00:01
Okay, we have these two lines of charge, infinite in line, and they're a distance d from each other, and we want to find the electric field at a height, y prime, above the two lines, and at the midpoint of that line d.
00:16
Okay, so there's two ways to approach.
00:19
We could use kelsa's law and find the electric field from the positive and the electric field from the negative, and then add them together.
00:32
Or we could integrate the one and then add the integration of the other.
00:40
Both ways would work just fine.
00:43
The integration method would be significantly harder.
00:46
So let's do gauss's law.
00:50
Okay, gauss's law.
00:58
So this is gauss's law, the electric field, the integral of the electric field with some area da is equal to q.
01:18
Enclosed over epsilon knot.
01:21
Okay, so if we think about these lines of charge going in and out of our page, the symmetry here is cylindrical.
01:34
So if we had this line of positive charge, then our gush and surface would look something like this.
01:54
Hopefully that makes sense.
01:56
So that would be our gowshoot surface.
02:01
Okay, so if we think about the electric field here, coming off of this positive charge, it's going to point radially outward, right? so in all directions, just radially outward, up and down and left and right.
02:22
And then rda from the, from the surface of this cylinder is going to point in that same direction.
02:32
So here could be our electric field and here could be our da.
02:39
So we could also write this as e in the r -hat direction, so the radial direction, dotted with da also in the r -hat direction.
02:54
So since they're both in the r -hat direction, we can just get rid of that dot product the other thing we can do is pull the e out because it's constant everywhere everywhere along this cylinder if you pick any point the electric field will be the same so we can pull it out so we get the integral of d a oops q enclosed over epsilon up okay so we basically get e a is q enclosed over epsilon knot okay q and closed that's our next thing to tackle so we have a line of charge so q and closed is the charge density times the length okay the length is infinite though so is this going to give us trouble it won't because the lengths will actually cancel so the area is the surface area of a cylinder not including the like end caps so 2 pi r l is the surface area of a cylinder excluding the end caps equals lambda l over epsilon not okay yeah so there goes our else so e is equal to lambda over two pi epsilon not r okay and this will be in the r hat direction with respect to that charge itself okay so there's our electric field now if we just think about how this electric field is going to look, we have positive, negative, and here's our measuring point.
05:15
So the positive electric field will point this direction, and the negative electric field will point sort of in this direction, right? so we can see that the y components will cancel, and that the net electric field should be in the plus x direction if we call this direction x then the electric field should be in the plus x direction okay so let's go ahead and get these two fields written down so let's call well let's do the so e1 is the positive charge okay we need to figure out what our r -hat vector is our height unit vector um here's our charge, here's our measuring point, this distance is y prime, this distance is d halves, r hat is the unit vector that points from a charge to the measuring point p.
06:32
So in this case, r hat is going to be the vector that points from the charge to the measuring point.
06:43
So d halves in the x and y prime in the y, divide by, the magnitude of the distance.
06:57
So square root d squared over four plus.
07:06
Okay, that's our unit vector r hat.
07:11
We know that our lambda is positive.
07:13
The other thing we need to figure out is what is our r.
07:17
So that's the like radius of our cylinder, right? so that's the same thing actually as this right here in the denominator.
07:26
It's just that distance from the charge to the measurement point.
07:30
So r is equal to the square root of d squared over 4 plus y prime squared.
07:41
Okay.
07:43
So e1 is then equal to lambda over two pi epsilon.
07:56
And then we get r hat.
08:00
So d halves x hat plus y prime y hat...