91Ó°ÊÓ

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,

$\begin{array}{l}E=\frac{2k\mathrm{λ}}{r}\sin \frac{\mathrm{θ}}{2}\text{ â¶Ä‰}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)\\ \mathrm{λ}=\text{linechargedensity},\\ r=\text{distanceofpointfromthecurve,}\\ \mathrm{θ}\text{=theangleofquadrant}\end{array}$

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

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

Now, for case b, the magnitude of the electric field at the center can be given using equation (i) as$(\mathrm{θ}=\mathrm{π})$:

$\begin{array}{c}{E}_2=\frac{2k\mathrm{λ}}{r}\sin \frac{\mathrm{π}}{2}\\ =\frac{2k\mathrm{λ}}{r}\end{array}$

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

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

Now, for case d, the magnitude of the electric field at the center can be given using equation (i) as:$(\mathrm{θ}=2\mathrm{π})$

$\begin{array}{c}{E}_4=\frac{2k\mathrm{λ}}{r}\sin \mathrm{π}\\ =0\end{array}$

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)

$\begin{array}{c}{E}_5=\left|\frac{2k\mathrm{λ}}{r}\sin \frac{3\mathrm{π}}{4}\right|−\frac{2k\mathrm{λ}}{r}\sin \frac{\mathrm{π}}{2}+\frac{2k\mathrm{λ}}{r}\sin \frac{\mathrm{π}}{4}\\ =\frac{4k\mathrm{λ}}{r}\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{2\sqrt{2}k\mathrm{λ}}{r}\end{array}$

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