00:01
In the problem, there are three charges that lie along a line, q1, q2, and q3 separated by the distance as shown.
00:10
Now, let's say there is a point p directly above q2 at a distance of 8 .0 centimeters with an electric field 4 .70 times 10 to the 6 newton per column.
00:26
Now, if q1 is equal to positive 4 .00 microculum, the question is, what would be the charges of q2 and q3 for this electric field to exist? now, we know that a positive test charge can be placed at a point p to describe the electric field there.
00:51
So let's place positive sign here and a positive sign for q1.
00:57
Since both of the charges are positive, the force between them would be propulsion.
01:05
Hence, the electric field would be directed upwards and to the right for q1.
01:18
Since q2 is directly below the point b, there won't be any x component for the electric field.
01:29
So it could be either upward or downward.
01:37
Now let's skip q2 and go to q3 first.
01:43
Now, since q2 doesn't have any x component, the only electric field that could counter this eastward electric field from q1 would be from q3.
01:55
So we need some force to the left.
02:00
And in order to have a force to the left, q3 and the positive test charge should repel each other.
02:09
And that is only possible if q3 is positive.
02:15
Hence, there will be an upward and to the left force from q3.
02:24
Now we've already figured out what should the direction of the electric fields be along the x component.
02:33
Now, if q1 and q3 both exert an upward electric field on the positive test charge, we are only left with q2 having a downward electric field, since the resultant electric field at point f is directly downwards.
02:55
In order to have that electric field for q2, it should attract the positive test charge.
03:03
And that would only be possible if q2 is negative.
03:10
Therefore, in order to have a directly downward electric field at point p, q1 should be positive, q2 should be negative, and q3 should be positive, which is case number 4.
03:37
Now, the next question is to calculate for the value of, q3.
03:43
To solve for the charge q3, let's go back to our diagram q1, q2, and q3.
03:53
The distance between q1 and q2 is 5 .0 centimeters, while the distance between q2 and q3 is 10 .0 centimeters.
04:04
The test charge is located directly above q2, and let's say that the test charge is positive.
04:13
Now, we know that q1 is equal to positive 4 .00 microculum, and q3 is still unknown.
04:24
But we do know that the electric field at point p by q1 is an upwards and eastward vector, while the electric field from q through at point 3 is an upward and westward vector.
04:51
And since the resultant electric field at point p is purely downward, or in other words, there is no x component to it, we know that the x components here at point p should negate each other.
05:07
Or in other words, the summation of electric field at point p in the x component should be equal to 0.
05:18
Now, there are only two charges that has a contribution to this x component here, because q2 only has a vertical electric field.
05:32
Now, at point p, we could get the x component of the electric field 1, and we could get the x component of the electric field.
05:43
And that should be equal to 0.
05:46
But since we know the direction of e3 is upward and westward, we could place a negative sign here.
05:59
And by the formula for electric field, we know that it is equal to kq over r squared, where r is the distance between the point or the test charge and the source charge.
06:18
So we'll have kq3 over r3 squared is equal to zero.
06:27
But remember that we are looking for the x components of the electric field.
06:33
So we'll have to include some factor in our electric field formula here.
06:39
And that would be cosines of an angle.
06:43
So we'll have kq1 over r1 squared, cosine tasease, 1 minus kq3 over r3 squared cosine theta 3 where theta 1 is the angle the electric field 1 makes with the horizontal and theta 3 is the angle that the electric field 3 makes with the horizontal.
07:15
Now how do we get this theta 1 and theta 3? now remember our soft.
07:21
Or our trigonometric functions, we know that tangent theta is equal to the opposite side of the triangle over the adjacent side of the triangle.
07:36
Let's say we want to look for theta 1, that would be the tangent inverse of the opposite side of theta 1 is the vertical line with length of 8 .00 centimeters.
07:52
While the adjacent side from q1 to q2 is 5 .00 centimeters.
08:02
Solving for this will get theta 1 as 57 .994 degrees.
08:14
The same can be done for theta 2, that is the tangent inverse of 8 .00 centimeters divided by the distance between q2 and q3, which is 10 .0 centimeters.
08:34
That gives us theta 2 is equal to 38 .66 degrees.
08:44
Now we have the angles for our electric fields and we have q1.
08:51
We need to get also r1 and r3.
08:56
Since r1 is the distance between q1 and point b we could get this distance using the pythagorean theorem that is r1 is equal to the square of the distance in y squared plus the distance in x squared for electric field one we have r1 is equal to squared of the distance of 8 squared plus 5 squared...