Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 25 - Capacitance - Problems - Page 742: 35a

Answer

$u = 9.15\times 10^{-18}~J/m^3$

Work Step by Step

We can find the magnitude of the electric field at $r = 1.00~mm$: $E = \frac{1}{4\pi~\epsilon_0}~\frac{\vert q \vert}{r^2}$ $E = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~\frac{1.6\times 10^{-19}~C}{(1.00\times 10^{-3}~m)^2}$ $E = 1.438\times 10^{-3}~V/m$ We can find the energy density: $u = \frac{1}{2}~\epsilon_0~E^2$ $u = \frac{1}{2}~(8.854\times 10^{-12}~C/V~m)(1.438\times 10^{-3}~V/m)^2$ $u = 9.15\times 10^{-18}~J/m^3$
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