Electric potential confusion?

Question

In Ernest Rutherford's scattering experiment, described in detail in an earlier homework problem in this chapter, alpha particles (with a mass of 6.68×10–27 kg and a charge of 3.20×10–19 C) come from a few milligrams of radium (to be very precise, its decay product radon). They leave with an initial speed of 2.25×107 m/s before being directed to strike a thin sheet of gold foil. What is the change in electric potential (watch the sign of your answer) that would be needed to accelerate an alpha particle, starting at rest, up to this speed? Ignore relativistic effects.

Answer 1

Energy = QV = mv^2/2. We want to solve for V.

V = mv^2/(2Q) = (-)5.28398E6 V. The sign is negative because a positively charged particle is accelerated in the direction of negative charge.