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Hi.
00:02
Here in this given problem first of all there is a cartesian plane having x -axis and y -axis.
00:16
Over this cartesian plane this is the origin at which a charge q1 has been kept.
00:23
Another charge q2 that is kept over the positive x -axis.
00:28
Suppose this is the point a at which this charge q2 has been kept and another charge the third charge q3 this is also kept over the same axis between q1 and q2 suppose this point is marked as b now the values given are q1 is 4 .10 nanoculum q2 is minus 2 .90 nanoculum and this q3 this is 2 .10 nanoculum and this is 2 .10 the distance of the second charge over the positive x -axis means oa, this is given as 20 .0 centimeter and in the first part of the problem, position of the third charge q3, this is given as plus 11 .0 centimeter.
01:33
So we have to find electrostatic potential energy of the system of the system of the these three charges and it will be given by electrostatic potential energy of q1 and q2, which is marked as, which is labeled as u1, then u -1 -3, then u -2, and then using the expression for electrostatic potential energy, this is given as k into q1 into q2 divided by the gap between them.
02:05
Here that is oa plus k.
02:09
Into q1 into q3 divided by the gap between them which is o b plus k q2 into q3 divided by the gap between them which is a b.
02:21
Now plugging in all the known values and taking this k as a common out and the value of k is 9 into 10 to the part 9.
02:31
Then here in the bracket for all these charges q1 and q2 and q3 the conversion factor is nano -culum.
02:42
For nanoculum the conversion factor is 10 to the power minus 9 each.
02:47
So it will become 10 to the bar minus 18.
02:50
It can also be taken as a common out.
02:54
So finally, leaving behind in the bracket that will be 4 .10 for q1 minus 2 .90 for q2 divided by oa and that is 20 centimeter or 0 .20 meter then for q1 again 4 .10 for q3 this is 2 .10 gap between them 11 centimeter or 0 .11 meter then the q2 minus 2 .90 again q3 2 .10 again divided by ab that will be 20 minus 11 and that is 9 so this is 0 .09 meter so finally here it is 9 into 10 dash minus 9 and in the bracket first term comes out to be equal to minus 59 .45 second term comes out to be equal to 78 .27 and the third term is 67 minus 67 .67.
04:03
So finally this electrostatic potential energy is minus 4 .4 into 10 dash minus 7 joules answer for the first part of this given problem here.
04:21
Now in the second part of the problem in this time we don't know that distance we don't know the position of this charge q3 but it should be such that the total electrostatic potential energy of the system having three charges should come out to be zero.
04:40
So suppose this distance now x, so a, when we will find we will use, it will become 0 .20 minus x...