00:01
So we have a problem that is dealing with electric fields and in this case the electric field effect of static charges on each other and on another charge that will be introduced as describing the problem.
00:19
So initially we have two charges.
00:23
So by describing the problem if this is the y -axis and this is the x -axis and this is the x -axis, there is a negative charge in the origin, which is qa, and there is a positive charge on a distance from it, qa plus.
00:53
And then the third charge is introduced at distance from both of them, which is unknown, and we need to find the distance.
01:03
This is qc.
01:06
And we need to see first we need to see if it is even in this situation of three charges is possible in order, if we say that these three charges must stay in the equilibrium, is it possible to even place a third charge in this manner and for the system of charges to stay at the force equilibrium? and of course, if it is possible, then in how many ways.
01:41
So initially these two charges, q8 and qb were in equilibrium because they exerted four.
01:48
Forces on each other that were on that were at equal magnetions and of course we know that the force is proportion to the field electric field and electric field itself is proportional to the square of a distance between the charges so there was a specific difference are where two forces that one force coming from the qa fields and another force coming from the qb fields became equal.
02:23
So the repulsive, the attractive forces actually became equal.
02:27
And they were basically the exactive forces that are equal.
02:32
And when the third charge is placed on the right side, now the force which both positive charges exert in this case, of course is greater than the force than just of ql and qc.
02:53
So if the charge is placed between qa and qb, there will be net force in qc, of course, because they will be pushing it from both sides in positive and negative direction.
03:10
So the only possibility for another equilibrium will be if the charge qc is placed like this on the side, in this case the left side of the negative charge.
03:30
The net force can also be zero in the x direction, which was the first case in the beginning.
03:39
So yes, it is possible, the answer for the first part of the problem, yes, it is possible.
03:45
And the way is only this way.
03:47
Every other way would cause no zero net forces and it will cause some motion, both horizontal and vertically.
03:57
Now we need to actually express this in mathematics and say that, the force ac, which is from point charge a to point charge c.
04:09
In this case, must be equal to the force b, c from b to c.
04:14
Of course, equal charges repel each other and heterogeneous charges attract each other.
04:22
So this force, we can substitute, in this expression, we can substitute the regular expression for electrostatic force.
04:36
Which is a trostatic constant times.
04:39
And now we're only looking at the intensity.
04:42
So intensity must be zero.
04:44
We'll already explain for the signs and the directions.
04:48
But in this case, since discharge will attract and discharge will repel, for this charge and the whole system to be in equilibrium, the repelling and the traction must be the same.
05:01
So we are looking only at the intensities of the force.
05:05
We are not looking at the signs of the force.
05:09
So basically qa, qc, x squared, which is the distance here, must be equal to the new distance, which is, we know this is distance 0 .15 meters or 15 centimeters.
05:30
Now this plus x will be the distance between qc and qb.
05:35
So the tristatic constant charges qb, and qc over x plus 0 .15 and all this squared...