Electric potential inside and outside sphere

[Ene Dene]

Put a charge displaced from origin by unit distance on z axis. Now, if you try to find out about potential in space, you can usualy read, "the potential inside a sphere is proportional to r^l, and outside a sphere is proportional to (1/r)^(l+1)". I can't understand what sphere? Is it a sphere around the origin where charge is on its surface? Is it a sphere around the charge, of radius which in magnitude equals the distance between charge and origin? Is it a sphere between origin and charge? I can't combine the position of sphere and r^l inside, (1/r)^(l+1) proportionality outside sphere. If someone could draw it and explain why does the potential depends on r^l inside, and (1/r)^(l+1) outside sphere, and also, what about the surface of sphere, and first of all, what sphere are we talking about?

 

[lightgrav]

Firstly, the "r" in the formula for potential (V = kQ/r)
is the *distance* from the center-of-charge, so
you might as well NOT displace the charge from the origin.

Second, the potential inside a NON-CONDUCTING sphere
that has been given UNIFORM CHARGE DENSITY
is V = kQr/R^2 .
I'm guessing that your book is pretending to not know
that we live in 3-d. With a 3-d situation l = 1 .

If your drawing doesn't have an "l" in it, then
you can't have an "l" in a formula derived about it!
If you want to call "l" the distance from the origin,
then this makes the drawing NOT symmetric,
so you'll never end up with a symmetric formula.

 

[Ene Dene]

I understand now...
I was talking about potential in spherical coordinates (I should have said so). If you have a charge on z asix displaced by unit value it will be proportional to:
1/|r -k |=Sum P(l)(Cos(theta))r^l, where sum goes from l=0 to infinity.

Where P is Lagrande polynom.


If you put a sphere around an origin, and charge on north pole of that sphere, than the potential is proportional 1/r^(l+1) outside that sphere, and to r^l inside sphere, where l is a order of multipole. (for exampe, dipole l=2).