00:01
Hi there.
00:02
So for this problem, we are told that positive charges are placed at the three corners of a rectangle as is shown in this figure.
00:12
So we need to calculate the magnitude and direction of the electric field at the fourth corner.
00:19
So that means the electric field at this point right here.
00:25
Now, we know that the electric field has the following form that it is.
00:32
Columns constant times the charge divided by the radius square, which is the separation between the two charges or the two points.
00:48
Now, in this case, what we have is three charges.
00:55
We're going to call the charges q1, q2, and q3.
01:02
To identify them.
01:05
Now, the charge q1, it has a magnitude of 6 nanocolums.
01:11
Recall that nano means tens to the minus 9.
01:15
The charge q3 is equal to 3 nanocoloms, and the charge q2 is 3 nanocoloms, and q3 is equal to 5 nanocoloms.
01:28
So, these three will produce an electric fuel that is upward in this point in red.
01:36
Now, the charge q1 will produce an electric field that is to the left.
01:45
This is the electric field produced by 1.
01:48
This imbctor notation, of course.
01:51
And the charge q2 produce an electric field that is upward.
01:59
That is the electric field produced by 2.
02:02
And finally, the electric field produced by the charge, george q3 is along this line that connects these two points.
02:11
So we will have that this is something like this.
02:15
So this is the electric field produced by the charge q3.
02:20
Now what we need to find is also this angle in here that we're going to call teta because that is the same angle in here that it meets with the horizontal.
02:30
So with that we can write the components of this vector of electric field e3.
02:36
Now, with that said, we know that the total electric field at that point is going to be the sum of all of these electric fields.
02:49
So we need to write each of them.
02:53
So we start with the electric field 1, and that is column constant times the charge q1, and this divided by the separation distance.
03:05
Now, we're going to call this separation distance a and this other.
03:08
Separation distance b.
03:10
Now, as you can see from the figure, that separation distance a is equal to 60 centimeters and the separation b is equal to 20 centimeters.
03:20
So in this case, we will have that the separation between the charge q1 and the point b where we are calculating the electric field is the distance a square, of course.
03:35
And then as we, as you can see from the figure, this is pointing to the left.
03:40
So if we said that to the left is negative, we need to add a minus in here because we are going to write an in vector notation because it only has a component in the x direction, in the minus s direction.
03:55
This is just simply in the x direction.
04:00
Now, something similar for the electric field produced by the charge q2.
04:07
So in this case we have columns constants times the charge q2 divided by the separation distance.
04:13
In this case, the separation distance is b.
04:17
So that is b to the square in and, well, in this case it is positive because it is upward.
04:23
If we said that upward is positive in the y direction.
04:27
So this is in the y direction because it is upward.
04:30
And finally, what we need to have in here is the following.
04:35
So the electric field 3 is going to be its magnitude.
04:41
It's going to be a colon constant times the charge q3, and this divided by the separation distance.
04:51
Now, to obtain the separation distance, what we can use is the pythagorean theorem to obtain this distance right here.
04:59
And for that, we obtain that distance will be the square root of a square plus b, square.
05:06
So we put that in there.
05:13
Now because we want to separate this, well, into components, we need to multiply this by each component.
05:21
So the x component is given by the horizontal component of this, which is given since the angle theta, we set it to the horizontal, then that the x component is given with respect to the cosine of theta.
05:38
And in this case, is negative because it is pointing to the left.
05:42
And for the y component, we can see from the figure that is pointed upward.
05:48
So it's going to be positive.
05:50
And that component is given by the sign of theta.
05:54
So sign of theta in the y component, in the y direction, of course.
06:00
Now, what we need to do first is to determine this angle theta.
06:04
But that angle theta, we can obtain it by trigonometry.
06:07
We know that by trigonometry, the tangent of theta is going to be equal to the y component of this triangle, which is 20 centimeters, divided by the x component of this triangle, which is 60 centimeters.
06:24
So we will have that that is 20 divided by 60s.
06:28
And then since we want the angle theta, we apply the tangent of minus 1 to both sides.
06:33
So using our calculator, we obtain a value of 18 .43 degrees.
06:54
So that's what we obtained from there.
06:58
Let me verify.
07:02
Yes, that's the angle.
07:04
Now, we can introduce this angle in here to obtain each company.
07:13
Of the electric field three.
07:19
Well, let me just put it first into the equation for the total electric field.
07:25
As you can see, as i said, you use this equation.
07:28
We just need to add together all of this.
07:31
So we can add component by component.
07:34
So, for example, and this only has only an ex component that is going to add up with the ex component of the electric field three...