A nonconducting solid sphere has a uniform volume charge...

A nonconducting solid sphere has a uniform volume charge density $\rho$ Let $\vec{r}$ be the vector from the center of the sphere to a gen- eral point $P$ within the sphere. (a) Show that the electric field at $P$ is given by $\overline{E}=\rho \vec{r} / 3 \varepsilon_{0}$ (Note that the result is independent of the ra- dius of the sphere.) (b) A spherical cavity is hollowed out of the sphere, as shown in Fig. $23-$ $60 .$ Using superposition concepts, show that the electric field at all points within the cavity is uniform and equal to $E=\rho \vec{a} / 3 \varepsilon_{0},$ where $\vec{a}$ is the position vector from the center of the sphere to the center of the cavity.

Transcript

00:01 All right.

00:01 So i remember seeing this one back in the day, and i was like, what? so you're given a sphere.

00:13 You're told it has a cavity, and you want to get, ultimately you want to get the goal, the field inside that cavity.

00:23 And so first you want to just get the field inside here and show that it can be written in this nice way row over three epsilon not.

00:33 Times r.

00:34 So let's start with gauss's law, so not e equals, but the flux is equal to the q enclosed over epsilon not.

00:44 And if we do some gaussian surface of radius r, then we can say e times pi r squared is equal to the charge enclosed over epsilon not.

00:57 The charge and closed is going to be the total charge, so let q be the total, q equal total charge.

01:10 And only a portion of that total charge is going to be enclosed.

01:13 In particular, it's the ratio of the radius of this sphere with radius r to the radius of the full sphere, which i'll call big r.

01:23 And so that's going to be, so it'll be four thirds pi, little r cubed, over four thirds pi, big r cubed.

01:30 And if you can't solve the four thirds pi, you'll get this.

01:33 And then if you also use the definition of density, cube, is so charges density times volume and then the volume and then the volume of the sphere is four -thirds pi r cubed so if you sub in this q cancelled this r oops i almost forgot my four here and then these fours will end up canceling you'll get the answer you'll get that e is equal to row over three epsilon knot and it's times r in the r hat direction which is because you a vector r so now i'll do this on another page uh you you make a cavity and you're like what's the field in the cavity and you have to show that it it's this constant which is kind of interesting um that it comes out a constant and maybe that's something that i would ponder with you if i were actually here in real life with you but um i'm trying to see it's not too long so um but but do ponder it maybe it'll be fun.

02:44 So anyway, so basically we know that the field and we know the field and without the cavity is this row over three epsilon knot times r.

02:54 We want to get it with the cavity.

02:55 So the trick to modeling the cavity is to say that it's the same thing as taking this full sphere and superimposing on it a sphere of negative charge.

03:06 Right, because the negative charges from this new sphere plus the positive charges that were already there, maybe i'll kind of draw this in.

03:15 We'll make a cavity.

03:17 And then this is my representation of the charges that were already in the sphere.

03:22 And i didn't draw it this way, but they should be the same density of charges.

03:28 So then we're in love with the task of finding the sum of the fields from the positives and some of the fields from the negatives to model the cavity behavior.

03:36 And so we already know, though, the shape of the electric field.

03:42 And the same.

03:42 And the in a spherical symmetric within a spherical symmetric chart distribution and if it's negative it's just going to be minus row over 3 epsilon not but with a new coordinate system so this i'm going to call this the r prime coordinate system where r prime refers to oh i wish i'd made this slightly bigger i'm still trying to learn how to use this how to totally predict with this oh my god, it's really bugging out.

04:15 Okay, let's see if i can draw a circle.

04:22 Okay, maybe over here is helpful.

04:27 Okay, so here's my sphere...

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