Answer to Question #199096 in Field Theory for Mallory

Question

What potential difference is required to accelerate a proton from rest to a speed of 5.72E7 m/s?

Answer 1

$$PE=KE,$$ $$q\Delta V=\dfrac{1}{2}mv^2,$$ $$\Delta V=\dfrac{mv^2}{2q}$$ $$\Delta V=\dfrac{1.67\cdot10^{-27}\ kg\cdot(5.72\cdot10^7\ \dfrac{m}{s})^2}{2\cdot1.6\cdot10^{-19}\ C}=1.7\cdot10^7\ V.$$