W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 101

b

R

ρ 0

b) Material is removed from the sphere to create a spherical cavity of radiusb=a/2 with center

atx=bon thexaxis (shown above). Show that the electric field inside the cavity isuniform

and equal to:

E~=ρ^0

~b

3 ǫ 0 =

4 πkρ 0 ~b

3

in magnitude (where~b=bxˆ).

Hint: By far the easiest way to attack this problem is to imagine that the “hole” is made up
of a sphere of uniform charge density−ρ 0 and radiusbthat issuperposedon the uniform sphere of
charge densityρ 0 and radiusa. In that way the two charge densities cancel and leave “the cavity”,
while you can easily find the fields using the results of part (a) with a bitof algebra. Also,draw
big picturesof the spheres. You have to add vectors in the hole! If you don’t make a big sphere
with a hole large enough to draw vectors in, it’s going to be really hard to visualize what’s going on
accurately enough to guide you when you try to add up the field. If you do a reallygoodpicture,
you may see thetrivialway to do the addition that actually makes this problem rathereasy(given
(a)) instead of a matter of adding up vector components the hardway!