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 2.4 cm, and the electric field within the capacitor has a magnitude of 1.8 x 10% v/m. What is the kinetic energy of the electron just as it reaches the positive pliate? KEpositive- Electric ield Electron

Answer 1

E = ΔV/Δs

E = 1.8 * 10⁶ V/m

Δs = 2.4 * 10^-2m

Solve for ΔV,

so ΔV = E Δ* Δs

ΔV = (1.8 * 10⁶ V/m)* (2.4 * 10^-2 m)

ΔV = 43200 V

(43200 J/C) * (1.6 x 10^-19 C) = 6.912 x10^-15 Joules