91Ó°ÊÓ

In the given problem, the electric field at the midpoint of any rod will be zero, considering the point as an axial point to the rod. But for the rod of given length when formed circle, the electric field is given by differentiating the electric potential of the given system.

Formulae:

The linear chare density of a material$\mathrm{λ}=\frac{q}{L}$ (i)

The electric potential at a point due to an individual charge $V=\frac{kq}{r}$,(ii)

The electric field at a point $E=−\frac{dV}{dx}$, (iii)

The electric field at the midpoint of each rod will cancel out as there is an equal charge acting on the other half of the rod. Thus, the electric field at the mid-point of the rods a and b is zero$(E_a=E_b=0)$.

Considering a charge element at the circumference of the circle, the charge value can be given using equation (i) as follows:

$\begin{array}{c}dq=\mathrm{λ}dx\\ =\frac{Q}{L/2}dx\end{array}$

Now, the electric potential at any point P at a distancecan be given using the above value in equation (ii) as follows:

$∫dV=∫\frac{kdq}{R}$

$\begin{array}{c}{V}_c=∫\frac{kQ}{RL/2}dx\\ =\frac{2kQ}{RL}\underset{0}{\overset{2\mathrm{π}R}{∫}}dx\\ =\frac{kQ}{RL}\left(2\mathrm{π}R\right)\\ =\frac{2\mathrm{π}kQ}{L}\end{array}$

Now, the electric field at point P for rod c can be given using equation (iii) as follows:

$\begin{array}{c}{E}_c=\frac{−d}{dx}\left(\frac{2\mathrm{π}kQ}{L}\right)\\ =0\end{array}$

Hence, the rank of the rods according to the magnitude of the electric fields is $E_c=E_a=E_b$.