# Earth, its charge, its potential and how objects discharge...

1. Sep 19, 2015

### fog37

Hello Forum,

When we take a small metal sphere of radius R1 an charge Q1= 1C and a larger metal sphere with R2=20*R1 and charge Q2=-2C. The spheres are initially separated. Being conductors, all the charge resides on the surface of the spheres. Sphere 1 has an electric potential V_1 (relative to infinity) proportional to Q1 and inversely proportional to R1. Same goes for sphere two which has potential V2. The potentials V1 and V2 are different and assume V1>V2.

Later we connected the two spheres via a metal wires. Charge will flow to make the potentials equal everywhere. The two spheres will eventually become, after a transient, a single equipotential conductor with potential V_final (which is different from V1 and V2). The net charge, Q1+Q2= -1 C, will spread on both spheres in a non-uniform way: the negative surface charge density is larger on the smaller sphere.

Now lets consider a charge conductor carrying charge Q_initial=+1C and planet Earth.
Planet Earth can be considered to be a huge spherical conductor of radius 6,371 kilometers that carries a net negative surface charge of about -1 nC/square meter and capacitance ~ 710 microfarad. Earth's surface is 5.1×108km^2 or 510.1 trillion m² which implies that its total negative surface charge is 5.7*10^5 C. Lots of negative charge!

When we ground the charged object, the object and Earth becomes a single equipotential conductor at the same electric potential V_final. Charge will move around and distributes itself. This equipotential conductor will carry a net negative charge since (Q_initial+Q_earth )~ Q_earth. The highest negative surface charge density will be on the "discharged" conductor which does not really get discharged at all. In fact, regardless of the sign and amount of its initial charge, a charged object that is grounded will eventually carry some negative charge and will not become completely discharged. Capacitance is C=Q/V and for Earth it is very small for Earth (only 710 microfarad). But Q_earth (5.7*10^5 C ) is so huge that V=Q/C for Earth does not change much when Earth receives or gives aways a little amount of charge. The final electric potential that the grounded object and Earth have together is very close (the same for all practical purposes) to the initial Earth's potential V_earth=(9*10^9)*(5.7*10^5 C )/(6,371*10^3) (relative to infinity where V is set to zero).

Since it is potential difference is the quantityy that matters, we often set V_earth=0 instead of using a point at infinity...

Are my observations correct?

I hope so, because I would like to pose another question after this...

thanks,
fog37

2. Sep 19, 2015

### Staff: Mentor

You don't have to assume that, you can calculate it.
If they are separated sufficiently. Otherwise the charge distribution will be non-uniform due to mutual influence of the spheres.
What is "not much"? If you connect something in your lab to Earth, the voltage relative to infinity doesn't change in any significant way because the charge stays very close to the surface. If you make a huge cable to somewhere in outer space, things can be different.

3. Sep 19, 2015

### fog37

Ok.

a) True. Assume that before being connected to each other, the sphere are so far that the charge each carries on its surface is uniformly distributed.

b) My point is: consider earth by itself with its huge negative surface charge and capacitance 710 microfarad. Its potential V_earth (relative to infinity) is the same at every point on its surface. Adding or subtracting some small charge from - 5.7*10^5 C does not change modify 5.7*10^5 C much which implies that V_earth= k Q_earth/R_earth remains constant.

c) once we connect via a thin wire (so we can neglect the charge that gets deposited on its surface) a charged object to earth, at equilibrium the object+earth become an equipotential system whose potential (relative to infinity) is equal to V_earth for all practical purposes.

4. Sep 19, 2015

### Staff: Mentor

Then the charge of earth is negligible because you have to take the atmosphere into account.