Question

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.8 cm, and the electric field within the capacitor has a magnitude of 2.6 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Answer 1

K = 4.2 10⁻¹³ J

Step-by-step explanation:

For this exercise we will use the principle of conservation of energy

starting point. Where the electron is released

Em₀ = U = e V

final point. On the positive plate

Em_f = K = ½ m v²

as there is no friction, energy is conserved

Em₀ = Em_f

e V = ½ m v²

K = e V

let's calculate

K = 1.6 10⁻¹⁹ 2.6 10⁶

K = 4.2 10⁻¹³ J