00:01
In this question we had given here potential difference is equal to positive 5 .0 volt.
00:15
Now the diagram in this question is, this is the diagram here.
00:19
Now the solution to this question is electron will experience a force in upward direction.
00:42
That is f is equal to q multiply by e opposite to field now in the a part of this question we have to find what is the potential difference experienced by a so 4 2 negative ion that starts from rest at the negative plate and accelerates to the positive plate so solution for a part is for so4 -2 negative, q is equal to minus of 2e, that is equal to minus of 2 multiply by 1 .6 multiply by 10 to the power minus 19 coulum, that is equal to minus 3 .2 multiply by 10 to the power minus 19 coulum.
01:36
So here, potential difference is equal to applied potential difference, that is, equals to 5 minus 0 which is equal to 5 volt.
02:01
This is the answer for a part.
02:05
Now for the b part we have to find what is the change in electric potential energy experienced by the electron as it travels from the negative plate to the positive plate.
02:19
So the solution for b part is change in potential energy for for electron q is equal to minus of e delta v is equal to vf minus vi that is equal to q multiply by vf minus vii so this is equal to q multiply by five minus zero that is equal to 1 .6 multiply by 10 to the power minus 19 multiply by 5...