Answer
$Q = 225~\mu C$
Work Step by Step
We can find the voltage at $t = 0.500~s$:
$V = 6.00+(4.00)(0.500)-(2.00)(0.500)^2$
$V = 7.50~V$
We can find the charge on the capacitor:
$Q = C~V$
$Q = (30~\mu F)(7.50~V)$
$Q = 225~\mu C$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 26 - Current and Resistance - Problems - Page 770: 85a
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