Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 24 - Electric Potential - Problems - Page 713: 44

Answer

$2.1\times 10^{-25}~J~~$ of work is required to bring the $+6e$ charge to the center.

Work Step by Step

We can write a general expression for the change in electric potential energy of the system when we bring in the particle of charge $+6e$: $\Delta U = \frac{1}{4\pi~\epsilon_0}~\sum \frac{(+6e)(q_i)}{r_i}$ We can find the change in electric potential energy of this system of charged particles: $\Delta U = \frac{1}{4\pi~\epsilon_0}~[\frac{(+6e)(+5e-2e)}{0.020~m}+\frac{(+6e)(-4e+4e)}{\sqrt{2}~(0.020~m)}]$ $\Delta U = \frac{1}{4\pi~\epsilon_0}~\frac{(+6e)(+3e)}{0.020~m}$ $\Delta U = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~\frac{(6)(1.6\times 10^{-19}~C)(3)(1.6\times 10^{-19}~C)}{0.020~m}$ $\Delta U = 2.1\times 10^{-25}~J$ We can find the work that is required to bring the $+6e$ charge to the center: $W = \Delta U = 2.1\times 10^{-25}~J$ $2.1\times 10^{-25}~J~~$ of work is required to bring the $+6e$ charge to the center.
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