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Q11Q

Expert-verifiedFound in: Page 652

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**${\mathbf{+}}{\mathit{Q}}$In Fig. 22-30 a, a circular plastic rod with uniform charge**

The rank of the five arrangements according to the magnitude of the electric field at the center of curvature is${E}_{5}>{E}_{2}>{E}_{1}={E}_{3}>{E}_{4}$ .

**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,

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

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

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

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

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

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

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

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

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

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)

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

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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