Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 20 - Electric Potential and Electric Potential Energy - Problems and Conceptual Exercises - Page 719: 41

Answer

$-(4-\sqrt{2})(KQ^2/a)$

Work Step by Step

We know that the electric potential energy for the given scenario can be expressed as $U=K[\frac{Q(-Q)}{r_{12}}+\frac{Q(Q)}{r_{13}}+\frac{Q(-Q)}{r_{14}}+\frac{Q(-Q)}{r_{23}}+\frac{Q(-Q)}{r_{24}}+\frac{Q(-Q)}{r_{34}}]$ $\implies U=KQ^2[\frac{-1}{r_{12}}+\frac{1}{r_{13}}-\frac{1}{r_{14}}-\frac{1}{r_{23}}+\frac{1}{r_{24}}-\frac{1}{r_{34}}]$ $\implies U=KQ^2[-\frac{-1}{a}+\frac{1}{\sqrt {2a^2}}-\frac{1}{a}-\frac{1}{a}+\frac{1}{\sqrt{2a^2}}-\frac{1}{a}]$ This simplifies to: $U=-(4-\sqrt{2})(KQ^2/a)$
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