00:01
Hello students, in this question we are given a figure in which there is a proton which is travelling with the speed of 2 .5 multiplied by 10 to the par 6 meters per second towards a very long positively charged plate with the charged density is equal.
00:18
Charged density 2 .4 multiplied by 10 to the minus 5 coulum per meter square.
00:25
So, we have to find the magnitude of the acceleration of the proton in meter per second square as it approaches the charge plate.
00:35
So let us see the figure.
00:36
Now this is the charge plate and this is the figure over here and the distance over here, the distance which is equal to 10 cm, which is equal to 0 .1 meters.
00:49
And we know the charge on proton is 1 .6 multiplied by 10 to minus 19 coulum and the mass of the amount.
00:56
Of proton is equal to 1 point mass of proton is 1 .67 where 1 .67 multiplied by 10 to the bar minus 27 k g so now we know that the electric field due to infinite plate of charge it is equal to e is equal to sigma upon sigma over 2 epsilon of so we have now the force on the charge particle due to elective field is q multiplied by e.
01:39
So from here the force will be equal to sigma multiplied by q over 2 epsilon 0 over here.
01:49
Now we know that force is equal to mass multiplied by acceleration.
01:56
So from here the acceleration will be force over mass...