Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 26 - Potential and Field - Exercises and Problems - Page 739: 45


$\displaystyle \vec{E}(\frac{d}{2}) = -\frac{2V_0}{3d}$

Work Step by Step

We can find an equation for the electric field by taking the negative derivative of $V(x)$ since $\displaystyle \vec{E} = -\frac{dV}{d\vec{s}}$ where $\vec{s}$ is the position vector. $\displaystyle V(x) = V_0\ln(1+\frac{x}{d})$ $\displaystyle \vec{E}(x) = -V'(x) = -V_0\frac{1}{1+\frac{x}{d}}(\frac{1}{d})$ $\displaystyle \vec{E}(x) =-V_0\frac{1}{\frac{d+x}{d}}(\frac{1}{d})$ $\displaystyle \vec{E}(x) =-V_0\frac{d}{d+x}(\frac{1}{d})$ $\displaystyle \vec{E}(x) =-\frac{V_0}{d+x}$ We are then asked to find the electric field halfway between the two electrodes, which is $x = \displaystyle \frac{d}{2}$ $\displaystyle \vec{E}(\frac{d}{2}) = -\frac{V_0}{d+\frac{d}{2}}$ $\displaystyle \vec{E}(\frac{d}{2}) = -\frac{V_0}{\frac{3d}{2}}$ $\displaystyle \vec{E}(\frac{d}{2}) = -\frac{2V_0}{3d}$
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