Question

Problem 7 Let us now consider a nonconducting sphere of radius R carrying a total positive charge Q uniformly distributed throughout its volume, and predict what the behavior of the potential is as a function of radial distance from the center of the sphere In class, we argued that the field points radially outward everywhere and found (using Gauss's law) that the magnitude of the electric field E(r) at a radial distance r < R from the center is given by E(r) = 0R3 while the field at r R is E(r) = In other words, the field is zero at the center, increases linearly in strength as we move away from the center as long as we are still within the sphere; Once we pop out of the sphere, the field gets weaker as 1r2 (a) Draw a qualitatively correct graph of the electric field strength as a function of r. Technically, we should keep track of the direction of E by using positive values for field pointing out and negative for field pointing in, but since the field all points out, we don't have to worry about that (b) Recall that the electric field strength tells us the rate of change of potential and points toward lower potentials "down".) If we move in the direction of the field, V should decrease (go down) and if we move against the field, V should increase (go up.) The slope of V(r) is determined by the magnitude of the field Taking the electric potential at the sphere's surface (at rR) to be equal to zero (ie setting our starting point at the surface,) draw a qualitatively correct graph of V(r), the potential as a function of radial distance from the center (c) Check that your V graph and E graph are consistent with each other at r 0, at T< R, at r R, and at > R. Also check that your V graph is consistent with our choice of V-0 at (d) Redraw your V(r) plot for the same source, if we set V0 at infinity (e) Draw a V(r) plot f if the charge Q were on a conductor (so the field E-0 nside the sphere)

Answer 1