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

Published by Pearson

# Chapter 26 - The Electric Field - Exercises and Problems: 18

#### Answer

The magnitude of the charge on each electrode is $14~nC$

#### Work Step by Step

We can use the expression for the electric field strength in a plate capacitor to find the charge on each electrode. $E = \frac{\sigma}{\epsilon_0}$ $E = \frac{Q}{A~\epsilon_0}$ $Q = E~A~\epsilon_0$ $Q = (1.0\times 10^6~N/C)(0.040~m)(0.040~m)(8.85\times 10^{-12}~C^2/N~m^2)$ $Q = 1.4\times 10^{-8}~C$ $Q = 14~nC$ The magnitude of the charge on each electrode is $14~nC$.

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