Kinetic Energy and Electric Potential

1. Mar 13, 2010

ihearyourecho

1. The problem statement, all variables and given/known data

Points A and B have electric potentials of 350 V and 160 V, respectively. When an electron released from rest at point A arrives at point C, its kinetic energy is KA.
When the electron is released from rest at point B, however, its kinetic energy when it reaches point C is KB=2KA

What is the electric potential at point C?
What is the kinetic energy KA?

2. Relevant equations

KA=qV

3. The attempt at a solution

I found the solution to the first part by doing:

2(VC-VA)=VC-VB
2VC-700=VC-160
VC=540

For the second part, I did
KA=qV
KA=(1.6022*10^-19)*(540-350)
KA=3.044*10^-17

This isn't right. Any help would be awesome :)

2. Mar 13, 2010

rl.bhat

3. Mar 14, 2010

ihearyourecho

Okay, well it says it wants the answer in something that looks like eV, is that the problem there?

4. Mar 14, 2010

rl.bhat

Energy in eV = energy in Joules/charge on electron.

5. Mar 14, 2010

ihearyourecho

So I would do 3.044*10^-17/1.60217646*10^-19?
Which would be KA=190?

6. Mar 14, 2010

ihearyourecho

190 is the correct answer. Thanks!