00:01
In this problem, we have two point charges located at the origin and x equals 4 meters.
00:10
And we want to know the electric field at p1 first and then p2, second.
00:17
So what we do with electric field, by definition, we look at the force on an imaginary unit positive charge at the location we are interested in.
00:25
So we just look at the forces.
00:28
So q1 being positive, that means it would be a problem.
00:31
Propulsive force.
00:33
So this is what we would define as e1.
00:35
Q2 is negative.
00:37
So it's attractive.
00:39
That also would point in positive x.
00:42
So there would be our electric field at p1.
00:47
Both pointing to the right, so your electric field would be to the right.
00:51
So let me just get the components, x components, good enough.
00:55
Strip off the sign and you have the magnitude.
00:59
This would be e1 plus e2, which is kq1 over.
01:09
This is going to be, now, remember, this is the distance from the charge to the point under observation.
01:15
So this would be xp1 squared plus kq2.
01:21
Now, we want this distance here.
01:25
This would be our, this is our r2, this is our r1.
01:31
So r2 is going to be x2 minus xp1.
01:37
So x2 minus xp1 squared.
01:42
And so at this point we can just put in our numbers, 8 .99 times 10 to 9, newton, meter squared, coulum squared.
01:54
And we have 10 times 10 to minus 9, koolomes over 2 meters squared plus 20 times 10 to the minus 9 koolomes over 2 meters squared also.
02:09
And this works out to be 67 .4 nudes per koulomb.
02:15
So that's the x component.
02:16
And that's the only component we have.
02:18
So that is the electric field at p1.
02:25
Part b wants it now at p2.
02:30
So we do the exact same thing on our imaginary unit positive charge.
02:35
Q1 is propulsive.
02:39
Now, don't read anything into my links and my arrows.
02:41
I'm just giving you a general direction.
02:45
This obviously should be smaller than that, but i'm not doing anything to scale.
02:51
Q2 is attractive.
02:53
So now that's this way to the left.
02:59
So now we're going to have subtraction.
03:02
So e net x now, so e net x is e1 minus e2.
03:16
E minus e2 is the x component of e2 vector.
03:22
So that's, and this will be the same type of thing.
03:26
So 8 .99 times 10 to the 9, new to meter squared, coulom squared.
03:33
We can factor that out.
03:34
Then we got q1, 10 times 10 to the minus 9, coolomes.
03:38
Now the distance, this would be the r1, so that's just again, this is xp2, xp2 squared, plus 20 times 10, or actually minus.
03:55
This is a minus sign, minus.
04:05
So minus 20 times 10 and minus 9.
04:07
Now this distance here would be the r2, and that would be xp2 minus x2.
04:16
So this would be xp2, which is 6, x2 is 4, so this is 2 meters squared, and i'll put in the number here, this is 6.
04:37
And this works out to be minus 42 .5 nons per coulomb.
04:43
So the electric field is magnitude 42 .5, but it points the negative x direction to p2.
04:52
So that's the first part.
04:55
Now, we have a question here where we now have the point under observation p3 is now on the y axis, 4 meters along the y axis, x2 is still at 4 meters, q1, but the charges are different, but both positive charges.
05:12
So, got a little work to do.
05:15
We think of our imaginary unit positive charge there, so both positive, so this is repulsive, this is e1, now q2, also positive.
05:30
So we'll have along the line connecting, we'll have this fourth here.
05:37
This is going to be e2.
05:40
This distance here is r2.
05:44
This distance here is r1, which is equal to yp3.
05:51
Now we're going to need this angle theta.
05:53
So we got a few things to get.
05:54
Let's get theta first.
05:57
Tangent theta is r1 over x2.
06:02
So theta is inverse tangent, you know, opposite or adjacent.
06:07
Inverse tangent, four meters, four meters, and this gives me 45 degrees.
06:15
R2, hyponus of the right triangle, is going to be x2 squared.
06:22
This is x2 here, this is x2, x2 squared plus r1 squared, and this is a square root, 4 meter squared plus four meters squared, and this is 4 times the square root of 2 meters.
06:41
I'm not going to calculate that as a decimal number because it's going to end up being squared, so it comes out nicely.
06:48
So those are all our extra things.
06:52
Now, this angle here, if this is theta, this is theta.
06:58
So we want e net x.
07:01
E net x is going to be just e2x, which is minus e2, cosine theta because here is e2y and here is e2x.
07:18
So it's in the negative x direction.
07:20
So this works out to be minus 8 .99 times 10 to 9, newton meter squared, coulome squared, 36 times 10 to minus nine coulomes, it's nano coolomes, so 10 minus nine, cosine 45 degrees over our 2 squared, which is, if you square this, you get 16 times 2, 32.
07:51
So 4 square root of 2 meters squared.
07:56
And this works out to be minus 7 .15 nons per cullo.
08:02
So that is the x component.
08:07
Now the y component, e -net y, we have.
08:14
E1 in positive y.
08:16
So we got e1 plus e2 sine theta, that is e2y.
08:23
And this would become 8 .99 times 10 to 9, newton meters squared, coulom squared.
08:34
And we got 60 times 10 to the minus 9 kuloms over 4 meters squared, the r1 squared, plus 36 times 10 of the minus 9 kuloms, minus 9 kouloms times sine of 45, which actually if i had sine 45, cosine 45 are the same, but we want to be correct about overriding and for general, over 4 times a square root of 2 squared.
09:07
And this works out to be 40 .9 nudence per koulon...