00:03
In this problem, for the first part, we need to find the distance x such that the rod is horizontal unbalanced.
00:17
Now for that, the torque at the pivot, let's call it p, at the pivot p must be equal to zero.
00:33
Now we have positive charges on either end and they both generate a repulsive force.
00:38
So the electromagnetic force, the electromagnetic force is acting upwards from both sides.
01:08
And the way w is acting downwards.
01:12
Note that both of the forces are fighting each other.
01:19
So let's call this f1.
01:24
This one is w and this is f2.
01:28
So f1 is generating a clockwise torque.
01:35
W is also generating a clockwise torque.
01:38
And f2 is generating an anti -clockwise torque about p.
01:44
So the sum of the torques must be zero, and the formula for torque is equal to the force times distance.
01:56
So the torque generated by the first force, we'll call it t sub 1, is f sub 1 times the distance between the force and the pivot, the perpendicular distance, the distance between the line of action and the force, and the pivot is l by 2.
02:16
Now the force is actually given by kulom's law, which is q1 times q2 times kulom's constant divided by the total distance between the charges, that is h squared.
02:36
Now first, the first charge is given as, so we're calculating f1, the first charge is q, the second is capital q, and the distance is h, so that gives f1 is equal to k times small q times capital q divided by h squared.
03:02
Similarly, we can calculate f2.
03:12
So we'll get f2 is equal to k times 2q times capital q divided by h squared.
03:22
That gives us 2kq divided by times small q divided by, divided by, times small q, divided by h squared.
03:35
That means our first torque is equal to k times q times capital q divided by h squared times l by 2 and that is equal to l k small q times capital q divided by 2 to which squared similarly second torque is also at a distance l by 2 from the pivot and we get the answer k l k q x x xxx divided by h square for the second torque now we can calculate the torque generated by w let's call it d sub w and that is equal to the distance between the line of action of the force w and the pivot that just turns out to be l minus x times w so that's l minus x times w so that's l minus x times w and all of these forces, rather all of these torques must be zero.
05:15
So the sum of the torques must be zero.
05:20
Now note that f1 and w generate a clockwise torque.
05:33
And clockwise torques are always negative.
05:39
So our total torque is actually minus tau sub 1, minus tau sub w, plus tau -s -s -2 is equal to zero.
05:56
Now we can go ahead and substitute in our values, so that'll be negative l, k times q times capital q divided by 2h squared minus l -1 -x times w plus l times k -q times capital q divided by h squared equal zero.
06:30
Now we can go ahead and do some simplification, and that will give us l times k times small q times capital q divided by h squared is equal to l minus x times w plus l times k times q divided by 2h squared.
06:58
Now we're going to go ahead and perform multiplication over here to isolate the x's...