A test charge q moves across a potential difference ΔV = 150 V. What is the change in potential energy ΔU for q = 1.0 nC?

• A. 1.5×10^-10 J
• B. 1.5×10^-7 J
• C. 1.5×10^-4 J
• D. 1.5×10^-13 J

Answer: (B)

Changing potential energy of a charge as it moves through a potential difference follows ΔU = q ΔV. A positive charge moving across ΔV = 150 V increases its potential energy by q × ΔV. With q = 1.0 nC = 1.0 × 10^-9 C, ΔU = (1.0 × 10^-9 C)(150 V) = 1.5 × 10^-7 J. The volt is joules per coulomb, so the units work out, and the sign is positive since the charge is positive and the potential increases. So, ΔU = 1.5 × 10^-7 J.