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

Published by Pearson

# Chapter 25 - Electric Charges and Forces - Exercises and Problems - Page 746: 23

#### Answer

The electric field has a strength of $4.5\times 10^4~N/C$ directed toward the bead.

#### Work Step by Step

We can find the magnitude of the electric field 4.0 cm from the bead: $E = \frac{k~q}{r^2}$ $E = \frac{(9.0\times 10^9~N~m^2/C^2)(8.0\times 10^{-9}~C)}{(0.040~m)^2}$ $E = 4.5\times 10^4~N/C$ An electric field points toward a negative charge. Therefore, the electric field has a strength of $4.5\times 10^4~N/C$ directed toward the bead.

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