00:01
Let's discuss this question.
00:02
So here we have a question that a point charge q1 is held stationary at the origin and second point charge q2 moves from point x, that is 0 .135 meter, y is equal to 0 to the point x that is 0 .270 meter to y that is equal to 0 .270 meter.
00:22
So here we need to find two things.
00:24
Firstly, the change in potential energy of the pair of charges and secondly how much work is done by the electric force on q2 so here for example if this is 0 .135 meter so here we will be having point q1 charge that is 2 .4 micropoleum and here we will be having q2 charge that is equal to minus 4 .3 micropulum so here we can say if this is the diagram over here and this will be y so here we can say q2 will be equal to minus 4 .3 microculum and here q1 will be equal to 2 .4 microculum so here we can say y is equal to under root of 0 .270 minus 0 square plus 0 .270 minus 0 .270 plus 0 .270 minus 0 .2 1 .135 square.
01:31
So here y will be equal to under root of 0 .09115.
01:38
So y is equal to 0 .301 meter.
01:42
So now here we can say at initial ui will be equal to k q1 q2 divided by r1 so here we can put down the values.
01:58
So q i will be equal to minus 9 multiplied by 10.
02:02
Multiplied by 2 .4 multiplied by 10 raised to minus 6 multiplied by 4 .3 multiplied by 10 to minus 6 divided by 0 .135.
02:14
So here we can say delta u will be equal to uf minus ui that is final minus initial.
02:26
Here we need to calculate uf so here final will be uf that is equal to kq1, q2 divided by r2.
02:44
So here we can put down the values.
02:46
So here, u -pinal is equal to minus 9 multiplied by 10 -rays to 9 multiplied by 2 .4 multiplied by 10 -rays to minus 6 multiplied by 4 .3 multiplied by 10 -rest -2 -1 jol...