Velocity of an electron and Conservation of energy

[kevnm67]

Homework Statement

A pair of charged conducting plates produces a uniform field of 12,000 N/C directed to the right, between the plates. The separation of the plates is 40 mm. In the figure, an electron is projected from plate A, directly toward plate B, with an initial velocity of 2.0 x 10^7 m/s.
A) The velocity of the electron as it strikes plate B is closest to:

Homework Equations

Conservation of energy

The Attempt at a Solution

Equation: qV= 1/2mv² My answer: 1.3x10^7m/s
I tried the equation above based on what I gathered from my book but I am suppose to get 1.5x10⁷ How do I use the conservation of energy to get this answer? Thanks for your help

 

[Doc Al]

How do I use the conservation of energy to get this answer?

If no external work were done, there would be no change in total energy. But work is being done, which increases the total energy. Set the work done equal to the change in total energy (which is the sum of PE and KE).

 

[Doc Al]

So Work=qED or (1.6x10^-19)(12000)(.04)= 7.68x10^-17?

Not relevant. (And I have no idea where you're getting the numbers from--they weren't in your problem statement.)

and W= KE+PE

This is the only one you need.

PE= kqq/r

Not relevant.

I tried the equation above based on what I gathered from my book but I am suppose to get 1.5x10⁷

FYI, this answer is not correct. Why do you think it is?

 

[Doc Al]

I got the first one saying W = qED and plugged it in being you mentioned work being done and I thought that's what you use. 12000 is the charged produced and 1.6x10^-19 is the charge of an electron.

Are you sure you haven't mixed up this problem with a different one? See your first post.

W = KE + PE

Again, for this problem (the one in your first post) this is all you need. You are given W and KE, just solve for PE. (That really should be W = ΔKE + ΔPE.)

 

[kevnm67]

I copied the wrong problem..sorry. I am going to edit and delete my previous posts.