00:01
So we have two charges q1 and q2 of the same magnitude, 3 .2 times 10 per minus 19 volumes.
00:07
Let's call this charge q.
00:10
And the other charge q3 has twice that much.
00:13
So this is 2 here.
00:15
We'll move this from this point, x equal to 0 to x equal to 5 meters.
00:19
We know that t is equal to 17 centimeters, which is this.
00:32
Now we want to find the maximum and minimum forces.
00:35
Now, let's say the distance is x and this is d.
00:42
So the force due to q1 will be in this direction and the force due to the other one will be in this direction.
00:54
Now, we'll consider the angle theta and because these two forces will have the equal magnitudes as these two charges have the same, as these two charges are the same, the net force will be 2 f pos theta, where f is the force you to each charge.
01:16
Now because each one is f, we can divide it into two components.
01:22
The total magnitude is f.
01:24
This will be f cost theta.
01:26
So we have f sine theta, this one.
01:30
And the two f sine thetas cancel each other out while the two f cost thetas add up to give us two f cost theta.
01:39
And that is it.
01:41
So we know that cost theta is equal to x by square root of x squared plus d squared and we use this so the net force on the charge q3 is simply 2f cost theta so that's 2 times f if you remember is k q by r squared so that's k q1 q3 so q3 is 2 q by r squared and notice that r is the distance of this hypotenous so r is equal to square root of x squared t squared so that's square root of x squared plus d squared times cos theta and cost theta is x by square root of x squared plus b so this is 4 k q squared x by x squared plus t squared whole power 3 by and that is it.
02:49
To see where the net maximum is and the net minimum is, we'll simply need to minimize and maximize this function...