00:01
Hi there, so for this problem, we are told that an electric diaple is formed from two charges.
00:08
Charges q, there are separated at distance that we're going to call the distance a, that is equal to 1 .3 centimeters.
00:21
Now, the diaple is at the origin, oriented along the y -axis.
00:28
Now, the electric field strength at the point xy equals to 0 centimeters and 10 centimeters.
00:44
So the magnitude of that electric field is equal to 370 neutrons per column.
00:54
So the question a of this problem is, what is the charge cube? so to calculate this, we use the fact that the point 010 is on the axial line of the diaple.
01:15
Then the electric field at any point in the axial line of the diaple is given by the following equation.
01:23
That is that the electric field is equal to two times columns constant times the radius of separatius.
01:36
Times the momentum, the dipole moment.
01:45
And this separated by a distance that is equal to the radius of separation square minus a to the square and this to the square.
02:01
So what we need to do in this case, first, we know that this radius of semitage of separation that in this case it has a value of one.
02:19
Let me see here.
02:25
So the point r in this case is the separation that we are given in here, that we are tall is 10 centimeters.
02:33
So that is 10 centimeters.
02:37
And then a is the separation between the charges in this diapult that we know is 1 .0.
02:45
Three centimeters apart.
02:49
Let me verify that in here.
02:53
Yes.
02:55
And then what we are going to do in here is to solve for the diapult moment.
03:05
So from there we can later obtain the charge.
03:08
So this is going to be this expression times the electric field.
03:14
And this divided by two times columns constant times the radius are.
03:18
So now we substitute all of these values.
03:21
So we have the radius.
03:25
Remember that we need to pass this from centimeters to meters...