We observe a spherical area with radius r in this area we...

We observe a spherical area with radius R. In this area, we have the constant electric field E = k r̂, where the origin is placed in the center of the sphere. There are no free charges outside of or on the edge of the sphere. 0) Illustrate the situation. 1) Determine the charge density, ρ(r), generated by the field using Gauss' law. 2) Determine the total charge in the area r < R, where r = |r|. 3) Determine the electric field for r > R. 4) Determine the potential V(r) for r > R and r < R with respect to r = ∞. 5) Determine the electric energy in the charge distribution and show that it is the same as the energy in the electric field.

Transcript

00:01 All right, so let's say we have a spherical region that has a constant electric field that points radially outwards everywhere.

00:08 So we could draw as many errors as we want, but they all have the same magnitude where e is equal to just some constant k times r.

00:17 So the charge density, according to gauss's law here, is going to be epsilon knot times the divergence of the electric field.

00:24 And the divergence in spherical coordinates is going to be 1 over r squared times r squared times.

00:30 The, or sorry, times the derivative with respect to r of r squared times the r -hat component of the electric field, which is just, you know, this constant k.

00:44 So this is going to be k -epsilon -0 times 1 over r squared times the derivative of that with respect to r -square.

00:51 So 2 -r.

00:52 So we could write this as k - epsilon -not over 2r.

00:55 That's our charge density.

00:57 And then next up, we want to find the total charge in this spherical in this area.

01:05 So this is part one, part two.

01:09 So q as a function of r, it's just going to be the integral of 4 pi r squared times row, dr.

01:19 We'll call these r prime, and this is from zero to r, something like that.

01:24 So this should come out to 4 pi, and they'll have the integral from 0 to r of r prime squared we'll have a k over two uh sorry this shouldn't be over two there should be the two should be on top anyway so this should be 8 pi k o times epsilon not i keep screwing up the units 8 pi k epsilon not and then we have r squared divided by r d r prime so if we do this this should be 4 pi k epsilon not times little r squared so that's our total charge as a function of r and then the electric field outside of this region well the total charge is just going to be that evaluated at capital r so it's 4 pi k epsilon not r squared the electric field outside of this region according to galsazza is just going to be the charge and cloak it's going to be that of a point charge so 1 over 4 pi epsilon not times our total charge, which is 4 pi k epsilon not, r squared, divided by r squared...

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