An electron is liberated from the lower of the two large parallel metal plates separated by a distance of 20mm. The upper plate has a potential of 2400V relative to the lower plate. Will an electron in an electric field move towards higher potential or lower potential? How long does it take to reach the upper plate?
Answers
Explanation:
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 x 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 ? 2400 J
= 3.84
? 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
= 3.84
? 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