Electric potential inside and outside sphere

In summary, the potential inside a non-conducting sphere with uniform charge density is proportional to r^l, where l is the order of multipole. This formula is derived in 3-dimensional space, where l = 1. However, if the charge is displaced from the origin on the z-axis, the potential becomes proportional to 1/|r - k| = Sum P(l)(Cos(theta))r^l, where P is the Lagrange polynomial and the sum goes from l = 0 to infinity. If a sphere is placed around the origin with the charge on its north pole, the potential outside the sphere is proportional to 1/r^(l+1) and inside the sphere is proportional to r^
  • #1
Ene Dene
48
0
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?
 
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  • #2
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.
 
  • #3
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).
 

What is electric potential?

Electric potential is the amount of electric potential energy per unit charge at a specific point in space.

How is electric potential calculated inside and outside a sphere?

Inside a sphere, the electric potential is given by V = kQ/R, where k is the Coulomb constant, Q is the charge of the sphere, and R is the distance from the center of the sphere. Outside the sphere, the electric potential is given by V = kQ/R.

What is the relationship between electric potential and electric field?

Electric field is the force per unit charge experienced by a test charge at a specific point in space, while electric potential is the amount of electric potential energy per unit charge at that same point. The relationship between the two is given by E = -∇V, where E is the electric field, and ∇ is the gradient operator.

What is the significance of the electric potential inside and outside a charged sphere?

Inside a charged sphere, the electric potential is constant and does not depend on the distance from the center of the sphere. This also means that the electric field inside the sphere is zero. Outside the sphere, the electric potential decreases with distance and the electric field becomes weaker as the distance increases.

How does the electric potential inside and outside a charged sphere affect the behavior of charged particles?

Inside a charged sphere, charged particles will experience no net force and will remain at rest. Outside the sphere, charged particles will experience a force towards the center of the sphere due to the electric field, causing them to accelerate towards the center.

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