00:01
We have three charged balls as shown in this drawing below.
00:06
The red one has charge of plus five and the black and blue charge of minus five.
00:12
We have been labeled a, b, and c.
00:15
So they're all a and b and b and c are one meter apart, b and c are one meter part, and a and c are two meters apart.
00:23
And we're asked about the size of the forces on the balls.
00:28
Okay, so option a, is that ball a experience the largest force.
00:56
Option b is ball b experience is the largest force.
01:12
Option c is ball b experiences no force.
01:23
No net force.
01:25
Option d is ball c experience is no force.
01:36
And option e, the forces on all three are equal.
01:57
So let's write down the formula for the force.
02:07
So the force on a charge due to another charge is equal to the constant k, which is 1 over 4 pi times epsilon not.
02:23
But we'll just write it as k here.
02:25
That's equal to k times the first charge, times the second charge, times r vector minus our prime vector all over the magnitude of r vector, vector, cubed.
03:01
So this may be slightly different than what you've seen before, but r vector is from origin to the place that you're measuring the force.
03:30
And our prime vector is from the origin to, well, i guess we could read it this way.
03:48
Let's write it this way.
03:52
So r vector is from the origin to q1, and our prime vector is from the origin to q2.
04:00
This force gives us the force from q2.
04:23
Yeah, so this formula would give us the force on q2, sorry, the force on q1 because of q2.
04:44
So let's set the charge a at the origin.
04:48
So then charge b would be at one meter, and charge c would be at two meters.
04:54
So let's just start with charge a.
05:01
So the force on charge a, okay, is equal to k and then q1 is a, so plus 5, and then let's do b first.
05:25
So q2 is b, so times negative 5, times r minus r prime vector.
05:33
So r vector goes from the origin to the red charge, which is 0, so 0 i had, and our prime vector goes from the origin to charge two so now we subtract them so subtract one i half because the black charge is at one meter okay and we're dividing by the magnitude of r minus r vector cubed so magnitude of zero minus one then we're going to add the force due to the blue charge so we have k q1 is 5, q2 is now the blue charge minus 5...