00:01
Okay, so for this problem, we have two charges.
00:04
One's at the origin.
00:06
I'll call this q1, and then another one is over here, 2 .4 meters away.
00:14
So i'll call this distance d, and it has a value of 2 .4 meters, and then q2.
00:23
And then the magnitude of q1 is 2 .5.
00:30
Times 10 of the minus 5 coulombs.
00:36
Coolums.
00:36
Oh my gosh.
00:40
Let me fix this.
00:42
See, make it look a little better.
00:46
Okay.
00:48
So that's the magnitude of q1, and then the magnitude of q2 is 5 times 10 of the minus 6.
00:56
Q2 is negative, but i'm just going to kind of put a negative here and a positive here.
01:00
So we can remember that.
01:03
And then just write down its magnitude.
01:06
And our goal is to find where the electric field is zero.
01:10
I'm just going to double check that that's the prompt.
01:12
Yes, indeed.
01:16
Okay.
01:17
So originally i was like, uh -huh, there's nowhere it's zero because i know the field due to a dipole, and this is basically a dipole, and it's not zero anywhere.
01:25
But alas, q2 is in q1 or not equal, so it's not truly a dipole.
01:32
It's a dipole plus a monopole, which does, which i did find a place where it's zero.
01:39
So let's try to examine like the candidates for where it could possibly be zero.
01:45
So let's note that q1 is bigger than q2.
01:49
And so the field due to q1 is going to be bigger than q2.
01:52
So let's kind of go through and figure out, first of all, in broad strokes, where the field is zero.
01:57
So it could either be to the left of q2 between the two or to the right of q2.
02:01
Let's eliminate the easiest one first.
02:04
So let's consider between here.
02:07
So i'm going to draw these vectors and green to represent the electric field due to each one.
02:11
So the electric field due to q1.
02:15
It's to the right because the electric field is always point away from point charges.
02:20
And then if they're positive, and then towards the point charge if it's negative.
02:25
And so that's from the positive charge.
02:28
And then here's the contribution from the negative charge.
02:32
Which is going to be a bit smaller because it's a smaller charge, but nonetheless to the right...