Calculate Potential Energy of Electron at a Parallel Plate Capacitor

[SpyIsCake]

Homework Statement

An electron with a speed of 5.0x10^6 m/s is injected into a parallel plate apparatus, in a vacuum, through a hole in the positive plate. The electron collides with the negative plate at 1.0x10^6 m/s. What is the potential difference between the plates?
Vi= 5.0x10^6
Vf=1.0x10^6[/B]
Me= 9.11x10^-31
q=1.60x10^-19

Homework Equations

Ke = 1/2 mv^2
Ee =q∆V
W=∆E

The Attempt at a Solution

I believe I should calculate the change in KE of the electron.
Since masses are the same, I can cancel them out and only use velocities.
(1.0*10^6)^2 - (5.0x10^6)^2 = -2.4x10^13

-2.4x10^13/(1.60x10^-19) = -1.5x10^32

That is obviously an incredibly high potential difference. The answer in the textbook is 68V.

What's the proper ay to solve this?

 

[mfb]

Work with units (a very good idea in general), then the error will be obvious.

 

[rude man]

Ke = 1/2 mv^2
Ee =q∆V

What are E and e? Does not look right. q∆V has units of energy.

W=∆E

The Attempt at a Solution

I believe I should calculate the change in KE of the electron.
Since masses are the same, I can cancel them out and only use velocities.

uh-uh

(1.0*10^6)^2 - (5.0x10^6)^2 = -2.4x10^13

This is 2 x ∆KE. Why 2?

-2.4x10^13/(1.60x10^-19) = -1.5x10^32
That is obviously an incredibly high potential difference. The answer in the textbook is 68V.
What's the proper ay to solve this?

How about final KE = initial KE - loss in potential energy.