Question

A solid sphere 10 cm in radius carries a uniform 40-?C charge distribution throughout its volume. It is surrounded by a concentric shell 20 cm in radius, also uniformly charged with 40 ?C. Find the electric field 5.0 cm from the center.

Answer 1

So let's see if there are some easy-to-remember rules that will help you with this problem. It turns out that Gauss' Law (and Gaussian surfaces S) can help you with all four situations, so first make sure you understand it.

There are 4 corollaries you can use when "volume symmetry" is present:

For a uniformly charged sphere, let R be its radius, and let r be the radius at which you wish to know the electric field Er (as well as the radius at which you create a Gaussian surface Sr, a sphere of radius r):

(C1.1) If r < R, the enclosed charge Qenc is proportional to the volumes of the two spheres [see Source 1], then:

(1) Qenc = Q * (r^3 / R^3), so

(2) Er = k * Q * r / R^3,

where:
k = 1 / (4??0)
Q = the charge on the full sphere


(C2) If r >= R, then

(3) Er = k * Q / R^2 [see Source 2]


(C3) Any charge outside r does not affect Er. [see Source 1]


For a uniformly charged shell:

Let Ri be the inner radius and Ro be the outer.

If r <= Ri, then (C3) applies.

If r >= Ro then (C2) applies.

If Ri < r < Ro,
then the charge outside r doesn't affect Er, so (C3) again.
The Qenc is proportional to volumes of the two shells, similar to (C1), except you have to subtract the volume of the inner hollow sphere:

(4) Qenc / Q = (Vr - Vi) / (VRo - Vi)
and
(C1.2):
(5) Er = k * (Q / r^2) * (r^3 - Ri^3) / (Ro^3 - Ri^3)


In summary, you have 4 corollaries to Gauss' Law to remember. It may seem complex, but, if you group them, moving from outside the charge inward, it becomes simpler:

Outside the charged volume: C2 (the inverse square rule)

Surrounded by the charged volume: C1.1 (sphere) or C1.2 (shell) (proportionality); and C3 (exclusion)

Inside the charged volume: C3 (exclusion)