0:00
Hi there.
00:01
So for this problem, we are asked about what is the magnitude of the net force on the charge q3 produced by the charges q1 and q2.
00:21
Now, we are told that the charges, all of these charges are the same and they have a magnitude of 5 .6.
00:32
27 times 10 to the minus 8 columns.
00:41
And the separation distance d is equal to 95 centimeters.
00:47
We can write this in meters as 0 .95 meters because we just simply divide this by 100.
00:58
Now with that set, we know that in this case we're going to assume that all of these.
01:05
Charges are positive so the force that are going to produce in the charge q3 are going to be repulsive so let's call this the force produced by two and the force produced by one is also repulsive now as you can see the the x component points to the same direction to the right however, the y component or the vertical component of this force, it is in the opposite direction, so that it will produce that the net force is going to be, is only going to have an x component because the vertical component is going to cancel.
02:05
So now with that set, the vertical, the, sorry, the horizontal component of two, for example, is given by columns constant times the charge q1 and q2 and q3, and this divided by the separation distance.
02:30
Now, the separation distance is this distance right here.
02:37
Now, to calculate that distance, we use the pythagorean theorem.
02:42
And for that, we know that this is just simply, in this case, is the square root of two times the distance d.
02:54
Because we know that this side measures the same side as this d.
02:59
So with that said, we will have that that is the same.
03:02
Separation distance which is the square root of two times the distance d and all of these to the square.
03:09
However, we know that since the vertical component cancels is going to cancel with the vertical component of the force f1, we are just going to be interested in this in the x component of this force and that x component because if we set this angle with respect to the horizontal, then this is going to be given by the cosine of theta.
03:43
Now, with that set, we need to determine the angle theta, but in this case, it's very easy to obtain, because we know that both sides of this measures d.
03:54
So we can use the tangent of theta, so that will be the distance d divided by the distance d, which is equal to 1.
04:02
If we solve for the angle theta, we just need to use the tangent of minus 1 of 1, that we know we produce an angle of 45 degrees...