00:01
Hello and welcome to this video's illusion of numerate.
00:03
So here we are given a small object with mass m and charge q and initial speed v0.
00:09
That is equal to 5 times of 10 to the power of 3 meters per second is projected into a uniform electric field between two parallel metal plates of length 26 centimeters.
00:22
So length of each plate is 26 centimeters.
00:27
Kind of like this.
00:29
The electric field between the plates is directed downwards and is having a magnitude d which is equal to 800 newton per coulomb.
00:39
Right.
00:42
Kind of like this.
00:43
This is v0 and the electric fields are directed down.
00:49
Right.
00:49
This is 26 centimeters and assume that the field is zero outside the region between the plates.
00:59
The separation between the plates is large enough for this object to pass between the plates without hitting the lower plate.
01:04
After passing through the field region the object is deflected downwards by a vertical distance d which is 1 .25 centimeters.
01:15
So this is the straight and here you have got deflected by a distance of 1 .25 centimeters.
01:27
And the collecting plate is 56 centimeters from the edge of the parallel plate.
01:31
So this distance is 56 centimeters.
01:37
Now you have to calculate the objects charge to mass ratio.
01:44
Now what you can do is you can first calculate the deflection in the electric field.
01:49
Right.
01:49
So here you have got let's say d1 that is equal to using the expression ut plus half at square.
01:56
So this is the expression of the deflection.
01:58
Let's say this is d1.
02:00
I'm taking it the first initial time.
02:02
Now this d1 will be equal to l the length of this plate over the speed.
02:11
Right.
02:11
U or v naught.
02:15
And also you have got the acceleration to be equal to the net force over the mass.
02:20
Right.
02:21
Now the net force is you know that it's charge times the electric field over the mass.
02:26
Right.
02:26
So there's a value of acceleration.
02:28
So using these two conditions you can plug in and find the value d1.
02:34
So d1 it's equal to this u will be u y rather which is zero because there is no initial vertically rounded velocity.
02:48
Right.
02:49
Plus half the acceleration is qe over the mass and t1 square is l by v naught whole squared.
02:57
Right.
02:57
And this will be around qe l square by 2m v naught squared.
03:12
Right.
03:12
So this is d1 you have got.
03:16
Now next you have to consider calculate the time it will take outside the magnetic field.
03:21
Right.
03:21
So this is t2 which is equal to outside length which is 56 centimeters is l.
03:30
Let's say capital l i take.
03:32
Right.
03:33
So this l by the velocity v naught itself.
03:37
Right.
03:38
Because this is the vertical horizontal velocity with which it will cover the entire distance.
03:43
Right.
03:44
And you have to consider calculate the vertical motion also...