Question

An electron is projected as in figure with kinetic energy K, at an angle 0 = 45° between two charged plates. The
magnitude of the electric field so that the electron just fails to strike the upper plate, should be greater than:​

Answer 1

Answer:

Let mass of electron be m , now

$k = \frac{1}{2} m {}^{2} \\ \\ u = \sqrt{2k \over m}$

Now component of velocity along y axis

$= usin45 = \sqrt{ \frac{2k}{m} } \frac{1}{ \sqrt{2} } = \sqrt{ \frac{k}{m} }$

Retarding Force on electron due to electric field

= q × E , where q is charge of electron

So retardation

$a = \frac{f}{m} = \frac{qE}{m}$

Final velocity = 0 , distance traveled = d

use 3rd law of motion

$2as = {v}^{2} - {u}^{2} \\ \\ 2 \frac{qE}{m} .d = - \frac{k}{m} \\ \\ E = \frac{ - k}{2qd}$

Or use energy conversation along y-axis

$qEd = \frac{1}{2} m {v_{y} }^{2}$