University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 24 - Capacitance and Dielectrics - Problems - Exercises - Page 810: 24.23

Answer

$u = 0.028 \mathrm{~J/m^3}$

Work Step by Step

The electric field $E$ equals the potential $V$ between the two plates divided by the distance $d$ between the two plates $$E = \dfrac{V}{d} = \dfrac{400 \,\text{V}}{0.005 \,\text{m}} = 8 \times 10^{4} \,\text{V/m} $$ Energy is the amount of work stored in the ca[acitor and the energy density $u$ is related to the electric field by $$u = \dfrac{1}{2} \epsilon_o E^2$$ $\epsilon_o$ is the permittivity of the free space and equals $8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2}$. Substitute to get $u$ \begin{align*} u &= \dfrac{1}{2} \epsilon_o E^2\\ &= \dfrac{1}{2} (8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2}) (8 \times 10^{4} \,\text{V/m})^2 \\ & =\boxed{ 0.028 \mathrm{~J/m^3}} \end{align*}
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