Answer
$\sigma = 6.7~\mu C/m^2$
Work Step by Step
We can find the surface charge density on the belt:
$\sigma~w~v = i$
$\sigma = \frac{i}{w~v}$
$\sigma = \frac{100\times 10^{-6}~A}{(0.50~m)(30~m/s)}$
$\sigma = 6.7~\mu C/m^2$
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: 78
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