## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) The rate that charge is lifted by the charge escalator is $0.50~C/s$ (b) $W = 1.5~J$ (c) $P = 0.75~Watts$
(a) $0.50~A = 0.50~C/s$ The rate that charge is lifted by the charge escalator is $0.50~C/s$ (b) We can find the work: $W = q~V$ $W = (1.0~C)(1.5~V)$ $W = 1.5~J$ (c) Since the current is $0.50~A$, it takes 2 seconds to lift $1.0~C$ of charge. We can find the power output: $P = \frac{Work}{time} = \frac{1.5~J}{2~s} = 0.75~Watts$