00:01
Hi, so let's go ahead and take a look at this problem.
00:03
So this problem, in this problem, we're giving a rod of a certain length, and it also has a particular charge, where in this case the charge is negative 22 microculums.
00:14
We need to determine the magnitude and the direction of the electric field along the axis of the rod at a point 36 centimeters from its center.
00:25
So first thing to do, let's go ahead and consider the electric field, direction.
00:31
So this point here, point p, that is what we are, that is the location that we are solving for.
00:38
So they give us that it's that a here is equal to 36 centimeters or 0 .36 meters, right? we also know that q is equal to negative 22 times 10 to the negative 6 coulams.
00:54
And since q is a negative charge, we know that the charges to the left of point.
01:00
P will incite an electric field pointing to the left.
01:04
And that's because it's along the axis as well, since this is an entire body that's to the left of p.
01:10
We know that all the electric field's going to point to the left, so we can just immediate know that e points to the left.
01:21
Next thing we want to consider is how we're going to solve for the electric field.
01:27
So let's go ahead and first look at the electric field equation, which i've written as kq over x squared.
01:33
You might have seen it written before as kk uv or r squared, but that's still the same.
01:37
I'm just using x to designate the length from any point on the rod to the point p.
01:45
So this could be x, this could be x.
01:49
It just matters what we're doing within the integration, but it's any point along the rod to the point p.
01:57
So to integrate this equation, first let's go ahead and implicitly differentiate it.
02:05
So this will kind of explain where we get the integral that we set up.
02:10
So if we take the derivative on both sides, first i'm going to take the derivative with respect to e, and then i'll take the derivative with respect to q.
02:19
So e, the derivative of that is 1 times de.
02:23
You can think of it that way.
02:24
And the derivative of q is also 1, but all that'll be implied.
02:28
So we're going to have k over x squared dq.
02:33
Now this is an equation that we can go ahead and integrate.
02:36
So if we were to take a general integral on both sides, de is equal to the integral dq.
02:45
Then the integral of de is just e, which is what we're solving for.
02:49
And now we have an integrable equation k over x squared dq.
02:57
So this is where we get our integral that we are going to use.
03:00
And now we just need to figure out what the limits are for x.
03:07
Also, you may notice that in this equation, we will actually have to do one more modification, that we don't have any variable of dq that we can integrate by, but we do have x.
03:20
So we know that lambda, the charge density, is a small change in charge over a small change in length.
03:28
That's what we can view this as.
03:30
So if we rearrange this equation, we'll have that dq is equal to lambda dx.
03:36
And if we go ahead and substitute that in here, we'll get the electric field is equal to k over x squared or k lambda over x squared dx.
03:50
So this will be the final equation that we're using, and these will be the limits that we're going to use.
03:57
We don't have any limits for charge necessarily.
03:59
We have limits for position which we are going to figure out, whatever a and b are...