Question 3.4: (a) Show that the average electric field over a spherical su...

Refer to Fig. 3.3, letting α be the angle between ᴫ and the z axis. Obviously, E _{\text {ave }} points in the -\hat{ z } direction, so

E _{ ave }=\frac{1}{4 \pi R^{2}} \oint E d a=-\hat{ z } \frac{1}{4 \pi R^{2}} \frac{q}{4 \pi \epsilon_{0}} \int \frac{1}{ᴫ^{2}} \cos \alpha d a .

By the law of cosines  ,

R^{2}=z^{2}+ᴫ^{2}-2 ᴫ z \cos \alpha \Rightarrow \cos \alpha=\frac{z^{2}+ᴫ^{2}-R^{2}}{2 ᴫ z} ,

ᴫ^{2}=R^{2}+z^{2}-2 R z \cos \theta \quad \Rightarrow \quad \frac{\cos \alpha}{ᴫ^{2}}=\frac{z^{2}+ᴫ^{2}-R^{2}}{2 zᴫ^{3} }=\frac{z-R \cos \theta}{\left(R^{2}+z^{2} -2 R z \cos \theta \right)^{3 / 2}} .

\text { (where } u \equiv \cos \theta) \text {. } The integral is

=\frac{1}{R}\left(\frac{1}{|z-R|}-\frac{1}{z+R}\right)-\frac{1}{2 R z^{2}}\left[|z-R|-(z+R)+\left(R^{2}+z^{2}\right)\left(\frac{1}{|z-R|}-\frac{1}{z+R}\right)\right] .

(a)   \text { If } z>R ,

=\frac{1}{R}\left(\frac{2 R}{z^{2}-R^{2}}\right)-\frac{1}{2 R z^{2}}\left[-2 R+\left(R^{2}+z^{2}\right) \frac{2 R}{z^{2}-R^{2}}\right]=\frac{2}{z^{2}} .

So

E _{\text {ave }}=-\frac{1}{4 \pi \epsilon_{0}} \frac{q}{z^{2}} \hat{ z } ,

the same as the field at the center. By superposition the same holds for any collection of charges outside the sphere

(b)   \text { If } z<R ,

=\frac{1}{R}\left(\frac{2 z}{R^{2}-z^{2}}\right)-\frac{1}{2 R z^{2}}\left[-2 z+\left(R^{2}+z^{2}\right) \frac{2 z}{R^{2}-z^{2}}\right]=0 .

So

E _{\text {ave }}=0 .

By superposition the same holds for any collection of charges inside the sphere  .