00:01
In this problem, you have charges all the same positive value q, 1 microculeome at a, e, c, the corners of a square of side 4 centimeters.
00:15
And the problem wants to know what would the charge you have to be at d so that the force on b would be zero.
00:25
So let's look at this.
00:27
Actually, before we do that, let me give you d is the diagonal.
00:31
So d is going to be square root, l squared, plus l squared.
00:40
So that's going to be l times the square root of 2.
00:47
Let's look at what goes on with the forces on a, b, a on b, and c on b.
00:58
Q is positive, so it's positive, positive.
01:01
It doesn't matter even if they were negative, negative.
01:03
The charges are the same at a, b, and c, so they're always going to be repulsive.
01:08
But in our case, we got positive, positive, repulsive.
01:12
So this would be f -a -on -b.
01:20
Now, c on b, again, that's q and q, two positives.
01:26
Again, but if it was two negatives, it would be the same thing.
01:28
So either way, it's repulsive.
01:33
So here it would be f -c -b.
01:38
So those are two forces.
01:39
Now, cbb, f -cb, f -a -b, and magnitude are exactly the same.
01:45
The charges are the same.
01:47
The distance between c and b and a and b are the same.
01:51
So they're exactly the same length.
01:54
So if f -a -b is 100 -nuton, so is f -c -b.
01:57
Not the same vector, but they're the same magnitude.
02:01
So if i were to look at this graphically, here would be using the triangle rule where i move f -c -b, so tail of fcb is at the tip of fab and add them together.
02:13
That's how we add vectors with the triangle rule.
02:19
So this would be fa plus c, i'll call it net.
02:27
It's not the whole thing, but it's the, but it is what a and c are doing on b.
02:34
Now, what is this angle in here going to be? if you've got a right triangle where the two sides are exactly the same, what's the angle? 45 degrees.
02:51
Isn't this, isn't this 45 degrees? because you got l and l.
02:58
So you got, you got fcb and fab the same.
03:04
So this is 45 degrees also.
03:07
So the net of a and c is along the diagonal.
03:15
So we need to have a force, we need to have a q that's going to knock that out.
03:21
Should it be positive? let's talk about sign first.
03:23
Should be positive? should be negative? if q, is positive, then it'd be positive, positive, that'd be repulsive.
03:29
So it would be along the same line as f net, a plus c.
03:33
Well, that's not going to cancel out.
03:35
It's got to be along the opposite to 180 degrees from that direction.
03:41
So q must be negative.
03:43
It's got to be an attractive force.
03:45
So we know q has to be negative.
03:48
So we need this called fdb.
03:57
That's what we need to have.
03:59
It's got to be the exact, that's the only way.
04:02
You got to be in the opposite direction to cancel out.
04:05
You can have the same magnitude doesn't do any good if you're in the same direction.
04:09
I'll be the exact opposite.
04:12
So, we know the sign.
04:15
So now we just got to look at the magnitude aspect of it.
04:17
It's got to be the same.
04:21
Fdb is going to be the magnitude.
04:26
The magnitude of a plus c net, which is the length of this vector.
04:33
That's what we have.
04:35
Now, could you do this where you look at fdbx and fdby and look at the connection we're there? certainly, it's going to lead you to the exact same answer.
04:50
But you don't need to go to the components here because it's along the same line.
04:57
You just, we found, we know from that information the q must be negative, but also we need the magnitude to be the same...