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 1.0 cm, and the electric field within the capacitor has a magnitude of 2.9 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate

Answer 1

K = 4.64 10⁻¹⁵ J

Step-by-step explanation:

For this exercise we can use conservation of energy

starting point. Next to the negative plate

Em₀ = U = e V

final point. Right when you hit the positive plate

Emf = K = ½ m v²

energy is conserved

Em₀ = Em_f

e V = K

The electric potential is related to the electric field

V = - Ed

we substitute

-e E d = K

let's calculate

K = -1.6 10⁻¹⁹ (-2.9 10⁶) 1.0 10⁻²

K = 4.64 10⁻¹⁵ J