$\mathbf{+}\mathit{Q}$In Fig. 22-30a, a circular plastic rod with uniform charge$\mathbf{+}\mathit{Q}$produces an electric field of magnitude Eat 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 dexcept the rod in the fourth quadrant has charge$\mathbf{-}\mathit{Q}$
. Rank the five arrangements according to the magnitude of the electric field at the center of curvature, greatest first.

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,

$\begin{array}{l}E=\frac{2k\lambda }{r}\sin \frac{\theta }{2}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)\\ \lambda =\text{linechargedensity},\\ r=\text{distanceofpointfromthecurve,}\\ \theta \text{=theangleofquadrant}\end{array}$

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

$\begin{array}{c}{E}_1=\frac{2k\lambda }{r}\sin \frac{\pi }{4}\\ =\frac{2k\lambda }{r}\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{\sqrt{2}k\lambda }{r}\end{array}$

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

$\begin{array}{c}{E}_2=\frac{2k\lambda }{r}\sin \frac{\pi }{2}\\ =\frac{2k\lambda }{r}\end{array}$

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

$\begin{array}{c}\left|E_3\right|=\left|\frac{2k\lambda }{r}\sin \frac{3\pi }{4}\right|\\ =\frac{2k\lambda }{r}\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{\sqrt{2}k\lambda }{r}\end{array}$

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

$\begin{array}{c}{E}_4=\frac{2k\lambda }{r}\sin \pi \\ =0\end{array}$

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)

$\begin{array}{c}{E}_5=\left|\frac{2k\lambda }{r}\sin \frac{3\pi }{4}\right|-\frac{2k\lambda }{r}\sin \frac{\pi }{2}+\frac{2k\lambda }{r}\sin \frac{\pi }{4}\\ =\frac{4k\lambda }{r}\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{2\sqrt{2}k\lambda }{r}\end{array}$

Hence, the rank of the arrangements according to the magnitudes of electric field is$E_5>E_2>E_1=E_3>E_4$ .