00:01
Hello, in the question we have given a point charge.
00:04
So let us call it as q1 which is 4 .6 microculum is fixed at origin a second point charge.
00:16
So let us call it as q2 which is 1 .3 microculum and this q2 has mass of 2 .8 into 10 raise to minus 4 kg.
00:28
And this is located this q2 is located at x is equal to 0 .240 meters so let us call this as x1 now we have to find out the potential energy of the pair of charges so we so in order to find out the potential energy so this is charge q1 and this is the charge q2 so they are separated by distance x1 which is this 0 .2 0 .2 040 meters now in order to find out the potential energy what we are going to consider is we are bringing this charge q2 and placing it over here so there will be some work done to bring this charge so that is nothing but the potential energy so it is q2 times v so this v is the potential of q1 so q2 is coming in the potential of q1 so it will be q2 k q1 divided by x1 so just put the value you will get the potential energy so we will move further because there are many parts in this question so the next part is what is this second charge is released from the rest so what is the speed when the distance from the origin is 0 .6 meters so now the distance has become 0 .6 so let us call this as x2 which is 0 .6 meters so so this is q1 now they they are like charges they are both positive charges so what will happen when when it is released from rest so it will repel it will repel and that is the distance so this distance is 0 .6 meters this is x2 we are calling so this is q2 and now we have to find out the speed so to find the to get the speed we will use the conservation of energy so we will use the conservation of energy in this case so what conservation of energy is so we know that conservation of energy is initial kinetic energy plus initial potential energy is equal to final kinetic energy plus final potential energy now initial kinetic energy is zero because though both of the charges are at rest so initial kinetic energy which is known so this is the initial kinetic energy sorry potential energy so this is the initial potential energy now final potential energy we don't know so let us rearrange first this equation so kinetic energy final will be initial potential energy minus final so final potential energy will be so k e f we can write it as half mv square kinetic energy is half m v square is equal to ui ui is how much k q1 q2 divided by x1 now final will be how much k q1 q2 divided by x2 so this is only the difference now what we can do is we can pull this common so it will be half mb square and this k q1 q2 we are taking commons it will be one upon x1 minus one upon x2 so if we plug the values so it will be half mv square is equal to k is 9 into 10 raised to 9 q1 is q1 is 4 .60 into 10 raise to minus 6 and this is 1 .3 10 raise to minus 6 divide into 1 upon this is 0 .24 x1 is 0 .24 and this is 0 .6 so if we compute this so this 9 into 4 .6 into 1 .3 is 53 .87 into 10 raise to minus 3 and this becomes 4 .16 minus 1 .6 calculating this we will get the answer as 1337 0 .71 7 .7 into 10 raise to minus 3.
04:39
This is what we get.
04:40
So this is half mv square.
04:42
So we will rearrange for v.
04:44
So we will be equal to under root of this 2 times 137 .7 into 10 raised to minus 3 divided by mass.
04:57
So if we put the value of mass which is 2 .8, so we will get this as 31 .37 .3 .7.
05:04
Meters per second so this is the velocity now next part in the next part they are saying that this charge so this is c part so in the next part they are saying that what will be the speed when the distance becomes six meters so now this x2 becomes six meters so what we are going to do is only we are going to put year six so it will be half m b square is equal to that a q1 q2 into 1 upon x1 minus 1 upon x2 it was so this we have already found out it is 53 .82 into 10 raised to minus 3 and this becomes 1 upon x1 so 1 upon 0 .24 minus this is 1 1 upon 6 so after calculating so rearranging if we rearrange and calculate as we have done earlier so we will be equal to under root of so we will be will turn out to be under root of 1537 .77...