If we apply the formula for the capacitance of a sphere (C= 4PEr, where P=pi, E= permittivity of free space, and R= radius of the earth), and apply the values we get a value of around 700 microfarad. If we now work out the potential on the surface of the earth (V=q/c) we get a fairly high value for even small values of q. Given these figures, why is it that the electric potential of the earth is taken as zero?
Electrical potential is defined only in terms of differences, that is, the potential difference between two points can be calculated, but there is no absolute definition of the electrical potential at one point (this is because the physically observable field, the electric field, is given by a derivative of the potential, so only differences are physical). The Earth is often chosen to be the zero point for convenience, much like choosing the origin of a coordinate system. It is not the only possible choice, and in fact the potential is often to be zero at r = infinity, so that the potential at any distance closer to a source charge at r = 0 will be negative. ______________ blitz.km
Thank you for the reply. However, I suppose my point with reference to the calculated value of potential on the earth's surface would also have some answer. I, of course, quite appreciate that we need a zero reference which could be any value of potential. Notwithstanding this I would like to understand the issue with regard to my calculation for the potential on the surface of the earth.
As far as I know, the Earth isn't a charged sphere, it should be more or less neutral for the most part so the voltage would be zero.
How about this, when you drop a charged balloon, does it go sailing into the air or plummeting at the earth at ridiculous speed? Neither happens so the field must not be high, and so the charge must be low.
I have read somewhere about the likely order of the charge on the earth's surface. Unfortuntely I am unable to lay my hands on the source! This data would hopefully help to resolve the doubt in my mind.
I am a podiatrist and am using an electrostimulation device on the feet of patients with diabetic neuropathy pain. I am told by the manufacturer that the device's frequency, 7.83 Hz, is the same as the electrical potential between the earth's suface and the atmosphere. I am trying to confirm this. Anyone have an answer? Thank you.
Reells, the device's frequency cannot be the same as any electric potential. The units don't match. I seem to remember that the earth's electromagnetic field resonates at 7.83 Hz (among other frequencies). I am not sure exactly what causes, this, however. Anyone?
Virtualetters, thank you very much. This (the electromagnetic field resonance) must have been what was meant by the spiel I read. It was just not stated in correct physics terminology. Whether it makes any difference in the treatment of a human is another matter. Possibly it does. Thanks again.
Potential of static charge is not a matter of difference from an arbitrary reference but is zero on a body where negative and positive charges balance. I suggest recall of early classical experiments with pith balls suspended on threads, gold leaf electroscopes, cat fur, amber rods etc, Wimhurst machine. induction of opposite charges, charging of simplecapacitors etc. Ever tear up some plastic and find it keeps stuck to your fingers? Back to basics. Darmog
If an uncharged conductor placed on an insulated stand is brought near a positively charged surface, equal and opposite charges are induced on the two ends of the conductor (near and far end from the charged surface). If the charged conductor is now earthed, the positive charge is neutralised by electrons flowing from the earth. Removal of the charged surface from the vicinity of the conductor as well as connection of the conductor with earth would then cause the negative charge on the conductor to spread throughout the conductor. This would leave the conductor with its surface negatively charged. The earth also has some net charge on its surface although I cannot recall the amount of such charge. Thus it is not true that the earth is considered to be at zero potential because it is charge neutral.
I would carefully examine the derivation of that particular equation and see whether it is appropriate to apply it to something like a planet. (To begin with, you wouldn't use the permittivity of free space, you need to use the permittivity of the Earth, which is around 10 times higher). Claude.
The permittivity of free space has been correctly applied. If you look at the method of deriving the potential of a charged sphere you will find that what I am saying is true. It is to be remembered that the charge residing on the surface of a sphere behaves as if it is all located at the centre of the sphere as a point charge even though this is actually not so. By application of Gauss's Law to a surface concentric with the sphere (of radius larger than that of the sphere) and calculating the potential on this outer sphere, it could be concluded that the permittivity to be considered for the potential is that of free space.
But again, granted if the formula you used is correct, you are assuming that the Earth must be fully charged. If I have a capacitor of 1 F, that does not mean that it has any charge built up or a voltage across its terminals. But this is still fairly moot because for the most part we arbitrarily choose the Earth to be zero potential. The main property of the Earth that makes it a ground as opposed to just some random object that is regarded as zero potential is that any feasible electrical device cannot strip or sink enough charges from/to the Earth to change its potential. It behaves, for all practical purposes, like an infinite resevoir of charge. What its actual potential is is immaterial for the most part though it is easy to see that it is more or less a neutral body from simple experiments liken Prologue mentioned. The Earth is not in true isolation, it can easily pull charges from the ionosphere or the various plasmas that surround the Earth if it had a significant net charge. Of course, the consistency of the Earth's voltage is mainly local, if you measured the voltage between hundreds of miles you can find a measurable voltage difference.
There is definitely a net charge on the earth's surface though this may not be uniformly distributed. However, this charge is not very significant. The issue may perhaps be now treated as closed on the understanding that this insignificant charge would not cause a potential of consequence to develop, and thus the earth's potential would for all practical purposes be considered as zero.
This is the fundamental Schumann resonance frequency of the ionosphere. See e.g. http://en.wikipedia.org/wiki/Schumann_resonances for more info.
If you are above the surface of the earth and below the clouds, there is a vertical electrical field of about 100 V/m. The earth is negatively charged by lightning strikes and the upper atmosphere retains a positive charge. The non-zero conductance between sky and earth gives the so-called "fair weather current" which would dissipate this charge imbalance fairly quickly if it were not being constantly renewed by thunderstorms. But the total potential difference between earth and clouds is typically a few hundred kV. The charge on a rubbed balloon might be on the order of a nanocoulomb (based on a few web pages I scanned), which would give a force of 0.0000001 newtons (0.1 micronewton) on the balloon due to the electric field. If the balloon weighs one gram, the gravitational force on it would be 9.8 millinewtons, or about 100,000 times greater than the electrical force. These numbers are crude, but I think they show that the failure of a charged balloon to levitate or plummet does not place a very tight constraint on the total electrical charge of the earth.