Electric potential inside and outside sphere

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?


Homework Helper
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.
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).

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