00:01
For this problem, we have two charges, charge one with a total charge of three microcouloams, and charge two with a charge of negative four microcoules, and we're given the x and y positions of each of those charges.
00:17
And first, we want to figure out what is the magnitude f and direction theta of the force between those two charges.
00:28
So we'll start by plotting out where these charges are located on the x, y, plane, just because that will give us a visual, which makes it easier to see what's going on and how these charges are related to one another.
00:47
So charge q1 is going to be over here in the first quadrant, and charge q2 will be over here in the first quadrant, and charge q2 will be over here in the.
00:58
Second quadrant.
01:02
In order to find the magnitude of the force, we need to use kulalms law, which is that the force will be 1 over 4 pi epsilon not, times the magnitude of q1, times the magnitude of q2, all divided by the distance between them squared.
01:22
In order to figure out the distance between them squared, we want to consider the triangle, the right triangle that's formed using q2 and q1, where one side has a length of delta y, which would in this case be one centimeter, and the other has a length of delta x, which in this case is negative 5 .5 centimeters.
01:51
Then the hypotenuse is d, which would be equal to the square root of delta x squared plus delta y squared by the pythagorean theorem.
02:06
Therefore, we can replace that in our expression for kulom's law and just simply divide the product of the charges by delta x squared plus delta y squared.
02:24
Now we have all of the information that we need.
02:27
You can go ahead and plug in all of the values that are known for these.
02:32
Variables and you would get that the magnitude of the force is 34 .5 newtons.
02:42
Next we want to figure out the direction and the direction is typically given by inverse tangent.
02:50
So theta is going to be tangent inverse of delta y over delta x.
02:56
And so again if you use a one centimeter for delta y negative 5 .5 centimeters for delta x, you get an angle of negative 10 .3 degrees.
03:08
So the magnitude of the force is 34 .5 newton's.
03:13
The angle is negative 10 .3 degrees.
03:17
Now we want to consider adding a third charge, q3, with a magnitude of four microculems.
03:30
And we want to place q3 such that, that the net charge felt by the, sorry, the net force felt by charge two is zero.
03:42
We already know that charge two is going to be attracted to charge one.
03:50
And so in order to balance that out, it needs to feel a force in the opposite direction from charge three.
03:58
And because charge two has a negative charge and charge three has a positive charge, be attracted to each other so we would expect charge 3 to be placed somewhere over here in the second quadrant.
04:15
But we want to figure out the actual x and y positions of charge 3.
04:24
I'm going to erase the previous work to make a little bit more space for this.
04:32
And there are a few different ways to go about figuring this out.
04:44
I'm going to consider the x and y components of the force separately because the sum of the forces on charge 2 in the x direction must be equal to 0 and the sum of the forces on charge 2 in the y direction must be equal to 0.
05:05
Excuse me.
05:06
If we are looking first in the x direction, charge 2 will feel a force from charge 1 and from charge 3.
05:14
So we know that the magnitude of the force felt from charge 1 in the x direction must be equal to the magnitude that it feels from charge 3 in the x direction.
05:28
And again, we'll use kulalm's law to find the magnitude of those forces.
05:33
So we would end up with 1 over 4 pi epsilon knot, q1, q2, and in this case we're just considering the x direction, so this would be over delta x squared...