Question

an electron is liberated from the lower of the two large parallel metal plates separated through 20mm. upper plate has a potential of 2400v realtive to lower plate. calculate acceleration of electron time taken to reach other plate velocity of electron

Answer 1

electric Field E = V/d
 V = 2400 V and d = 20 mm = 2 x 10^-2 m

Hence E = 1.2 x 10^5 V/m
So force on electron F = E x e = 1.2 x 10^5 x 1.6 x 10^-19 N 
                                 F = 1.92 × 10^-14 newtons

Hence acceleration a = F/m 
mass of electron = 9.1 x 10^-31 kg
Plug and come out successfully

Electrostatic potential energy acquired = e x V = 1.6 x 10^-19 x 2400 J
                                                              = 3.84 x 10^-16 J


Kinetic energy = 1/2 x m x v^2
m = 9.1 x 10^-31 kg
Equating K E = P E ,you can get v
1/2 x 9.1 x 10^-31 x v^2 = 3.84 x 10^-16
v^2 = 8.44 x 10^14
v =  2.9 x 10^7 m/s

To get the time use S = ut + 1/2 a t^2
S = 2 x 10^-2 m

and here v= u
and a = F/m = (1.9 x 10^-14) / (9.1 x 10^-31) = 2 x 10^16 m/s^2

now i hope uh will do it ur own

i hope it will help you
regards

Answer 2

Answer:

Here's your answer

Explanation: