

I like how thoroughly you worked out your question.

(I am not sure exactly why $$\vec E$$ couldn't be something nonzero that happens to have this line integral be zero)

The reason why $$\vec E$$ cannot be something non-zero yet integrate to zero, is that that kind of $$\vec E$$ would have to have positive and negative parts, and that would imply there being charges outside. You can also directly prove that the $$\vec E$$ field has to be zero everywhere outside because you can apply Gauß's law. You know by symmetry that the $$\vec E$$ field must also be spherically symmetric. Another argument is via Laplace's equation, where the solutions must always be incredibly boring, smooth interpolations, no waving of any sort.

It seems to mean that positive charge is replaced by no charge at all. Does this happen because electrons flow in to the outer surface?

Yes.

Does the charge on the inner surface mean there are more electrons there than on the outer surface?

Yes, and the excess electrons on the inner surface must still equal the charge $$q$$ on the inner ball, no more, no less.

What determines the amount of electrons on a surface that gives neutral charge?

Errm, don't you already know? Electrostatic balance. The average density of ions = average density of electrons, in some approximate sense.

So my question is: how does this happen, and why? i.e., What does it mean that "charge is drained" from the metallic shell?

What does it mean to ground the outer surface? Specifically, what does it mean to drain off charge?

I think it is better to give a heuristic argument. It is equivalent to having a wire connect, or temporarily expanding the spherical shell, so that the external shell radius is now infinite. Ground is akin to a gigantic metallic shell of infinite radius, centred on our experiment. Then you can immediately see why it is the case that only the external charges get cancelled if you ground a thing.

• Visualizing the metallic shell as having infinite radius would explain why the outer charge, which is now at infinite radius, has zero potential and it also explains why the charge on the inner surface is not affected (at least according to the basic properties of conductors). However, it is not very clear why this particular interpretation is insightful. The analogy seems to be that the outer charges move very far away. In the real situation is it not clear how this happens though.
– xoux
Commented May 5, 2023 at 9:41
• The argument for that to happen is quite annoying. Consider that you put merely a wire to connect the imaginary infinitely large metallic sphere at infinity to a spot on the shell. Then the charges near the spot will see a path that has a parallel component to the E field, and thus be accelerated outwards. Then, the charges on the other side of the shell will push charges away from their neighbours, thereby replenishing the spot's charges, and this continues until the outer surface is drained of charge. Commented May 5, 2023 at 9:45
• An even more outrageous thing happens if the spot connected is inside the shell. The E field that has a component along the wire will still do stuff, and then the wire, if insulated, will form a capacitor plate with charges trying to bridge the gap between the insulation of the wire and the outer surface of the shell. But the wire itself connects the puncture of the shell and the rest of the intact shell, leading to a lack of tolerance for voltage difference between these two places, all connected by metal. Charge will thus flow and cancel the charges, in agreement with earlier. Commented May 5, 2023 at 9:50

The system, sphere and shell, will try and reach a minimum potential energy state.

In this example the electric potential energy of the system is stored in the electric field with the energy density (energy per unit volume) is equal to $$\frac 12 \epsilon_0 \,E^2$$.
Any charge on the outside of the spherical shell will create an electric field and there will be energy stored in the field which is outside the shell.
Earthing the shell removes the charges on the outside of the shell, the electric field outside the shell becomes zero and hence the energy stored in the system is reduced to its minimum possible value.