Answer
(a) The rate that charge is lifted by the charge escalator is $0.50~C/s$
(b) $W = 1.5~J$
(c) $P = 0.75~Watts$
Work Step by Step
(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:
$P = \frac{Work}{time} = \frac{1.5~J}{2~s} = 0.75~Watts$
