University of Central Florida (UCF) PHY2054 General Physics with Calculus II Practice Exam

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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?

1.5×10^-10 J

1.5×10^-7 J

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.

1.5×10^-4 J

1.5×10^-13 J