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: 51a

Answer

$1.05\times 10^4\ V/m$

Work Step by Step

It is given that: Capacitance of capacitor $C =100\ pF = 100 \times 10^{-12}\ C$ Area of plate $A = 100\ cm^2 = 100\times 10^{-4}\ m^2$ Dielectric constant $k=5.4$ Potential difference across capacitor $\Delta V =50\ V$ Formula for the Electric field between two plates of a capacitor separated by distance d is: $E = \frac{\Delta V}{d}$ Formula for the capacitance having area A and separation d is: $C = \frac{\epsilon_oA}{d}$ This can be rearranged as: $d = \frac{\epsilon_oA}{C}$ We substitute expression for $d$ in Electric field E and solve; $E =\frac{VC}{k\epsilon_oA}$ $E=\frac{(50V)( 100\times 10^{-12} F)}{(5.4)(8.85\times 10^{-12}C^2/Nm^2)(100\times 10^{-4}m^2)}$ $E =1.05\times 10^4\ V/m$
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