College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 17 - Problems - Page 655: 56

Answer

(a) The potential difference between the plates is $0.050~V$ (b) The plate with the positive charge is at a higher potential.

Work Step by Step

(a) We can find the potential difference between the plates: $\Delta V = \frac{Q}{C}$ $\Delta V = \frac{0.75\times 10^{-6}~C}{15.0\times 10^{-6}~F}$ $\Delta V = 0.050~V$ The potential difference between the plates is $0.050~V$. (b) The electric field points from the positive charges to the negative charges. Since the electric field points from places of higher potential to places of lower potential, the plate with the positive charge is at a higher potential.
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