Impact speed of an electron in a parallel plate capacitor

[ShilpaM]

Homework Statement

The electric field strength is 2.50×10⁴ N/C inside a parallel-plate capacitor with a 1.20 mm spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?

Homework Equations

F = eE = ma
v_f² = v_i² + 2ad

The Attempt at a Solution

First, I used F = eE to figure out the force acting on the electron and then divided the force by the mass to get the electron's acceleration in the electric field:

a = eE/m = (1.60×10^-19 C)(2.50×10⁴ N/C) / (1.67×10^-27 kg) = 2.3952×10¹² m/s²

Then I used a kinematic equation to figure out the final velocity:

v_f² = v_i² + 2ad (v_i = 0 since the electron starts at rest)
v_f = √(2ad) = √(2(2.3952×10¹² m/s²)(1.2×10^-3 m)) = 7.58×10⁴ m/s

But apparently this is not the answer. Can anyone please tell me what I'm doing wrong? I've checked over my work several times. Any help would be much appreciated!

 

[gneill]

You've used the mass of a proton rather than that of an electron.

Note that you could also calculate the energy imparted to the electron by traveling through the electric field for the given distance. That'll end up as its kinetic energy...