Iii the volume charge density rhomathrme within a sphere of...

(III) The volume charge density $\rho_{\mathrm{E}}$ within a sphere of radius $r_{0}$ is distributed in accordance with the following spherically symmetric relation $$\rho_{\mathrm{E}}(r)=\rho_{0}\left[1-\frac{r^{2}}{r_{0}^{2}}\right]$$ where $r$ is measured from the center of the sphere and $\rho_{0}$ is a constant. For a point $\mathrm{P}$ inside the sphere $\left(r< r_{0}\right),$ determine the electric potential $V .$ Let $V=0$ at infinity.

Transcript

00:01 Okay, so we're doing chapter 23, problem 21 here.

00:04 So it says the volume charge density row e within a sphere of radius r -not is distributed by row e as a function of r is row not, r or one minus r over r -not squared.

00:25 So this is our function for row e, and where r is measured from the center of the sphere and row not is a constant.

00:33 Okay.

00:34 For a point p inside the sphere, inside the sphere, so that means r less than r not, determine the electric potential v, and v at infinity equal zero.

00:49 So we first need to find the electric field.

00:52 Since the charge distribution is spherously symmetric, we know we can use galsh's law, where e...

01:00 .d .a.

01:00 Is just e.

01:01 Times the surface area of a sphere and this equals the charge enclosed over epsilon not so the charge and closed now given as four -thirds pi r cubed times row e of r all over epsilon not the total volume four -third pi are not cubed okay but this is actually not the case actually we can't just take the ratio of the volumes here times the volume density because it's not constant.

01:45 If it was constant, we could do that, but now it depends on r.

01:48 So actually, we have to take another step here.

01:53 So we need to calculate what charge and closed is.

01:56 And this is given when we don't have a constant row e of an integral, so we want to integrate from zero to r, row e of a function of r d.

02:10 Okay, so let's do that.

02:12 So charge and closed is integral from zero.

02:14 0 to r.

02:15 If we change dv to d r, we're left with 4 pi r prime squared.

02:22 And now we have row not 1 minus r pime squared over r not squared the r.

02:32 Okay, so let's solve this.

02:35 We have 4 pi row not.

02:38 And now we have r squared on the first becomes r cubed over 3.

02:44 And then we subtract off r to the fourth.

02:49 So this is r fifth r to the 5 over 5 r not squared.

02:56 So this is what our charge enclosed is.

02:58 So let's take this back into our gauces law equation, and we should see that e is now 1 over 4 pi epsilon not, 1 over r squared times 4 pi row not, r cubed over three minus r to the fifth over five r not square so let's cancel out these four pies and some of these rs are going to cancel out and we're left with the answer of of our e field being row not over epsilon not and we have r over three minus r cubed over five r not so that's our e.

04:07 Another thing we need to find out is what the total charge is, which is the same as integrating what we had before from zero all the way to r0.

04:17 If we do that, we see something very similar to this equation, but it all simplifies out to being 8 pi row not, r0 cubed all over.

04:34 We're almost there.

04:35 We almost have what we need...

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