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 743: 50

Answer

$C = 4.55 \times 10^{-11} F $

Work Step by Step

Let , $C_{1}=ε_{0}\frac{A}{2}\frac{k_{1}}{2d} =ε_{0}A\frac{k_{1}}{4d}$ $C_{2}=ε_{0}\frac{A}{2}\frac{k_{2}}{d} =ε_{0}A\frac{k_{2}}{2d}$ $C_{3}=ε_{0}A\frac{k_{3}}{2d}$ Note that $C_{2}$ and $C_{3}$ are effectively connected in series, while $C_{1}$ is effectively connected in parallel with the $C_{2} - C_{3}$ combination. Thus, $C= C_{1} + \frac{C_{2} C_{3}}{C_{2} + C_{3}} = ε_{0}A\frac{k_{1}}{4d} \frac{ε_{0}A\frac{k_{2}}{2d} ε_{0}A\frac{k_{3}}{2d} }{ε_{0}A\frac{k_{2}}{2d} + ε_{0}A\frac{k_{3}}{2d} }$
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