Question 3.47: Show that the average field inside a sphere of radius R, due...

E _{\text {ave }}=\frac{1}{\left(\frac{4}{3} \pi R^{3}\right)} \int E d \tau ,  where  E =\frac{1}{4 \pi \epsilon_{0}} \frac{q}{ᴫ^{2}} \hat{ᴫ } ,

so    E _{ ave }=\frac{1}{\left(\frac{4}{3} \pi R^{3}\right)} \frac{1}{4 \pi \epsilon_{0}} \int q \frac{\hat{ ᴫ }}{ᴫ^{2}} d \tau.

(Here r is the source point, d \tau is the field point, so ᴫ goes from r to d \tau.) The field at r due to uniform charge ρ over the sphere is E _{\rho}=\frac{1}{4 \pi \epsilon_{0}} \int \rho \frac{\hat{ ᴫ }}{ᴫ^{2}} d \tau . This time d \tau is the source point and r is the field point , so ᴫ  goes from d \tau to r , and hence carries the opposite sign. So with \rho=-q /\left(\frac{4}{3} \pi R^{3}\right) , the two expressions agree: E _{\text {ave }}= E _{\rho} .

(b) From Prob. 2.12:

E _{\rho}=\frac{1}{3 \epsilon_{0}} \rho r =-\frac{q}{4 \pi \epsilon_{0}} \frac{ r }{R^{3}}=-\frac{ p }{4 \pi \epsilon_{0} R^{3}} .

(c) If there are many charges inside the sphere, E _{\text {ave }} is the sum of the individual averages, and p _{\text {tot }} is the sum of the individual dipole moments. So E _{ ave }=-\frac{ p }{4 \pi \epsilon_{0} R^{3}} .  qed

(d) The same argument, only with q placed at r outside the sphere, gives

E _{ ave }= E _{\rho}=\frac{1}{4 \pi \epsilon_{0}} \frac{\left(\frac{4}{3} \pi R^{3} \rho\right)}{r^{2}} \hat{ r } (field at r due to uniformly charged sphere) =\frac{1}{4 \pi \epsilon_{0}} \frac{-q}{r^{2}} \hat{ r } .

But this is precisely the field produced by q (at r) at the center of the sphere. So the average field (over the sphere) due to a point charge outside the sphere is the same as the field that same charge produces at the center. And by superposition, this holds for any collection of exterior charges.