Q85P Negative charge -Q is distribute... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 4: Q85P (page 719)

Negative charge -Q is distributed uniformly around a quarter-circle of radius a that lies in the first quadrant, with the center of curvature at the origin. Find the x- and y-components of the net electric field at the origin.

Short Answer

Expert verified

The x- and y- component of the net electric field at the origin is ${鈭�}_{x} = {鈭�}_{y} = \frac{Q}{2 蟺^{2} {鈭�}_{0} a^{2}}$

Step by step solution

Formula

The formula for the quadrant, with the center of curvature at the origin

Therefore, for x- and y- components;

$\overset{鈫�}{dE} = \frac{碍位 (诲胃)}{a} \cos 胃 \hat{(i)} + \frac{碍位 (诲胃)}{a} \sin 胃 \hat{(j)}$

Solution

$鈬� d {\overset{鈫�}{鈭�}}_{x} = \frac{kk诲胃}{0} 肠辞蝉胃 (+ \hat{x}) \{\begin{cases} d {\overset{鈫�}{鈭�}}_{y} = \frac{kh诲胃}{0} 蝉别苍胃 (+ \overset{鈫�}{J}) \\ p^{蟺 / 2} k诲胃 \end{cases} = {鈭�}_{0}^{蟺 / 2} \frac{k诲胃}{a} 肠辞蝉胃 (+ \hat{x}) + \frac{kh}{a} {鈭�}_{0}^{蟺 / 2} 肠辞蝉胃诲胃 (\hat{x}) {鈭�}_{0}^{蟺 / 2} \frac{k诲胃}{a} 蝉别苍胃 (+ j) = \frac{kh}{a} {鈭�}_{0}^{蟺 / 2} 蝉别苍胃t胃 (j) = \frac{kh}{a} ({蝉别苍胃I}_{0}^{蟺 / 2}) (+ 位) 鈬� \frac{kh}{a} (+ 位) \frac{kh}{a} (- {肠辞蝉胃I}_{0}^{蟺 / 2}) (- j) 鈬� \frac{kh}{a} (+ j) k = \frac{1}{4 蟺 {鈭�}_{o}} {鈭�}_{x} = {鈭�}_{y} = \frac{2 Q}{4 蟺^{2} {鈭�}_{o} a^{2}} = \frac{2 Q}{2 蟺^{2} {鈭�}_{o} a^{2}}$

Hence, the x- and y- component of the net electric field at the origin is

${鈭�}_{x} = {鈭�}_{y =} \frac{2 Q}{2 蟺^{2} {鈭�}_{o} a^{2}}$

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91影视

Negative charge -Q is distributed uniformly around a quarter-circle of radius a that lies in the first quadrant, with the center of curvature at the origin. Find the x- and y-components of the net electric field at the origin.

Short Answer

Step by step solution

Formula

Solution

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