
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/R^{3}
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/R^{3} ò r dr
where the limits of the integral are from r to R.
Integrating gives:
V(R)  V(r) = [kQ/2R^{3}] (R^{2}r^{2})
V(R) is simply kQ/R, so:
For r < R, V(r) = [kQ/2R] (3r^{2}/R^{2})
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/R^{3}, 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:
V_{A}  V_{B} = kq/r_{A}  kq/r_{B}
Here V_{A} = V(r), V_{B} = V(1.5R) = 2kQ/R, q=+Q, r_{A} = r, and r_{B} = 1.5R.
For r < R, V(r) = [kQ/6R](7+3r^{2}/R^{2})
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/2R^{3}] (R^{2}r^{2})
Solving for V(r) gives:
V(r) = 5kQ/3R + kQ/2R  kQr^{2}/2R^{3}
V(r) = 10kQ/6R + 3kQ/6R  3kQr^{2}/6R^{3}
For r < R, V(r) = [kQ/6R](7+3r^{2}/R^{2}) oo_supktmYJRFmAP9ifMD6M6dVjjV8Ar18ADTNi9co=.html