Figure initial kinetic energy (1/2) m v^2
(ok because speed well below that of light)
volts * charge = energy = (1/2) m v^2
I guess you can divide v by two and v^2 by 2
A proton has an initial speed of 3.9×105 m/s.
What potential difference is required to bring the proton to rest?
What potential difference is required to reduce the initial speed of the proton by a factor of 2?
What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 2?
1 answer