Why is Earth's potential taken to be zero?

Question

This is not a duplicate question for many similar questions, I have tried finding an answer to this everywhere but to no avail. So here is my problem: Why is Earth's potential taken to be zero?

Now I know that electric potential is not an absolute quantity, it's only the change in potential that is significant so we can set a reference anywhere according to our convenience. And that's what I thought too, we simply set Earth's potential to be zero instead of infinity for convenience.

Now here's the issue: my professor gave us the following problem:

"there are two thin concentric conducting shells of radius $a$ and $b$ ($a<b$), and the inner shell is given a charge $q$ on its outer surface (charge induction will take place on outer shell) and also the outer shell is now earthed. Find the final charges on the shells".

For the solution, he says that the potential of the outer shell is to be made zero as it is earthed. Using this and the fact that the potential on the surface of a shell is given by KQ/r, where Q is the charge on shell and r is the radius, he finds the solution.

Now my problem is that isn't the formula for the potential of a shell on its surface (i.e., KQ/r) derived with the assumption that the reference is set at infinity? That means my professor is setting the reference at ground level but using the formula derived with reference at infinity and still getting the correct answer? I think I am surely missing something. There are some questions in the textbook that have similar questions and use similar methods. What seems to be the issue here?

Answer 1

The Earth is often set to zero because in most scenarios, it is large enough relative to everything else to be approximated both as being everywhere and as an infinite charge sink. Therefore, over time all charges will eventually equalize to the potential of the Earth and the potential of the Earth remains unchanged in the process. Therefore it is convenient to set it to zero especially when a problem explicitly involves the Earth. It doesn't have to be set to zero, but you might as well.

Now, when a problem does not explicitly involve the Earth you may choose to involve the Earth due to its infinite charge sink property. By connecting the outer shell to the Earth you are fixing the outershell to the potential of an infinite charge sink. That means that no matter what you do to it, that potential will remain unchanged. Now that is true whether or not the potential of your infinite charge sink is zero. However, as mentioned in the first paragraph it's convenient for it to be zero but it doesn't have to be. The "infinite charge sink" part matters more here than the "set to zero" part.

If your problem involves a body significantly larger than the Earth (say, the sun), then that would probably be set to be your zero since it better approximates an infinite charge sink.

If your problem involves the Earth and another similar sized body then you can no longer treat approximate the Earth as an infinite charge sink since all objects present have enough charge to affect each other's potential. In that case, I guess you need to find a better way to set your zero.

Answer 2

Now my problem is that isn't the formula for potential of shell on its surface i.e KQ/r derived with the assumption that reference is set at infinity? That means my professor is setting the reference at ground level but using the formula derived with reference at infinity and still getting the correct answer?

In problems like this you should consider "the Earth" to be infinitely far away. That is, the problem is not taking place near a conductive plane but is taking place in empty space, and the reference for voltage is that it is zero at infinity/earth/ground. If something is "earthed" in this context, that doesn't imply changing the reference, but it means that the earthed object can acquire/lose charges as needed to maintain the same voltage as infinity. (I.e. it's connected to infinity/earth/ground and that earth serves as an infinite reservoir of charge.)

The reason you can reuse the the $kQ/r$ formula in this problem is a combination of several things. First, the superposition principle lets you compute the potential due to two charge distributions by summing the potential due to each one alone. Second, the outer sphere forms a Faraday cage for the inner one, so the inner sphere sees the same fields (not necessarily the same potential) as in empty space, so it takes on the spherically uniform charge distribution that makes $kQ/r$ valid for the inner sphere. Finally, considering the potential due to the inner sphere, the outer sphere is already on a surface of constant potential, so the charges on it do not need to rearrange away from uniform either, making $kQ/r$ valid for the outer sphere.

Just to be clear: there is no a priori issue with having both a conductive, uncharged ground plane and infinity set to be the zero voltage reference. The issue is solely in the validity of assuming the charge distributions are spherically uniform. In fact, the answer to this question in the presence of a ground plane is the same, but more reasoning is required. You need to say that, since the outer sphere is at the same voltage as the conductive plane, there is no field outside two spheres and therefore the conductive plane can be ignored. Then apply the reasoning for empty space.

Answer 3

Here is a simplified reason for he says that the potential of the outer shell is to be made zero as it is earthed.
As ever reality is much more complex but the general mechanism explained below is correct.

In your arrangement there is a sphere of radius $b$ with a charge $Q$ on its outer surface which has a capacitance of $4\pi\epsilon_0b$ and a potential $\dfrac {Q}{4\pi \epsilon_0b}$ relative to infinity which is taken to be the zero of potential.
You are connected, "earthing", it to another sphere of radius $R_{\rm Earth}$ which has a capacitance of $4\pi\epsilon_0R_{\rm Earth}$ and, if uncharged, a potential of zero relative to infinity.

So what happens when both capacitors are connected together?
The is a redistribution of charge to equalise such that the potentials of both objects is the same, $\dfrac {Q_{\rm sphere,final}}{4\pi \epsilon_0b}=\dfrac {Q_{\rm Earth,final}}{4\pi \epsilon_0R_{\rm Earth}}\Rightarrow \dfrac{Q_{\rm sphere,final}}{Q_{\rm Earth,final}} =\dfrac{b}{R_{\rm Earth}}$ and charge is conserved $Q = Q_{\rm sphere,final} + Q_{\rm Earth,final}$.

Putting in some numbers $b = 6.4\,\rm cm$ and $R_{\rm Earth} = 6.4\,\rm Mm $ shows that the potential of the Earth does not change by very much, ie effectively stays as zero, and the charge on the outer surface of the sphere becomes very much smaller than $Q$ as $ \dfrac{Q_{\rm sphere,final}}{Q _{\rm Earth,final}} =\dfrac{6.4\times 10^{-2}}{6.4 \times 10^6}=10^{-8}$.

Thus the basic idea in the situation described by the OP is that the Earth can be though of as a (infinite) source/sink of charges and the potential of the Earth can be assumed to be constant and arbitrarily chosen to be zero.

Answer 4

Preface: I am not quite understanding how your professor's interpretation of this problem is valid. It is very strange to me to give a spherical shell (which usually implies infinitesimally small thickness) an interior and exterior surface charge distribution. Thus, I will interpret the problem as I see it, and you can be the judge.

Observe that the potential at $b$ is given by the superposition of the electric potential as a result of the charge on the inner shell and the charge on the outer shell:

$$\Phi(r=b) = -\frac{1}{4\pi \epsilon_0}[\int_{O}^b\frac{q}{r^2}dr+\int_O^b\frac{q_{out}}{r^2}dr]=0.$$

Where O is where you set your potential to 0. No matter what you set your reference potential to, you are taking a potential difference so the constant will cancel. You're left with:

$$\frac{1}{4\pi \epsilon_0}[\frac{q}{b}+\frac{q_{out}}{b}]=0 \implies q_{out}=-q.$$

Thus, the use of the formula for electric potential with reference at infinity can be used with no contradiction.