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: 46


$\vec{E}_{net}(2,2)$m$ = 100$ N/C $\theta = 143^{\circ}$ counterclockwise of the $x$-axis

Work Step by Step

Find functions for $\vec{E}(x)$, the electric field vector on the $x$-axis, and $\vec{E}(y)$, the electric field vector on the $y$-axis. We can do this by differentiating $V = (150x^2-200y^2)$Volts since $\vec{E} = \displaystyle -\frac{dV}{d\vec{s}}$ where $\vec{s}$ is the position vector. $\displaystyle \vec{E}(x) = -\frac{dV}{dx}[150x^2-200y^2] = -300x + 0$ $\displaystyle \vec{E}(y) = -\frac{dV}{dy}[150x^2-200y^2] = 0 + 400y$ Now we can find the electric field at the point (2,2) meters $\vec{E}(2)_x = -600$ N/C $\vec{E}(2)_y = 800$ N/C To find the net electric field we can take the magnitude using both components: $\vec{E}_{net}(2,2)$m$ = \sqrt{\vec{E}(2)_x^2 + \vec{E}(2)_y^2} = \sqrt{(-600)^2+(800)^2} = 1000$ N/C To find the angle we take in the inverse tangent of the components. Since the $x$-component is negative and the $y$-component is positive, $\vec{E}_{net}$ must be in quadrant $II$. Therefore, we must add $90^{\circ}$(This measurement is counterclockwise of the $x$-axis) $\displaystyle \tan(\theta) = \frac{\vec{E}(2)_y}{\vec{E}(2)_x} + 90^{\circ}$ $\displaystyle \theta = \tan^{-1}(\frac{800}{600}) \approx 53^{\circ} + 90^{\circ} \approx 143^{\circ}$
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