Why is the answer to this electricity question negative?

[srk999]

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 V_B-V_A.

Here is what I did:
$$V = \frac{U}{q}$$
$$U = qV = \frac{1}{2}mv^2$$
$$V_B - V_A = \frac{m(v_B^2-v_A^2)}{2q}$$

Gives me 8.5V

My prof's answer says it is -8.5V. Why is it negative?

 

[Doc Al]

My prof's answer says it is -8.5V. Why is it negative?

Because B is at a lower potential than A. The electric field points from higher to lower potential.

That equation should be ΔKE = -qΔV.