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Expert-verified Found in: Page 652 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # ${\mathbf{+}}{\mathbit{Q}}$In Fig. 22-30a, a circular plastic rod with uniform charge ${\mathbf{+}}{\mathbit{Q}}$produces an electric field of magnitude E at the center of curvature (at the origin). In Figs. 22-30b, c, and d, more circular rods, each with identical uniform charges , are added until the circle is complete. A fifth arrangement (which would be labeled e) is like that in d except the rod in the fourth quadrant has charge${\mathbf{-}}{\mathbit{Q}}$ . Rank the five arrangements according to the magnitude of the electric field at the center of curvature, greatest first. The rank of the five arrangements according to the magnitude of the electric field at the center of curvature is${E}_{5}>{E}_{2}>{E}_{1}={E}_{3}>{E}_{4}$ .

See the step by step solution

## Step 1: Understanding the concept of wave

The electric field at the center of curvature in a given quadrant depends on the half-angle of the quadrant. Using this data, the electric field in all three cases is calculated and compared to get the required rank value.

The total electric field at the center of curvature of a circular plastic rod,

$\begin{array}{l}E=\frac{2k\lambda }{r}\mathrm{sin}\frac{\theta }{2}\text{\hspace{0.17em}\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)\\ \lambda =\text{line charge density},\\ r=\text{distance of point from the curve,}\\ \theta \text{= the angle of quadrant}\end{array}$

## Step 2: Calculation of the rank according to electric field magnitudes at the center of curvature

Now, for case a, the magnitude of the electric field at the center can be given using equation (i) as$\left(\theta =\frac{\pi }{2}\right)$:

$\begin{array}{c}{E}_{1}=\frac{2k\lambda }{r}\mathrm{sin}\frac{\pi }{4}\\ =\frac{2k\lambda }{r}\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{\sqrt{2}k\lambda }{r}\end{array}$

Now, for case b, the magnitude of the electric field at the center can be given using equation (i) as$\left(\theta =\pi \right)$:

$\begin{array}{c}{E}_{2}=\frac{2k\lambda }{r}\mathrm{sin}\frac{\pi }{2}\\ =\frac{2k\lambda }{r}\end{array}$

Now, for case c, the magnitude of the electric field at the center can be given using equation (i) as$\left(\theta =\frac{3\pi }{2}\right)$:

$\begin{array}{c}|{E}_{3}|=|\frac{2k\lambda }{r}\mathrm{sin}\frac{3\pi }{4}|\\ =\frac{2k\lambda }{r}\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{\sqrt{2}k\lambda }{r}\end{array}$

Now, for case d, the magnitude of the electric field at the center can be given using equation (i) as:$\left(\theta =2\pi \right)$

$\begin{array}{c}{E}_{4}=\frac{2k\lambda }{r}\mathrm{sin}\pi \\ =0\end{array}$

Now, for the additional fifth arrangement where the fourth quadrant has charge, the magnitude of electric field using equation (i) can be written as: (due to a negative quadrant we can get the net field as the sum of three case that is field due to three quadrant, field due to negative quadrant, and field due to the additional quadrant)

$\begin{array}{c}{E}_{5}=|\frac{2k\lambda }{r}\mathrm{sin}\frac{3\pi }{4}|-\frac{2k\lambda }{r}\mathrm{sin}\frac{\pi }{2}+\frac{2k\lambda }{r}\mathrm{sin}\frac{\pi }{4}\\ =\frac{4k\lambda }{r}\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{2\sqrt{2}k\lambda }{r}\end{array}$

Hence, the rank of the arrangements according to the magnitudes of electric field is${E}_{5}>{E}_{2}>{E}_{1}={E}_{3}>{E}_{4}$ .

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