Electrical potential energy:

$\overline{){\mathbf{U}}{\mathbf{=}}{\mathbf{q}}{\mathbf{V}}}$

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{E}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}}^{{\mathbf{2}}}}$

U_{0} = qV_{0}

U_{f} = qV_{f}

KE_{0} = (1/2)mv_{0}^{2}

KE_{f} = [(1/2)mv_{0}^{2}]/3 = (1/6)mv_{0}^{2}

A proton has an initial speed of 3.9 × 10^{5} m/s.

What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 3?

_____ kV