00:01
Hi there, here we have to solve the following problem.
00:04
First in equation a, we have to determine magnitude and direction of the electric field at point z.
00:11
Let's solve this problem.
00:15
Q1 is negative 2 times 10 power by negative 5 columns and q2 is 8 times 10 power by negative 6 columns.
00:32
So in the distance between points x and y is 60 centimeters, the distance between points x and y is 60 centimeters.
00:37
The distance between points y and that is 30 centimeters.
00:42
Let's solve it.
00:44
At point z, yeah, let's first sketch this system.
00:53
And let's show the direction of the field created by its charged point z.
00:59
Charge x is negative, that's why it will create the field towards it, so that is x.
01:05
And charge y is positive, it will create the field.
01:12
To the opposite direction so basically it will create to the field to the left to the right sorry and here is let's direct the axis to the right let's label this axis as x1 therefore is that here equals to e y minus e x which is k q 2 divided by d, y, z squared minus k absolute value of q1 over d xz squared.
02:03
Then this is that is k constant multiplied by q2 squared, which is 8 times 10 power by negative 6 columns over 30 over yz squared.
02:24
So this distance is 0 .30 meters squared, minus 2.
02:31
Times 10 power by negative 5 columns over square distance which is 0 .90 meters squared.
02:46
Let's complete this calculation.
03:33
That is 5 .8 times 10 power by 5 newton per column and that is directed to the right.
03:57
That is answer to question a.
03:59
Now it's time to move on to question b.
04:02
In question b, a test charge of one microcolum with a positive sign experiences an electric force of 6 times 10 power by negative 6 0 to the right.
04:34
First we have to determine the strength of the field at this point...