The electric field strength is 240 times 104 NC inside a par

The electric field strength is 2.40 times 10^4 N/C inside a parallel-plate capacitor with a 1.50 min spacing. An electron is released from rest at the negative plate. What is the electron\'s speed when it reaches the positive plate? Express your answer with the appropriate units.

Solution

V = E*d = 2.4*10^4*1.5*10^-3 = 36 V

Using energy conservation method:

KE1 + UE1 = KE2 + UE2

0.5*m1*v1^2 + qV1 = 0.5*m1*v2^2 + q*V2

V2 = 0 (potential)

v1 = 0 (velocity)

v2^2 = 2*q*V1/m

v2 = sqrt(2*1.6*10^-19*36/(9.1*10^-31)) = 3.56*10^6 m/sec

Let me know if you have any doubt.