Introduction to Electrodynamics 4e

Published by Pearson Education
ISBN 10: 9332550441
ISBN 13: 978-9-33255-044-5

Chapter 2 - Section 3.2 - Electric Potential - Problem - Page 83: 22

Answer

$$V=-{2k\lambda}\ln[{\frac{s}{a}}]$$ $$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{2k\lambda}{s}\hat s=-\vec E$$

Work Step by Step

We know that Electric field due to long straight wire is $$\vec E=\frac{2k\lambda}{s}\hat s$$ Potential(V) in the form of electric field $$V=\int\limits_{a}^{s}-E. ds$$ In this case, we cannot set the reference point at infinity, since the charge itself extends to infinity. $$V=\int\limits_{a}^{s}-[\frac{2k\lambda}{s}\hat s]. \vec ds$$ $$V=-\int\limits_{a}^{s}\frac{2k\lambda}{s} ds$$ $$V=-{2k\lambda}ln{s}\Big|_a^s \ $$ $$V=-{2k\lambda}\ln[{\frac{s}{a}}]$$ In this form, it is clear why a = 1 would be no good—likewise the other “natural” point, a = 0 $$ \mathbf{\vec \nabla} \mathbf{V}=\frac{\partial V}{\partial s} \hat s+ \frac{\partial V}{\partial \theta}\hat \theta + \frac{\partial V}{\partial \phi} \hat \phi $$ Potential is dependent only on s So $$ \mathbf{\vec \nabla} \mathbf{V}=\frac{\partial V}{\partial s}\hat s$$ $$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{\partial ({2k\lambda}\ln[{\frac{s}{a}}]) }{\partial s}\hat s$$ $$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{\partial ({2k\lambda}\ln[{\frac{s}{a}}]) }{\partial s}\hat s$$ $$ \mathbf{\vec \nabla} \mathbf{V}=-2k\lambda\frac{a}{s}\frac{1}{a}$$ The gradient of potential is $$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{2k\lambda}{s} $$ $$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{2k\lambda}{s}\hat s=-\vec E$$
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