#### Potential and Charged Spheres

Note that in the simulation above, the insulating sphere has a net charge +Q and the conducting sphere has a net charge -5Q. Drag the red circle left or right to plot field or potential as a function of r.

For a charged sphere of radius R, what is the potential due to the sphere? If the charge is symmetrically distributed, the potential outside a sphere of charge Q is the same as that from a point charge Q placed at the center of the sphere. This is what happened for electric field, too.

V = kQ/r

What happens inside the sphere? If the sphere is a conductor the potential is the same everywhere throughout the conductor, and is equal to the value of the potential at the surface:

V = kQ/R.

If the sphere is an insulator with a uniform charge density, the potential is not constant because there is a field inside the insulator. We showed previously that the field inside a uniformly charged insulator is:

E = kQr/R3

Starting from some point a distance r from the center and moving out to the edge of the sphere, the potential changes by an amount:

DV = V(R) - V(r) = -òE·ds = -òE dr = -kQ/R3 ò r dr

where the limits of the integral are from r to R.

Integrating gives:

V(R) - V(r) = -[kQ/2R3] (R2-r2)

V(R) is simply kQ/R, so:

For r < R, V(r) = [kQ/2R] (3-r2/R2)

#### Concentric Spheres

Now we'll put the two cases together. The insulating sphere at the center has a charge +Q uniformly distributed over it, and has a radius R. The concentric conducting shell has inner radius 1.5R and outer radius 2R. It has a net charge of -5Q.

What is the electric field as a function of r?

What is the electric potential as a function of r?

Let's try the field first:

For r > 2R, E = 4kQ/r and points toward the center

For 1.5R < r < 2R, E = 0

For R < r < 1.5R, E = kQ/r, directed away from the center

For r < R, E = kQr/R3, directed away from the center

The potential is a little trickier:

For r > 2R, V(r) = -4kQ/r
The potential looks like the potential from a -4Q charge.

For 1.5R < r < 2R, V(r) = -4kQ/2R = -2kQ/R
The potential inside the conducting shell is constant, and is equal to the value it has at the outside of the shell.

For R < r < 1.5R, V(r) = kQ/r - kQ/1.5R - 2kQ/R
From the previous region, we know that V(1.5R) = -2kQ/R. We also know how the potential changes as we move closer to a point charge:

VA - VB = kq/rA - kq/rB

Here VA = V(r), VB = V(1.5R) = -2kQ/R, q=+Q, rA = r, and rB = 1.5R.

For r < R, V(r) = [-kQ/6R](7+3r2/R2)

From the previous region, we have V(R) = -5kQ/3R.

We derived previously that the potential difference between a point on the edge of an insulating sphere and a point inside is:

V(R) - V(r) = -[kQ/2R3] (R2-r2)

Solving for V(r) gives:

V(r) = -5kQ/3R + kQ/2R - kQr2/2R3

V(r) = -10kQ/6R + 3kQ/6R - 3kQr2/6R3

For r < R, V(r) = [-kQ/6R](7+3r2/R2)