A proton (m = 1.7 * 10–27 kg, q = +1.6 * 10–19 C) starts from rest at point A and has a speed of 40 km/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference VB-VA.
Here is what I did:
V = \frac{U}{q}
U = qV = \frac{1}{2}mv^2
V_B - V_A...