00:01
In this problem, we're asked to find the resultant force on a point charge due to two other point charges.
00:08
So to begin, i've kind of drawn out the situation here and also some equations that we're going to want to be using.
00:17
So first of all, let's kind of think about what our expected result is going to be here, right? so let's start with this two microculum charge.
00:28
What kind of force is it going to produce on the seven microculum? well, we expect, because they're both positive, right? it's going to be pushing, the two microcrum is going to want to push it away from itself, right? so we expect it to kind of be in this direction that i've drawn here.
00:48
Now, what about our negative point charge, negative 4 microculum? well, since it's negative, and we're talking about positive point charge, kind of force that has on a positive point charge, it's going to want to push it towards itself.
01:02
And i've kind of over -exaggerated this vector here, just to demonstrate the sense this is a negative 4 microculum, right? the magnitude is greater than that 2 microculem, so it will have a large effect on it, right? keeping in mind this equation down here for force, the magnitude of the charge has a direct effect on our force, right? and we can make that kind of overarching claim because the r is the same in both these cases, right? they're both .5.
01:40
So what kind of, when we find the resultant force, right? so when we add these two forces together, what do we kind of expect? well, something sort of like that, right? going out in the positive x direction and a little bit down in our y direction.
02:01
So now that we kind of know what we're expecting, let's go ahead and solve the problem.
02:08
Before we solve it though, i did want to point out this little drawing i made here.
02:14
Basically, since we're going to be finding the x and y components is easy to kind of start off by actually finding those vectors for our radius.
02:24
So our radiae are going to be 0 .5, right? but then the y magnitude will be around 0 .433, and you can find that using some really simple trig.
02:36
And then our x will be 0 .25 because it's an equilateral triangle, right? so it's just half of that 0 .5.
02:43
So make sure you keep this in mind.
02:46
Similarly, for the negative charge, it would be the same, but our 0 .25 would be negative, right? all right? so, let's begin.
02:57
So force net on, let's just call it 7 microculum equals force.
03:12
So we have the 2 microculum on 7 micrculum plus our force from the negative 4 microculum on the 7 micrachulum.
03:26
And again, very important.
03:28
These are all vector quantities, right? so let's start with the force from the two microculum.
03:40
So following our equation right here, we have our k out in front, whose value is 9 times 10 to the 9th, which we'll plug in later, times q1, so that can be our 2 microculum, so 2 times 10 to the negative 6, times 7 times 10 negative 6.
04:03
And then we're dividing by the magnitude of our radius, so it was 0 .5 squared.
04:10
And then we have this bit right here, our r hat.
04:14
So r hat is equal to the vector r, right, divided by r magnitude.
04:21
So i'm going to split it up into i and j hats because i think that's the easiest to see makes most sense.
04:30
So we have 0 .25i -hat plus 0 .433j hat.
04:40
Now this is a rounded value.
04:42
So when you're actually calculating at the end, you're going to want to plug in the actual value, not the rounded value, but for sake of writing this out, i'm just using the rounded value...