I know its the potential differences that matter and generally we define the zero of the electric potential according to our convenience. I would like you to look at this standard problem:

enter image description here

Charge $-Q$ is given to the inner (conducting) shell and the larger outer (conducting) shell is earthed . And the question asks to find the net charge on both the shell. In the standard solution the electric potential of the outer shell is taken to be zero saying that since it is earthed and potential of the earth is zero so electric potential at the surface of the outer shell must also be zero. Then the potential due to the charges on the surfaces of the shells is equated to zero by considering Zero potential at infinity. I want to know how is potential zero both at the earth and at infinity. Also, I want to know how to solve such problems without using the concept of "potential." A similar question can be found here: Why the electric potential of Earth is zero?

The top voted answer says that the earth is just a reference point.But then I would like someone to solve the above problem by considering the earth to be at some arbitrary nonzero potential $V$. (Also, don't forget that the outermost surface of the outer shell will also be charged). It would be great if you can provide me links to reliable sources about the zero potential of the Earth.

P.S :- This is not a homework question.


The diagram above has a very important feature.
It is the connection between the Earth and the outer conducting shell.
Assume that the Earth is a conducting sphere and has some net positive charge on it.
This will mean that the outer shell connected to it will also have some positive charge on it but the wire between the outer shell and the Earth means that they are at the same potential.

Now what is the value of that potential?
You can call it zero and then infinity will be at a negative potential or you could have infinity at zero potential and then the outer shell and the Earth would be at the same positive potential.

Starting from the beginning with no Earth connection.

For simplicity let the zero of potential be infinity and the Earth and the inner and outer shells having no net charge on them also having a potential of zero.
If the Earth does have a charge in what follows the only change is that the potentials all change by an amount equal to the initial (non-zero) potential of the Earth.

Move a charge of $-Q$ from the Earth to the inner conducting shell which leaves a net charge of $+Q$ on the Earth.
Charges on the inside of the outer shell will redistribute themselves as in the diagram and there will be a net charge of $-Q$ on the outside of the outer shell.
That negative charge will not reside evenly on the outside of the outer shell rather there will be more near the Earth as there will also be more positive charge on the Earth in that region.

In terms of potential relative to infinity the potential of the outer shell has decreased and that of the Earth has increased but not by the same amount.
Because the Earth is so big relative to the outer shell the redistribution of the charge on the Earth is greatest near the outer shell and negligible on the other side of a diameter.
So the potential of the Earth relative to infinity has increased hardly at all.
What you cannot do is assume that the Earth is an isolated sphere and apply the $+Q=C_{\text{Earth}} \Delta V$ formula to find the change in potential of the Earth.
The Earth is not isolated, it is under the influence of the negatively charged outer shell and this in turn means that charges are not uniformly distributed on the surface of the outer shell or the Earth.

Now connect the negatively charged outer shell to the positively charged Earth.
Charges will flow until the potential difference between the outer shell and the Earth is zero when the net charge on the outside of the outer shell and the Earth is zero.
The potential of the outer shell and the Earth will be zero.

The redistribution of charge occurs locally and the area of the local region is very, very much smaller than the surface area of the earth.
So a 10 F capacitor with one terminal earthed when charged with $+1$ coulomb will leave a charge on the Earth of $-1$ coulomb close to it thus disturbing the local potentials but the potential of the Earth as a whole hardly at all.

I did want to try and make the answer more quantitative but this interesting article “Electrostatics of two charged conducting spheres” made me realise that finding and then applying the capacitance of two unequal sized spheres is not a trivial matter.

  • $\begingroup$ "For simplicity let the zero of potential be infinity and the Earth and the inner and outer shells having no net charge on them also having a potential of zero." Why fixing the zero of the potential implies earth's potential is $0$? $\endgroup$ – MarcoCiafa Jul 9 '20 at 22:56

So, the potential is not necessarily zero both at the ground and at infinity. First off, there's nothing in physics which forces the potential to be zero at infinity; it just happens to be a nice way to think of the Coulomb potential. Potential energies are not absolute numbers; there is a constant of integration that enters into them!

Second, there is nothing in physics which forces the surface of the Earth to be the same potential as points in the distant cosmos; the Earth could retain some electric charge while the solar wind sends some of it out into distant space, for example.

However, we do know that this must be bounded, because if the Earth does have a net charge, it must be pulling in the opposite charge back towards it with a relatively strong force, and trying to expel other ions from our ionosphere etc. We know that the electrostatic force is actually relatively powerful even on human scales, and is certainly much much more powerful than the gravitational force holding the Earth together. If you want, you can try to see how big this would be. We know that free electrons are not repelled from Earth's surface into space or sucked into the ground appreciably; let's say we know this to within 1% or so. The Coulomb force of two electrons at a distance of Earth's radius is $5.67 \cdot 10^{-42} \text{ N},$ which we can divide by the electron's mass to find $6.23\cdot 10^{-12} \text m/\text s^2.$ The gravitational acceleration is almost $10~\rm{m/s^2}$, a little over $10^{12}$ times larger, so we have an upper limit of around $10^{10}$ surplus elementary charges distributed in any spherically-symmetric way about the Earth -- any more than that and we would have noticed (within 1%) that free electrons tilt towards the ground or else to the sky.

That gives a maximum surplus charge of only $10^{10}$ elementary charges, or $1.6 ~\rm{ nC}$ for the entire Earth, so the difference in electric potential from infinity from that would only be about $2.3~\rm{ \mu V},$ or less if there is some sort of opposite-charge cloud around the Earth (which would also increase the push on the electron!)

  • $\begingroup$ Your estimate for the charge of the planet is way, way off. Of course there is a rather large system of charged particle belts around the planet. $\endgroup$ – CuriousOne May 20 '16 at 3:11

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