A proton passes through a potential difference of 350 V. Find it's kinetic energy and velocity. What's wrong with my work?

Here's my work:

#DeltaU_e=qDeltaV=-W_e=-DeltaK=-(K_f-cancel(K_i))=-K_f#

#K_f=-qDeltaV=-(1.6*10^{-19})(350)=-5.6*10^{-19}#

So #K_f = -5.6*10^{-19} J#

This doesn't make sense though as I don't think kinetic energy can be negative. And if it is negative then what would the velocity be imaginary which definitely doesn't make sense

What am I doing wrong?

1 Answer
Jan 31, 2018

#Delta V# is negative for protons as protons go from high potential to low potential.

Explanation:

I realized what I did wrong. All my math was correct, but at the final step where I plugged my values into #K_f = -q DeltaV#, I didn't realize that #Delta V# was negative.

This is because protons go from high potentials to low potentials. #Delta V = V_f-V_i = V_{low}-V_{high} = 0 - 350 = -350V#

So when you plugin the values, you get:

#K_f = -q DeltaV = -(1.6*10^{-19})(-350) = 5.6*10^{-19}J#

Thanks for trying to help :P