00:01
Hi, here in this given problem there are two parallel plates charged parallel plates.
00:11
Upper plate positively charged, lower plate negatively charged, having electric field, a uniform electric field between the plates like this.
00:28
A proton starts from here at this point exactly in middle of the two plates and this point is marked as 0 .0 and then here this is along the x -axis and vertically this is the y -axis.
00:53
This is the velocity vector for this proton.
00:58
That velocity v -0.
01:01
It is given to be 4 .90 into 10 -dhist to the power 6 meter per second.
01:10
And surface charge density of the plates, that is 4 .90 into 10 -dage to the power minus 9 kulum per meter square.
01:26
X is 0 over here at the point from where the proton starts moving.
01:33
In the first part of the problem we have to find magnitude of this electric field and simply that electric field between two charge plates is given by an expression sigma by epsilon 0, sigma that is 4 .90 into 10 dash per minus 9 divided by epsilon not 8 .854 into 10 dash per minus 12 column square per newton meter square.
02:01
So solving this we get electric field to be equal to 553 .4 newton per column, which becomes the answer for the first part of this given problem.
02:17
Then in the second part of the problem, we have to find electrostatic force experienced by the electron, by the proton.
02:32
So that electric force, sorry, here it is electron.
02:41
This is the proton.
02:43
So electric force experienced by beta, we should be.
02:55
Look into this problem in the second part, calculate the magnitude of the electric force acting on the proton.
03:04
Yes, this is the proton experienced by proton and that will be given by simply using an expression, product of the force is equal to product of the charge with the electric field.
03:19
So that is 1 .6 into 10 dash per minus 19 into electric field 550.
03:26
So, finally this force comes out to be equal to 8 .85 into 10 -tash -the -power minus 17 newton, which is the answer for the second part of this problem over here...