A proton (m = 1.7 * 10(adsbygoogle = window.adsbygoogle || []).push({}); ^{–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:

[tex]V = \frac{U}{q}[/tex]

[tex]U = qV = \frac{1}{2}mv^2[/tex]

[tex]V_B - V_A = \frac{m(v_B^2-v_A^2)}{2q}[/tex]

Gives me 8.5V

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

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# Homework Help: Why is the answer to this electricity question negative?

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