A thin circular disk of radius r has charge q and surface...

Name: a thin circular disk of radius r has charge q and surface charge density of r 0r w here 0 is constant a find an expression for 0 b find the electric eld at a point z above the center of the 46322
Uploaded: 2022-07-28T20:20:10-08:00
Duration: 10 min 47 s
Channel: Gregory Devenport
Description: a thin circular disk of radius r has charge q and surface charge density of r 0r w here 0 is constant a find an expression for 0 b find the electric eld at a point z above the center of the 46322

A thin, circular disk of radius R has charge Q and surface charge density of (r) = 0r w here 0 is constant. a) Find an expression for 0. b) Find the electric eld at a point z above the center of the disk.

Transcript

00:01 We're trying to find the electric potential on the x -axis as a function of z for this charged disk as a uniform charge q distributed over it.

00:16 So i've drawn sort of a view from the side and then also a top -down view here.

00:23 So we're going to call the radius of the disk, the total radius, big r.

00:28 I'm going to call a little...

00:32 Changing radius of the disk r prime.

00:36 We're going to call the distance from some point on the disk to the place we're measuring.

00:42 We'll call that just little r.

00:45 So if big r, r prime, and r.

00:49 Okay, our equation for the potential is going to be kq over r.

01:10 But the contribution from each little piece of charge on the disk will be it will contribute differently to the potential because for example if you pick a point out here it's some distance away from the point p call that point p but if you pick a point here towards the center it's a different distance away right so the distance is changing so we have to integrate here.

01:43 So this is going to be k terms of the integral of dq over r.

01:51 Okay, what is dq for a disk? it's sigma d a which is sigma times 2 pi r prime, dr prime.

02:07 So a differential area da is going to be draw circle here.

02:16 So a little differential area would be like a little this little red donut type thing.

02:25 So it has length 2xr and has width dr.

02:31 Dr prime.

02:32 So there's our differential area.

02:36 Let's figure out what our distance r is.

02:39 That's from the disk to the point p.

02:41 So at any given time it's going to be equal to z squared plus r prime squared inside of a square root so we just use triangles for that can see that here on this top drawing let's draw it here so here's here's our triangle the hypotenuse is length r the height is z it's constant and then the little leg that's red is that distance r prime which is changing okay, so we should have everything we need, i think.

03:23 Oh, then the other thing i guess we can do is q is equal to sigma times a, right? total area is not 2 pi r.

03:38 It's pi r squared.

03:42 Okay, that's kind of a lot of things, but let's plug all those in.

03:46 So v is equal to k times the integral of d q which is sigma 2 pi r prime d r prime all over r which is square root of z squared plus r prime squared oh going back real quick we solved for what q is equal to what i really wanted was to solve for what sigma was which is q over pi big r squared.

04:25 So now we can say the potential is k times you can pull the sigma out.

04:31 Sigma is q over pi r squared.

04:36 We can pull the two pi out.

04:38 It's a constant and then we're left with r prime, d r prime, over the square root of z squared plus r prime squared.

04:50 We're integrating with respect to r prime from zero to big r.

04:59 Okay.

05:00 Let's cancel a few things.

05:02 Pies cancel there.

05:05 So the potential is k, 2kkk over r squared.

05:12 And then this integral is not super fun.

05:18 Probably the best way is just to look it up in a table or something like that...