00:01
In this question for part a, first of all, the potential energy at point b, let's call it dl.
00:08
It is equal to u2 minus u1.
00:11
This is equal to k, q1 times q2 over d2 minus k, q1 times q2 over d1.
00:25
Okay, let me take kq1 q2 common.
00:33
We will have 1 over d2 minus 1 over d1.
00:37
Putting the values we have this is equal to 8 .99 times 10 to the power 9 times 0 .3 times 10 .50 the power minus 6.
01:01
Just a second yes minus 4 .7 times 10 % to the power minus 6 over it 1 over 0 .07.
01:20
Okay so now simplifying this will give us 1 .81 now does the work done in moving the charge from point p to point r is 1 .81 jules? this would be the answer for the first part.
01:46
The total potential energy from the system of these charges let's go to you the formula for that will be k q1 q3 over d1 with respect to 3 plus k q1 q4 that is 2 with respect to 4 plus k q3 q4 over d3 with respect to 4 putting the values we will have u that is equal to 8 .99 times dendousity power 9 times 3 .3 times minus 2 .35 over 0 .052 square the 0 .013 square the 0 .013 square matter plus 3 .3 times minus 2 .35 over 0 .05 plus 0 .013 times times just a second minus 2 .35 times minus 2 .35 over 2 .35 over twice of 0 .01 so you times 10 sari kava minus 6.
03:43
Or simply finding this will give us u to be exactly minus 1 .98.
03:49
Okay this will be the answer for the second part...