00:01
Hello, for the system shown in the screen, we have to calculate the following parameters which are listed here.
00:08
But first, prior to the calculation, we have to calculate the geometry of the system.
00:13
As we know, these distances from 1 to 2 and from 2 and 3 are l, as well as the distance from 0 .2 to point p.
00:21
So now let's modify this drawing a little bit.
00:25
So we'll connect points p and one and we'll connect points three and point p.
00:34
And we'll also draw a line through the point p which is perpendicular to the line 2p.
00:41
So now we have to, we will label several angles.
00:46
We'll label this angle, let's lay angle one, this angle two, this angle will be angle two, this angle three, then angle four, and then angle five.
00:55
So now we have to calculate these five angles.
00:59
Although it looks a little bit complicated, it's not in reality.
01:04
So let's look at the triangle 1 -2 -p.
01:09
Sorry.
01:10
Yeah, from triangle 1 -2 -p, we see that the angle 1 -2 -p is 90 degree, and the distances between points 1 and 2, and distances between points 2 and p are the same.
01:27
Same and equal to l.
01:29
Therefore, the angle 4, which is this guy, equals to 45 degree.
01:37
And the distance between points 1 and p equals to l multiply by square root of 2.
01:48
Identic, yeah, in the same way from triangle 3 to p, where angle 3 to p is 90 degree, this guy.
02:04
Sorry, the distance between points 2 and 3 equals to the distance between points 2 and p, which is l.
02:21
And according to the same principle, angle 5 is 90 is 45 degree.
02:30
And the distance between points 3 and p equals to l multiplied by square root of 2.
02:37
So these are two important parameters which we need to know for the later calculations.
02:44
So this guy is l multiplied by square root of 2.
02:47
So is this length.
02:50
Okay, we found two angles.
02:54
We have to find three more.
02:57
So angle 3 plus angle 4 equals to 90 degree because this is perpendicular.
03:06
So therefore angle 3 3 plus angle 4 is 90 degree therefore angle 3 equals to 90 degree as well sorry to 45 degree as well so now we have to calculate angles 2 and angle 1 how would we do it so basically in a in the following way the angle p 3 2 equals to the angle 2 because they are basically angles formed by two parallel lines and one common line.
04:03
So it equals to 2.
04:05
And from the triangle p3 2, this angle is 45 degree.
04:12
So it means that angle 2 is 45 degree.
04:19
So angle 2 equals to 45 degree and angle 1 equals to 90 minus 4.
04:28
Angle 2, which is 45.
04:33
So let's summarize our finding for now, which is important for this task.
04:40
Angle 1 equals to angle 2, equals to angle 3, equals to angle 4 equals to angle 5, equals to 45 degree.
04:51
So now we did a very important thing.
04:54
We calculated the geometry of the system.
04:57
Now let's redraw it and we can proceed answering this question so basically now we can calculate the electric field again so here are our points one two and three and this is point p so uh charge first charge is negative second charge is positive third charge is positive so now let's uh let's call let's let's show each individual component of the electric field at point p.
05:35
The first charge is positive, it's negative, therefore the electric field created by the first charge will be directed to this charge.
05:43
So this is e1, e2 is, yeah, second charge is positive, therefore the field will be directed from the second charge, so it will be directed upwards.
05:57
And the third charge is positive as well.
06:03
Therefore, it will create a field which is directed from the search charge.
06:11
So this is e3.
06:14
Now we have to introduce the coordinate axis.
06:18
So we can simplify our life a little bit and direct x to the left.
06:27
Just as we see, both e3 and e1 have a component which is directed to the left.
06:33
But the choice of the direction of x -axis is not principal here because it will only change the sign but not the absolute value and y -axis is directed upwards so now let's write down the formula which control the total field at point p this is a vector sum of the individual electric field created by each charge or if we talk sorry if we if we are interested in the x projection, we have to take the x projections of the first component, of the second component, and of the third component.
07:24
So as we can observe, the second component is perpendicular to x -axis, therefore its contribution will be 0.
07:33
So this guy equals to 0, and this equation simplifies to e1x.
07:39
Plus e2x, sorry, e3x.
07:43
Now let's calculate e1x and e3x.
07:49
E1x equals to e1 multiplied by, so we need this angle, which is angle 3.
08:05
So, x component of e1, this length, equals to e1, multiplied by cosine of angle 3.
08:13
Cosine of angle 3, which is e1, multiply by cosine of 45 degree, which is e1 divided by square root of 2.
08:29
Right.
08:32
E3 equals to e3 x equals to e3 multiplied by cosine of angle 2, because we need this length, or it equals to e3, multiplied by cosine of 45 degree, which is e3 over square root of 2.
09:04
So therefore, ex at point p equals to e1 over square root of 2 plus e3 over square root of 2, which is 1 over square root of 2, multiply by the sum of e1 and e3.
09:25
Now we have to calculate e1 and e3.
09:29
E1 equals to k, a larger constant, multiplied by q1, divided by the distance between points 1 and p squared.
09:46
Or it equals to k multiplied by, divided, multiplied by q1, right, multiply by this guy squared, which is 2l squared.
10:03
In the same manner, e3 equals to k multiplied by q3 divided by the distance, by square distance between points 3 and point p, which is also 2 multiplied by square root of 2.
10:27
So this guy equals to kq3 over 2l square.
10:32
Now we can complete the calculation.
10:36
So, x at point p equals to 1 over square root of 2, multiply by the sum of kq3 over 2l square plus, sorry, kq1 over 2l square plus kq3 over 2l square.
11:01
So it equals to kk2 over 2l square.
11:07
2 square root of 2 multiplied by l squared and multiplied by the sum of q1 and q3.
11:17
Let's calculate this number.
11:25
K equals to 9 multiplied by 10 power by 9 newton's multiplied by meter squared divided by coulum squared over 2 squared of 2, multiplied by 0 .2 .00 meters squared multiplied by the sum of the charges which are 2 and 2 and 6.
11:52
So here, yeah, i have to state that we take an absolute value of q1 because we've already taken into account the direction of the field...