00:01
All right, so for this problem, we have this setup.
00:10
So we have two thread connected with two balls.
00:14
And the angle between the thread and the vertical is theta.
00:21
So the mass of the ball is m and the charge is q1.
00:25
So this is also m and the q2.
00:28
The information we know about the setup is that we know the mass.
00:36
M is 8 grams and the theta is 20 degrees and l l which is the length of the thread which is 0 .5 meters so for part a we want to show the free body diagram of the of the ball and we want to label all the forces acting on the ball so we just so for simplicity we just look at the right one we just look at this one as you can see that for this ball we have the gravity clearly and we have the column force because these two balls they carry the same the same kind of charge so the column force is opposite to each other so it's pointing to right direction and this is the force on the thread so let's say this is f t this is f c and this is f g and for part b we want to find out the the the column force so if we look at this diagram, we see that ft is opposite to the combination of fc plus fg.
01:50
So in that case, we know that fc over fg is tangent theta.
01:56
So we're, this is the data.
02:00
Okay.
02:01
So we know that we can find out fc equal fg times tangent theta.
02:08
And fg is simply the gravity.
02:10
So this is m g times tangent theta.
02:13
We already know the mass and we know the theta.
02:16
So just plug in the two values and see that's fc for 0 .02854 newtons.
02:26
And we also want to know the folds on the thread, ft.
02:36
Again, we just look at this free body diagram.
02:39
We know that's fg over ft.
02:44
Equal cosine data.
02:47
Okay, so just to plug in the video for fg and the theta.
02:51
So i can see that ft equal 0 .0834 newt okay and for part c, it's right here, for part c, we want to get some information about q1 and q2.
03:14
I'll say that at this moment we cannot determine the value for q1 and q2.
03:19
However, we can still obtain some information.
03:22
So we already have the magnitude of the column force over here, right? so we know the expression for column force is fc equal 1 over 4 pi epsilon times q1, q2, divide by the separation.
03:39
Here is the separation.
03:40
So just utilize the triangular knowledge.
03:44
And you can see that separation equal.
03:47
2l times sun theta with squared.
03:53
Okay, so this value equal 0 .02854, as we found previously, okay? so we can solve for q1 times q2 by using this expression...