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: 47

Answer

$\vec{E}_{net} = 40$ N/C $\theta = 27^{\circ}$

Work Step by Step

We first need to find an equation for the electric field of the region of space. We can do this by taking the derivative of $V(x,y)$ two different times-once with respect to $x$ and once with respect to $y$-to find the $x$ and $y$ components of the electric field. Know that $\displaystyle \vec{E} = -\frac{dV}{d\vec{s}}$, where $\vec{s}$ is the position vector. $\displaystyle \vec{E}_x(x,y) = -\frac{d}{dx}[\frac{200}{\sqrt{x^2+y^2}}] = -\frac{d}{dx}[200(x^2+y^2)^{-1/2}]$ $\displaystyle \vec{E}_x(x,y) =-[-100(x^2+y^2)^{-3/2}(2x+0)] = \frac{200x}{(x^2+y^2)^{3/2}}$ --- $\displaystyle \vec{E}_y(x,y) = -\frac{d}{dy}[\frac{200}{\sqrt{x^2+y^2}}] = -\frac{d}{dy}[200(x^2+y^2)^{-1/2}]$ $\displaystyle \vec{E}_y(x,y) =-[-100(x^2+y^2)^{-3/2}(0+2y)] = \frac{200y}{(x^2+y^2)^{3/2}}$ We can now find the values for the electric field components at the position (x,y) = ($2$m,$1$m) $\displaystyle \vec{E}_x(2,1) = \frac{200(2)}{(2^2+1^2)^{3/2}} \approx 35.8$ N/C $\displaystyle \vec{E}_y(2,1) = \frac{200(1)}{(2^2+1^2)^{3/2}} \approx 17.9$ N/C Now we can take the magnitude of these components to find $\vec{E}_{net}$ $\vec{E}_{net} = \displaystyle \sqrt{(\frac{400}{5^{3/2}})^2+(\frac{200}{5^{3/2}})^2} = 40$ N/C We can find the angle counterclockwise to the $x$-axis by: $\theta = \displaystyle \tan^{-1}(\frac{\vec{E}_y(2,1)}{{\vec{E}_x(2,1)}}) = \tan^{-1}(\frac{17.9}{35.8}) = \tan^{-1}(\frac{1}{2}) \approx 27^{\circ}$
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