- #1

rasanders22

- 8

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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 heres 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 cant find any reference to it in my book.

Thanks