Figure 1: Uniformly charged sphere of radius R. Source point at 7' and field point at FWe can get to the final result (Eq. (11)) through some slightly different considerations2Q???RFigure 2: Sphere of radius R containing point charge q, located at source point F. The field point is r(a) (5) Find the electric field at a point 7 inside a sphere of R carrying uniform charge density p. Do not use Gauss' law, use direct integration (set up the integral but don't evaluate it). See Fig. (1).(b) (5) Now suppose have the same sphere but forget for the moment that it has a charge density. Let's suppose we put a point charge q at the position F in the sphere (this was our field point in part (a) but here it is the source point because q lives there, see Fig. (2)). What is the average electric field at 7', (F'), inside due to the point charge q at 7. Hint: use Eq. (1) but you have to make some modifications to the field point and source point. Plug in E for a point charge and the volume for a sphere of radius R. Then take all the constants out of the integral. You don't have to evaluate the integral.(c) (5) What must the charge density p be for your answer to part (a) to be equal to your answer for part (b). If you pick the appropriate charge density then the average electric field of q at r

The average electric field over some volume V is given by (E) = JE)d'r. (1) Where E(F) is the electric field at the field point, not the source point. If we have some charge density p(r') occupying a volume V, we can write the electric field at the field point F by adding up the contribution of each infinitesimal bit of charge according to Coulomb's law E( F ) = K ( P(F) G-FBy. (2) This is the electric field that goes into Eq. 1: (EY = VK PRE-F (3) Switching the order of integration and factoring out a minus sign from the numerator yields (4) Let's define an auxiliary charge density Paux (F) = -1, then Eq. 4 can be rewritten as (5) The term in the square brackets looks very much like Eq. (2). But where Eq. (2) gives the electric field at due to some charge distribution over the volume V, Eq. (5) gives the "electric field" at F' due to some "charge distribution" (Paux) over the volume V': Eaux(F') = k PaIx (F) (P - FB,. (6) With the definition of the auxiliary field given by Eq. (6) we can rewrite the average field in terms of the auxiliary field: (E) = VP(F) Eaux ( F)d3 1. (7) If our volume "V' is a small spherical region of radius R and keep into consideration that Paux = -1 is uniform, we can take advantage of symmetry and use Gauss' law to find the auxiliary field inside our spherical region, i.e. we pick a spherical Gaussian surface which is a sphere with radius r' such that r'