Interactive Solution 18.51 at offers help with this problem...

Interactive Solution 18.51 at offers help with this problem in an interactive environment. A solid nonconducting sphere has a positive charge $q$ spread uniformly throughout its volume. The charge density or charge per unit volume, therefore, is $\frac{q}{\frac{4}{3} \pi R^{3}}$. Use Gauss' law to show that the electric field at a point within the sphere at a radius $r$ has a magnitude of $\frac{4}{3} \pi \epsilon_{0} R^{3}$. (Hint: For a Gaussian surface, use a sphere of radius $r$ centered within the solid sphere. Note that the net charge within any volume is the charge density times the volume.)

Interactive Solution 18.51 at offers help with this problem in an interactive environment. A solid nonconducting sphere has a positive charge $q$ spread uniformly throughout its volume. The charge density or charge per unit volume, therefore, is $\frac{q}{\frac{4}{3} \pi R^{3}}$. Use Gauss' law to show that the electric field at a point within the
sphere at a radius $r$ has a magnitude of $\frac{4}{3} \pi \epsilon_{0} R^{3}$. (Hint: For a Gaussian surface, use a sphere of radius $r$ centered within the solid sphere. Note that the net charge within any volume is the charge density times the volume.)

Key Concepts

Uniform Charge Distribution

A uniform charge distribution means that the charge is evenly spread throughout a given volume, resulting in a constant charge density. This uniformity allows for straightforward integration or multiplication by volume when calculating the total charge enclosed within any part of the distribution.

Spherical Symmetry

Spherical symmetry indicates that the properties of the system, including the electric field and the charge density, depend solely on the distance from the center of the sphere, not on the direction. This symmetry is crucial because it ensures that the electric field has the same magnitude at every point on a spherical Gaussian surface of a given radius.

Gauss's Law

Gauss's law is a fundamental principle stating that the net electric flux through a closed surface is equal to the total electric charge enclosed divided by the permittivity of free space. This principle is critical in solving problems with symmetrical charge distributions, as it simplifies the calculation of electric fields.

Gaussian Surface

A Gaussian surface is an imaginary closed surface chosen to match the symmetry of the charge distribution, facilitating the use of Gauss's law. For a spherical charge distribution, selecting a spherical Gaussian surface simplifies the calculation of the electric flux since the field has a constant magnitude over the surface.

Transcript

00:02 In this problem, we have a charge sphere, got a positive charge on it.

00:08 We know the radius of that sphere is capital r, and it's uniformly distributed the charge, so we can just calculate the density.

00:17 This is a charge per unit volume, density, row, charge q over the volume of a sphere, four -thirds pi r cubed.

00:26 We're going to need that in a few minutes.

00:29 Now, a problem like this is a gaussian law problem, because it has symmetry to it.

00:34 So let's draw gaussian surface, which looks like a circle, but it's really a sphere of radius r.

00:44 Our goal is to find the electric field angitude at any point that is a radius little r from the center.

00:54 Now, from the symmetry of this problem, the electric field at all points on that sphere, gaussian sphere, are going to be e.

01:08 And the fact that the charge is positive, that's outward pointing electric fields.

01:18 Now, galser's law involves the flux.

01:22 Let's talk about the flux now.

01:24 This is the left -hand side.

01:25 The total flux is a summation e .i.

01:32 C .a .i.

01:36 What does this mean? this is a summation over all, when we break up the surface, gaussian surface, into patches, flat patches where the electric field is uniform on that patch.

01:53 Now you say flat, how can you get flat with a sphere? well, what you have to do with a sphere is you must effectively break it up into an infinite, very tiny, tiny, almost zero area.

02:08 An infinite number of those.

02:10 That's technically where calculus would come in, but don't let that bother you at all.

02:14 Don't let the infinity or hearing the word calculus, we can do it.

02:18 We can do the same problem here without having to worry about any technical details, mathematical details.

02:27 So let me look at four distinct patches of the infinite collection.

02:35 There's going to be one here, one here, one here.

02:45 And one here.

02:46 I have four patches i'm going to look at.

02:49 And they will tell me everything i need to know.

02:52 What's important is that i can factor out the e and the cosine turn as you're going to see, and i'll just be able to get the area of the gaussian sphere.

03:02 That's the whole goal of this.

03:03 So let me look at each of the patches in turn.

03:09 Here is going to be a spot for number one, number two, number three, and number four.

03:15 Let's learn about these individual pieces, what they tell me and what they give me.

03:23 So in patch number one, which at the top, here is my area vector normal to the patch, and i have my electric field vector, and what will be the angle between the two? trivially, you can see 5 -1 is 0.

03:43 Now remember something, area vectors, normal vectors on a closed surface, which gaussian surfaces are, outward pointing, not inward pointing.

03:53 So that's our situation in patch one, patch two.

03:59 Here is our area vector.

04:05 Here is our electric field vector.

04:07 The angle between the two, still.

04:11 That's good.

04:12 Patch number three, here is our area vector...