Show that the average electric field over a spherical surface, due to charges outside the sphere, is the same as the center of the sphere.
I don't quite understand what they mean by the ''average'' electric field. I could interpret this as the sum of all magnitudes of the electric field pointing in all directions divided by the total field, but then that seems to just equal $1$. Or I could include the directions, but then that would sum to zero I think.
My attempt:
Ok so I gave it another try and here's what I know. Based on an example in the book (Griffiths, Intro to Electrodynamics, 2nd ed.), I know that the average potential over a spherical surface with a charge outside the sphere is $V=kq/z$. I also know that the average electric field is given by $E=\frac{\Delta V}{\Delta r}$. I think I should be able to relate the average potential and the average electric field using $E=-\nabla V$, but I'm not sure.