WEBVTT
00:00:02.110 --> 00:00:12.570
Point charges π one equals positive 50 microcoulombs and π two equals negative 25 microcoulombs are placed 1.0 meters apart.
00:00:13.090 --> 00:00:22.990
What is the magnitude of the force on a third charge π three equals positive 20 microcoulombs placed midway between π one and π two?
00:00:23.520 --> 00:00:31.630
We want to solve for the magnitude of the force on charge π three, what weβll call magnitude πΉ sub π three.
00:00:32.100 --> 00:00:44.940
Weβre told the distance between π one and π two, 1.0 meters, which weβll call π, as well as the values of all three charges π one, π two, and π three.
00:00:45.630 --> 00:00:49.450
We can start our solution by drawing a sketch of this situation.
00:00:49.930 --> 00:00:56.680
We have three charges π one and π two spaced a meter apart, a distance weβve called π.
00:00:57.190 --> 00:01:04.920
In the middle of that distance, charge π three is placed, π over two away from both π one and π two, in between them.
00:01:05.370 --> 00:01:12.910
We want to solve for the net force magnitude thatβs acting on charge π three created by π one and π two.
00:01:13.500 --> 00:01:21.890
We know that this magnitude will equal the force of π one on π three plus the force of π two on π three.
00:01:22.500 --> 00:01:32.530
As we consider the force between charges π one and π three, if we look at the signs of both of those charges, we see theyβre both positive.
00:01:32.850 --> 00:01:35.060
More importantly, they have the same sign.
00:01:35.340 --> 00:01:37.500
And therefore, they will repel one another.
00:01:37.960 --> 00:01:48.970
If we draw in force vectors, which show which way the force on π one and π three points from their mutual force, we see that π one pushes π three away.
00:01:49.490 --> 00:01:53.020
Now letβs consider the force of π two on π three.
00:01:53.400 --> 00:01:56.840
Looking at those charges, we see that π two is a negative charge.
00:01:56.840 --> 00:01:58.580
And π three is positive.
00:01:58.990 --> 00:02:01.860
That means the force between them will be attractive.
00:02:02.380 --> 00:02:07.230
So π two will be pulled towards π three by some force amount.
00:02:07.610 --> 00:02:10.980
And likewise, π three will be pulled towards π two.
00:02:11.510 --> 00:02:15.710
We donβt know the relative magnitudes of the forces acting on π three.
00:02:15.970 --> 00:02:29.010
But we do know that they point in the same direction, which means that their total, the magnitude πΉ sub π three, will indeed be equal to the sum rather than the difference of those two forces.
00:02:29.730 --> 00:02:52.630
In order to calculate πΉ sub π one π three and πΉ sub π two π three, we can recall Coulombβs law, which says that the electrostatic force between two point charges π one and π two equals the product of those charges times Coulombβs constant π all divided by the distance between the charges π squared.
00:02:53.260 --> 00:03:06.280
If we start out with πΉ sub π one π three and write out Coulombβs law for those two point charges, we see itβs π π one π three divided by π over two quantity squared.
00:03:07.000 --> 00:03:14.690
In this solution, weβll treat π as exactly 8.99 times 10 to the ninth newton meters squared per coulomb squared.
00:03:15.210 --> 00:03:35.040
If we then write out the force of π two on π three, then as we look at the expressions for these two forces, we see that they both involve factors of π and are divided by π over two quantity squared, meaning that, in our equation for the magnitude of πΉ on π three, we can factor those terms out.
00:03:35.640 --> 00:03:44.560
And looking further at this expression for the magnitude of the force on π three, we see that π three, the charge value, is common to both terms.
00:03:44.560 --> 00:03:46.470
So we can factor that out as well.
00:03:46.980 --> 00:03:55.130
Now that we have this simplified expression for the magnitude of the force on π three, before we plug in, we want to do one last thing.
00:03:55.570 --> 00:04:00.460
Recall that we figured out that both forces on π three act in the same direction.
00:04:00.980 --> 00:04:04.890
That means we indeed want π one and π two to add together.
00:04:05.490 --> 00:04:16.330
Since π one and π two have opposite signs, to guarantee that they do add together, letβs put absolute value bars around those two charge values.
00:04:16.710 --> 00:04:35.770
Now when we plug in to solve for magnitude of πΉ on π three, plugging in for π π one π two π three as well as distance, when we calculate this value, we find that, to two significant figures, it equals 54 newtons.
00:04:36.450 --> 00:04:40.180
Thatβs the total magnitude of the force on π three.