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 1.5 cm, and the electric field within the capacitor has a magnitude of 2.5 x 106 v/m. what is the kinetic energy of the electron just as it reaches the positive plate?

6 x 10⁻¹⁵ J

Explanation:

d = distance between the plates = 1.5 cm = 0.015 m

E = magnitude of electric field between the plates of the capacitor = 2.5 x 10⁶ V/m

q = magnitude of charge on the electron = 1.6 x 10⁻¹⁹ C

Force on the electron due to electric field is given as

F = q E

F = (1.6 x 10⁻¹⁹) (2.5 x 10⁶)

F = 4 x 10⁻¹³ N

KE₀ = initial kinetic energy of electron at negative plate = 0 J

KE = final kinetic energy of electron at positive plate = ?

Using work-change in kinetic energy

F d = KE - KE₀

(4 x 10⁻¹³) (0.015) = KE - 0

KE = 6 x 10⁻¹⁵ J