2. A Positive electric charge Q is distributed uniformly...

2. A Positive electric charge Q is distributed uniformly throughout the volume of an insulating solid sphere with radius a (Figure 2). By using Gauss Law, find the magnitude of the electric field at a point P, a distance r from the center of the sphere where: a) r < a (10 points) b) r > a (10 points)

Transcript

00:01 So in this question, we have an insulating sphere, and it's uniformly filled with positive charge, such that the total charge is q.

00:16 And we're going to use gauss's law, which tells us that the integral of the electric field through a surface, the total integral of the electric field through the surface, is equal to the charge enclosed by that surface, divided by epsilon naught.

00:33 So this is the charge enclosed.

00:40 Since we're going to be considering gaussian surfaces which are spheres, and this whole situation is spherically symmetric, we know that the electric field at each point on this sphere is going to be the same.

00:56 And that means that this integral is going to simplify to just 4 pi r squared times the magnitude of the electric field, because we can just add up the same electric field at all points on the surface of the sphere, and this is the area of the sphere, equals q enclosed over epsilon naught.

01:23 So now the charge enclosed up to a radius r, let's think about this.

01:32 So the charge density is going to be the total charge divided by the volume of the sphere, which is four thirds pi a cubed because the sphere is of radius a.

01:47 So now the charge enclosed up to a radius r for r less than a is going to be the integral of row, and then we're going to have 4 pi r squared d r from zero to r.

02:08 So we're integrating a dummy variable up to r.

02:10 And this gives us 4 thirds pi r cubed multiplied by row, which is q divided by 4 thirds pi a cubed...

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