Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 22 - Electromagnetic Waves - Problems - Page 641: 3

Answer

$1.7\times10^{15}\frac{V/m}{s}$

Work Step by Step

On page 627, the displacement current is derived as $I_D=\epsilon_o\frac{\Delta \Phi_E}{\Delta t}$. The electric field flux is given by EA, where A is the area of the capacitor plates. $$I_D=\epsilon_o\frac{\Delta \Phi_E}{\Delta t}$$ $$ I_D =\epsilon_o A\frac{\Delta E}{\Delta t}$$ The current in the wires leading to the plates is equal to the displacement current. Solve for the rate of change of the electric field. $$ \frac{\Delta E}{\Delta t}=\frac{I_D }{\epsilon_o A}$$ $$=\frac{3.8A}{(8.85\times10^{-12}C^2/(N \cdot m^2))(0.0160m)^2}$$ $$=1.7\times10^{15}\frac{V/m}{s}$$
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