of theinwardfield contributed by the outside sphere is|E−|=
{
0, r<b
k q/r^2 , r>b.
In the region outside the whole capacitor, the two fields are equal
in magnitude, but opposite in direction, so they cancel. We then
have for the total field|E|=
0, r<a
k q/r^2 , a<r<b
0, r>b,
so to calculate the energy, we only need to worry about the region
a<r<b. The energy density in this region isdUe
dv=
1
8 πkE^2
=
k q^2
8 πr−^4.This expression only depends onr, so the energy density is con-
stant across any sphere of radiusr. We can slice the region
a< r < binto concentric spherical layers, like an onion, and
the energy within one such layer, extending fromrtor+ drisdUe=dUe
dv
dv=dUe
dv
(area of shell)(thickness of shell)= (
k q^2
8 πr−^4 )(4πr^2 )(dr)=
k q^2
2
r−^2 dr.Integrating over all the layers to find the total energy, we haveUe=∫
dUe=
∫bak q^2
2r−^2 dr=−
k q^2
2
r−^1∣
∣∣
∣
ba
=k q^2
2(
1
a−
1
b)
608 Chapter 10 Fields