Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 29 - Potential and Field - Exercises and Problems - Page 865: 58

Answer

$ 1.33\;\rm pF$

Work Step by Step

As the author told us (see the hint), we can think of both parts as two capacitors that are connected, but we do not know if they are connected in series or in parallel. Let's assume that the upper plate is the negative one while the other is positive. We know that for the two in-series capacitors, the positive plate of the first capacitor is connected to the negative plate of the second one. But here the two plates are negative since the upper one is one unit and the lower one is one unit as well. We also know that for the two in-parallel capacitors, the positive plate of the first capacitor is connected to the positive plate of the second one. And this is the case here. Thus, this system can be considered as a combination of two in-parallel capacitors. Hence, $$C_{eq}=C_1+C_2$$ where $C=\dfrac{A\epsilon_0}{d}$ $$C_{eq}=\dfrac{A_1\epsilon_0}{d_1}+\dfrac{A_2\epsilon_0}{d_2}$$ where $A_1=A_2=A$ $$C_{eq}=A \epsilon_0\left[\dfrac{ 1}{d_1}+\dfrac{1}{d_2}\right]$$ Plug the known; $$C_{eq}=(0.01^2)(8.85\times 10^{-12})\left[\dfrac{ 1}{1\times 10^{-3}}+\dfrac{1}{2\times 10^{-3}}\right]$$ $$C_{eq}=\color{red}{\bf 1.33\times 10^{-12}}\;\rm F$$
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