Law of conservation of energy:

$\overline{)\begin{array}{rcl}{\mathbf{K}}_{\mathbf{i}}\mathbf{+}{\mathbf{U}}_{\mathbf{i}}& {\mathbf{=}}& {\mathbf{K}}_{\mathbf{f}}\mathbf{+}{\mathbf{U}}_{\mathbf{f}}\\ \frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{{\mathbf{v}}_{\mathbf{i}}}^{\mathbf{2}}\mathbf{+}\mathbf{q}{\mathbf{V}}_{\mathbf{i}}& {\mathbf{=}}& \frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{{\mathbf{v}}_{\mathbf{f}}}^{\mathbf{2}}\mathbf{+}\mathbf{q}{\mathbf{V}}_{\mathbf{f}}\end{array}}$

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

What potential difference is required to bring the proton to rest?

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