Question

The electric field strength is 2.30×104 N/C inside a parallel-plate capacitor with a 0.700 mm spacing. An electron is released from rest at the negative plate. Part A What is the electron's speed when it reaches the positive plate? Express your answer with the appropriate units.

Answer 1

The electric potential energy of the electron = eEd
Where e is the charge of electron, E is the electric field and d is the distance between the plates.
E = 2.30 x 10⁴ N/C, d = 0.7 mm = 7 x 10^-4 m
Electric potential energy is converted into kinetic energy of electron.
eEd = 1/2 mv²
Where m is the mass of electron and v is the final velocity
v² = 2eEd/m
v = SQRT[2eEd/m]
Substituting values,
v = SQRT[(2 x 1.60 x 10^-19 x 2.30 x 10⁴ x 7 x 10^-4) / (9.11 x 10^-31)]
= SQRT[(51.52 x 10^-19) / (9.11 x 10^-31)]
= SQRT(5.66 x 10¹²)
= 2.38 x 10⁶ m/s