00:01
Hello there.
00:02
We are given two point charges here that are placed along the x axis.
00:07
Q sub 1 is 0 .2 meter from the origin and charge 2, a negative charge is 0 .3 meter from the origin.
00:15
We are to determine the net electric field due to these two point charges at point p here, which is at coordinate 0 .04 meters.
00:28
Neither.
00:30
Okay.
00:31
So already shown in the figure here are the electric field due to charge one and electric field due to charge two.
00:40
So these are the required unknowns again.
00:47
The components of the total field at point p and the magnitude and direction of the total field at point p.
00:57
Okay.
00:58
So first things first, if we look at the definition of the magnitude of an electric field due to a point charge here, we need to determine the distance between the source charge and the point of evaluation.
01:13
So that distance is symbolized by letter r here.
01:18
So in this case, r sub 1 here is the distance between source charge 1 and point b.
01:24
We have a right triangle here whose space is 0 .2 meter and the height is 0 .4 meter.
01:32
Therefore, we can just use pythagorean theorem to determine r sub 1 and r sub 2.
01:38
So let's do it very quickly.
01:40
So r sub 1 here by pythagorean theorem, we just get the sum of the squares of the two sides, so that would be 0 .2 squared plus 0 .4 squared, and this gives us 0 .45 for r sub 1.
01:54
We do the same thing for r sub 2.
01:57
So that would be 0 .3 .3.
02:00
Squared plus 0 .4 squared will give us 0 .5 meter.
02:07
Let's determine the two angles here, angle phi and angle theta, using the trigonometric function tangent.
02:17
So let me just scroll a little higher.
02:21
So we know that phi is just inverse of 10 and then the opposite of phi is 0 .4 and then the adjacent is 0 .2 so angle phi is given as 63 .4 degrees above the horizontal and then we'll do the same for angle theta we'll use inverse tan so the opposite is still 0 .4 adjacent side this time is so this gives us 53 .1 degrees.
03:00
So everything is ready now.
03:02
We can now focus on the components of the electric fields here.
03:07
So electric field due to one is in the first quadrant, hence it has positive x components.
03:15
So let's show the components now.
03:17
This is now the x component of field one.
03:21
And then we have the y component of field one so this must also be angle five let's go this time to field two field two field two is in the fourth quadrant so we know that it has positive x component but it will have a negative y component so for part a now we can get the summation so we have the summation so we have the summation of the field x component that's the unknown so if we look at our free body diogen i'm not free body diagram if we look at the diagram of our fields here we only have two components in the x axis going towards right so this summation of all the field in the x component we just add e sub 1x plus electric field 2 x okay and then same thing for the y component.
04:30
So for the y component, let's write it here.
04:35
Summation of the y component.
04:37
So this will just be one positive.
04:40
So the summation of the field y component, we have positive y component for the first field, negative y component for the second field.
04:52
So we just need to determine now the magnet.
04:55
Of the electric field here, but if we further simplify this, e sub 1x, this would be the hypotenosis e sub 1, and then e1x is adjacent to phi, so this will be cosine pi plus same thing here, hypotenosis, e sub 2, and then we get the cosine of angle theta.
05:22
Okay, we know phi, we know theta, bit more, but we do not know the field strength yet, e sub 1 and e sub 2.
05:31
The same thing here.
05:33
Y is opposite to alpha.
05:35
So this would be field strength 1 times sine alpha minus field strength 2 sine alpha.
05:45
So let's determine first field strength 1.
05:49
This is just k magnitude of q sub 1 over r sub 1 squared.
05:55
The electrical constant here, k is of course equivalent to this constant, 4 pi epsilon subnot.
06:03
This is just, we'll use in one significant figure, 9 times 10 to the 9th.
06:09
And then 40 nanoculum, replace nano with powers of 10, so that's 10 to negative 9, divided by r sub 1.
06:18
This is 0 .45 squared.
06:22
So that field strength 1 is equivalent to positive 1 ,228 neutons per column.
06:32
So i'll show this now as a vector.
06:36
And the direction is 63 .4 degrees.
06:42
So magnitude, direction.
06:44
Let's go to field strength 2.
06:46
So electrical constant is 9 to the 9th.
06:50
And then magnitude of negative 20 is positive 20 nano, couloms divided by r sub 2, that's 0 .5 squared.
07:02
So this gives us as a vector.
07:05
This is 720 new tons per column at negative 53 .1 degrees as shown here.
07:14
Theta is 53 .1 degrees below positive x -axis...