# Back of the Envelope Calculation of Earth's Charge

1. Jul 25, 2009

### Nirgal

I've been reading some E&M text and came across an interesting electrostatic problem that went something like "find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere". The answer is (3Q^2) / (64*$$\pi$$$$\epsilon$$*R^2). Where Q is the total charge on the sphere, R is the radius, and $$\epsilon$$ is the permitivity of free space.

This is a rather beautiful problem, and an interesting theoretical exercise would be to apply it to the earth and figure out just how strong, roughly, the earth is resisting the separation, due to the electric force, of it's northern and southern hemisphere's.

My question: what is a nice way to guess the charge on the earth? The only variable in the solution to the problem that gives me any trouble is the charge. I am not sure if it is necessary to even consider whether or not most of the charge is gathered along the surface or rather if it is speckled throughout the volume.

The reason I am not sure if it is necessary is because I don't think the estimation will be off more than a factor of 10 if we consider the earth as a uniformly charged sphere and not a conductor. The work to build a spherical conductor is about 3/5 or 4/5 the work to build a spherical insulator which means the difference between the forces which pushes the hemispheres apart in both cases cannot be much different.

Reiteration of question: what is a nice way to guess the charge on the earth?

2. Jul 25, 2009

### Staff: Mentor

The net charge of the earth is roughly zero.

3. Jul 25, 2009

### Nirgal

I don't think that is true but you helped me out nonetheless because your reply made me realize how to guess the charge. There are several reasons why we know the earth's charge is not zero. One reason is that there is lightning, which means there is a potential difference between the earth and the clouds. That potential difference is a result of the electric field produced by the earth which is induced by the charges in the earth. We actually know the magnitude of this Electric field. It's about 100 Volts per Meter. You can measure it by taking a bucket of water, cutting a whole in the bottom and allow the water to pour out of the bucket, then measure the potential on the bucket. Therefore we can use Gauss's law law to guess the charge on the earth. E = (Charge Enclosed) / (permittivity of space). (Charge Enclosed) = E*(permittivity of space). (Charge Enclosed) = 100 * 9e-12 = 9e-10 Coulombs.

How wonderful.

The force of separation between the hemisphere's is therefore about 4e-23 Newtons. Thats about equal to the weight of a 4e-25 gram creature. I read somewhere that the weight of a small virus was about 10e-15 grams. The ratio of the two weights is roughly 1 billion.

That means the force of separation of the two hemisphere's on the earth is about 1 billion times less than the weight of a single virus.

4. Jul 25, 2009

### Staff: Mentor

Ok, well I guess I figured the atmosphere was part of the earth for the purpose of this thought experiment. In any case...
No, it isn't. The charge difference is a temporary result of friction between particles in thunderstorms, kinda like rubbing your feet on a carpet.

5. Jul 25, 2009

### Nirgal

That is incorrect. The 100 V/m Electric field is independent and has nothing to do with the thunderstorm. You can measure it on a clear day or a stormy day. It is a result of the total charge on the earth. One perspective for the reason we have lightning is that lightning acts as a mechanism which neutralizes the charge build-up on the earth. Since the earth is negatively charged the electric field points inwards and causes positive ions to build up on the surface. The earth becomes less negative because of these positive ions and therefore the charges that build up in the clouds become more negative with respect to the earth. In order to reduce this charge build up, the potential difference between the earth and the clouds reaches a threshold and the air breaks down which you see as lightning strike.

Nobody knows exactly how charge separation occurs in the clouds, but it is very unlikely that it can be related to "friction". The charge separation you get when rubbing your feet on a carpet is an electrostatic effect. Charge separation in a thunderstorm is an electrodynamic effect. One model is that the negative ions in the droplets of water which rain down from the cloud, move towards the back of the droplet (because of the direction of the electric field of the earth), which causes the front of the droplet to become positive. Because negative ions move easier than positive (their lighter usually), freely floating negative ions will attach themselves to the front of the droplet while postive ions will try to reach the back, but miss because their too heavy, and wind up floating up towards the top of the cloud. This creates a separation of charge in the cloud itself, positive at the top and negative at the bottom. But this is just one theory out of many.

6. Jul 25, 2009

### Staff: Mentor

No. You can google this and find plenty of confirmation:

You're confusing (reversint) the cause and the effect.

7. Jul 26, 2009

### Nirgal

That report is not saying that the electric field exists because of the thunderstorms. It is saying that thunderstorms provide the mechanism to move charges against this electric field to the earth. Which is exactly as I described. You might be confused because they said "Thunderstorms are the generators" but they are not describing thunderstorms as generating the electric field. It is obvious because of the statement "there must be an "electrical generator" somewhere which is pumping electrons from air to ground, AGAINST the electric field. The "electrical generators" are a result of this electric field of the earth and must combat it.

8. Jul 26, 2009

### Per Oni

For some time I’ve been thinking about why the earth is negatively charged.
In my opinion the solution is not all that complicated.
Considering the vast clouds of charged particles being blown from eg the sun in solar winds. It’s the lighter particles, electrons which are moving quicker and easier towards the surroundings and the earth. They will move easier through the atmosphere via water droplets.

Even if the earth on the whole has a net negative charge (or was initially neutral) then a local electron cloud will still induce a positive spot underneath and therefore be attracted to earth. This way earth becomes more and more negatively charged until forces come into work which balance the field to your quoted value. Therefore, ultimately it’s the sun which is the source of this energy.

9. Jul 26, 2009

### vin300

10. Jul 27, 2009

### Nirgal

Well, if you want me to give you the math in detail I will but for now this will do.

First figure out what the electric field is inside a uniformly charged sphere. Call it E.
Call the total charge on the sphere Q.
The total volume of the sphere V.
The charge density is Q/V.

You need to know what the force is and in which direction it is pointing to separate the two hemispheres, or in other words, what is the component of the force that acts to separate them.

The force per unit volume on a sphere is is (Q/V)E

We only need to find the force due to a single hemisphere, because that will be the force that is pushing the other hemisphere away.

If you position the sphere so that it is at the origin and the Northern Hemisphere is pointing in the positive z direction and the Southern Hemisphere is pointing in the negative z direction then the component of force that we want to look at (because we're going to just look at the northern hemisphere) is pointing in the positive z direction and so you must multiply your force per unit volume by cos(th)
force per unit volume = (Q/V)Ecos(th)

Multiply the force per unit volume by a differential volume element such as dV so that
dF = (Q/V)Ecos(th)dV is the differential force element pointing in the z direction.
Now integrate over the northern hemisphere, so you get something like

F = Int( (Q/V)Ecos(th)*r^2sin(th)(dphi)(dtheta)(dr) )

Where I have used spherical coordinates

That is essentially how you do it. Again, if you really want to see it in detail I can show you, it might be kind of hard to read those pseudo-equations.

11. Jul 28, 2009

### Per Oni

(Charge Enclosed) = E*(permittivity of space)*surface area
= E*(permittivity of space)*4*PI*(e6)^2 = 1.11e4 Coulomb.

The formula to work out the force has charged squared so that the answer isn't virus like anymore!

12. Jul 28, 2009

### maverick_starstrider

How is the earth's total charge not zero? If it were even the slightest fraction of a fraction of a fraction of a percent unbalanced would we not induce the moon to come plummeting towards us? EM force is amazingly strong relative to gravity. I always took lightning storms and the like to be a result of local charge inbalances and by no means indicative of a global charge imbalance.

13. Jul 28, 2009

### lalbatros

Nirgal,

Could you calculate how much should the earth be charged so as to get an attracing force on the moon comparable to gravity.
Assume the moon is neutral globally, but has some permittivity like the permittivity of sand.

I would be even more curious to see the result expressed in units like electron/atom ionation.

lal

14. Jul 28, 2009

### Nirgal

lalbatros, that sounds like an interesting problem, I'll look into it and report back. Good idea!

responding to Per Oni: You're right. I forgot to include the surface area in Gausses Law. Thanks for the heads up. That certainly changes the results a bit and makes it alot more interesting.

15. Jul 29, 2009

### Per Oni

Before you get too carried away working on this one remember that although the surface
has a large E field, at about 50 km altitude this field is close to zero. The reason is that high up there are still clouds of +ve ions which are shielding earth's negative charge.

We are sort of living in between the plates of a giant capacitor! A giant inverted atom.

I also looked up earth's radius: r=6356 km, with surface area: A=5.101 e14 m2

16. Jul 29, 2009

### lalbatros

Don't worry, this is not a difficult problem.