If protons are accelerated from rest in a Van de Graff accelerator through a potential difference of $1.00 \mathrm{MV}$, what is their resulting speed?

Question

Quizplus · Accepted Answer

The kinetic energy gained by the proton is equal to the electric potential energy, $ qV $, where $ q $ is the charge of the proton and $ V $ is the potential difference. Using the relation $ \frac{1}{2}mv^2 = qV $ and solving for $ v $, the speed of the proton is approximately $ 1.38 \times 10^{7} \mathrm{~m} / \mathrm{s} $.

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If Protons Are Accelerated from Rest in a Van De $1.00 \mathrm{MV}$