00:01
Okay, so we have a non -conducting rod of length l, and it has a charge negative q uniformly distributed upon its length.
00:10
So we're going to put that on a coordinate system, and from 0 to l, this is our rod, and it is negatively charged.
00:23
Okay, so let's write down our givens to begin with.
00:28
We are given that l is 8 .15 centen.
00:33
Meters, we need to put this in si unit, so it's 0 .0 815 meters.
00:41
And you're given that negative q is negative 4 .23 times 10 to the negative 15 colognes.
01:01
So part a is we need to find the linear charge density of the rod.
01:09
Let's recall that the linear charge density is denoted by lambda, so lambda is equal to, in this case, negative q over l.
01:22
Plugging in the given values for q and l that we just discussed, you get negative 5 .19 times 10 to the negative 14 colognes per meter.
01:39
This is your answer for part a.
01:44
Okay, so part b, it wants the magnitude of the electric field produced at point p a distance a from the rod.
01:57
And it gives us a value for a, but let's just discuss this in general first.
02:03
So part b, let's go and look at our diagram one more time.
02:12
So it wants us to look at the electric field at point p, and it says it's a distance a from the rod.
02:22
So this point here is going to be l plus a.
02:26
And it wants the electric field at that point.
02:29
Remember that this is a rod of charge.
02:32
So for us to find the electric field, let's find it in this small infinitesimal area.
02:42
So this is dx.
02:44
And then we can integrate and find it over the whole rod.
02:50
So let's also notice that we are calling this our positive x -axis.
02:57
So the electric field is going to be along the x -axis, and there is no y component.
03:06
So we don't have to worry about the y component.
03:11
Okay, so let us find the magnitude of the electric field.
03:15
We're actually going to find the electric field how it is, and then we're going to determine the magnitude in the direction from that.
03:21
You'll understand what i'm saying in just a second.
03:23
So dx is the infinitesimal length of the rod.
03:27
It is at, of course, since we called it dx, this is x.
03:34
So that is x, i point dx.
03:39
And we know that dq is lambda dx, since lambda is equal to dq over dx as we said before so the electric field this small component dq this can be thought of as a point charge so we can use our equation for a point charge so this is the infinitesimal portion dx it's one over four pi epsilon not and then q, dq, which is lambda dx, over the distance squared.
04:44
So let's look at what this distance should be.
04:49
So this distance needs to be the distance from the point, from the infinitesimal point to p.
05:02
So you need this distance...