# Potential difference to bring a proton to rest

## Homework Statement

A proton has an initial speed of 3.5×10^5 . What potential difference is required to reduce the initial speed of the proton by a factor of 2?

KE=.5*m*v^2
EPE=q*V

## The Attempt at a Solution

I did KE=.5(1.67x10^-27)(3.5x10^5/2)^2 and then divided that by the charge of a proton to get V, and I got 160 V which is wrong. Then I tried squaring the velocity first, then dividing it by 2 and I got 320 V, which is also not right. I'm stuck.

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this is just conservation of energy

electric potential starts at zero plus kinetic energy starts at some KEi

this must be equal to some electric potential energy plus half of KEi

this is just conservation of energy

electric potential starts at zero plus kinetic energy starts at some KEi

this must be equal to some electric potential energy plus half of KEi
Thank you so much! I didn't even think of that. I got the answer.