5. You are given a spherically symmetric volume charge...

5. You are given a spherically symmetric volume charge distribution located in free space, with\ $ ho = 3r^2$ for $r < a$, and $ ho = 0$ for $r > a$.\ (a) Use $Phi(r) = int_{r'} frac{ ho(r')dV'}{4piepsilon_0|r - r'|} $ to calculate the potential at the center of the charge distribution.\ (b) Now calculate the electric field $mathbf{E}$ from Gauss' law and then obtain the potential at the center of\the charge distribution by evaluating $- int mathbf{E} cdot dmathbf{l}$ where the limits range from $infty$ to 0.

Transcript

00:01 All right, so let's say we have a charge distribution of 3r squared, and this is for r less than some distance a.

00:08 And so it's a spherical charge distribution, so we want to find the potential at some location.

00:17 So we'll call it 5r, and we know this should be the integral of row of like r prime times like 1 over 4 pi epsilon not, and this is over r minus r prime.

00:30 Taken as a vector, and this is integrated over this volume space.

00:35 So this is going to be 1 over epsilon not.

00:38 We can do the integrals.

00:41 We only need to do the integral of the radio coordinate.

00:43 The other angular coordinates are just going to come out to 4 pi.

00:45 So that's why i've eliminated the 4 pi there.

00:47 And then we're going from 0 to a of 3r prime squared over r minus r prime times r prime squared.

00:58 The reason being is like our volume element, as i wrote it, is going to be 4 pi r prime squared d r prime.

01:06 So that's why we have that.

01:08 So if we write this, this is 3 over epsilon knot times the interval from 0 to a of r prime to the fourth over r minus r prime.

01:19 D r prime.

01:23 So let's just define a variable u that is r minus r prime.

01:28 Then du is equal to negative dr prime and our integral is going to become negative 3 over epsilon not times the integral so when r prime is zero this is just from r to r minus a and we can switch the limits and get rid of the negative signs this is r minus a to r and then this is going to be you or sorry r minus u to the fourth over u d u and so we do this integral sorry let's go back a step because it's actually there's we're only evaluating this at the center so phi at zero is just going to be three over epsilon knot times the integral from zero to a of r prime to the fourth over r prime with a negative sign d r prime right so this becomes r prime to the third and so this is negative 3 over 4 epsilon not times a to the fourth.

02:38 So that's our potential at the center...

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