00:01
Hello students, in this question a negatively charged particle is moved from the positive plate to the negative plate with the help of a constant force.
00:10
The distance between the plates is given as 17 .8 mm which will be equal to 0 .178 m.
00:20
The constant force acting on the charge is given as 9 .31 n and the charge of the object is given as minus 8 .3 mc which will be equal to minus 8 .3 into 10 to the power of minus 3 c.
00:42
The voltage applied across the plates is given as 7 .25 v.
00:49
We have to find the change in potential energy, change in kinetic energy and the change in total energy.
01:26
We know that the electric force acting on the charged particle fv will be given as q times e.
01:35
Here q is the charge of the particle e is the electric field.
01:39
Electric field is given by the formula e is equal to v over d.
01:44
Let us consider this equation 1 and let us consider this equation 2.
01:48
Substituting the value of equation 2 in equation 1, we get f is equal to q capital v divided by d.
01:58
This is equation 3.
02:02
Now we know that the net force acting on the charged particle will be equal to the constant force minus the electric field.
02:20
Work done on the charged particle will be equal to the net force acting on the particle times the displacement of the particle...