Question

Two very large parallel metal plates, separated by 0.20 m, are connected across a 12 V source of potential. An electron is released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what speed does the electron reach? (e=1.60×10−19 C)

Answer 1

Solution :

Given :

d = 0.2 m

V= 12 V

thus electric field between the plates will be given by, E = V/d = 12/0.2 = 60 V/m

And, the acceleration of the electron (a) = q E / m = { (1.6 x 10^-19)(60) } / ( 9.11 x 10^-31) = 1.05 x 10¹³ m/s²

Here, Displacement of the particle is : (s) = 0.20 m - 0.10 m - 0.050 m = 0.05 m

According to work energy principle :

kinetic energy gained (KE) = work done by potential

(1/2) m v² = (1/2) m ( 2 a s )

v² = (2)(1.05 x 10¹³ m/s²)(0.05 m) = 1.05 x 10¹² m²/s²

v = 1.025 x 10⁶ m/s