Physics 2 question velocity of electrons and voltages

[rasanders22]

Prob statement:
An electron is to be accelerated from a velocity of 1.00×10^6 m/s to a velocity of 9.90×10^6 m/s . Through what potential difference must the electron pass to accomplish this?


equations:
v=U/q0
1/2mv^2
U=K (this is where I have a question)

Well I already have the answer. But here's how I got it.

Ki=.5(9.11*10^-31)(1*10^6)^2=4.55*10^-19J
Kf=.5(9.11*10^-31)(9.9*10^6)^2=4.45*10^-17J

Vi=Ki/(-1.6*10^-19)=-2.843V
Vf=Kf/(-1.6*10^-19)=-278.125V

V=Vi-Vf=275.282V But the voltage is negative because the electron is flowing against the + to - field (atleast how i explained it to myself)

Well my first question is how does V=U/q0 turn into V=K/q0. I found one example where my instructor used this to determine the voltage but I can't find any reference to it in my book.

Thanks

 

[gneill]

Not sure what your Vi and Vf business is all about. Basically you want to add a certain amount of KE to what the electron already has (your Ki). So you have a desired ∆KE which is Kf - Ki.

The electron will gain energy falling through a potential difference: ∆KE = -q∆V. Since the required energy gain is positive in this case, and since q is negative, then ∆V should be a positive potential change (the electron should be accelerated by a positive potential).