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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 # Figure 22-32 shows three rods, each with the same charge Q spread uniformly along its length. Rods a (of length L) and b (oflength L/2) are straight, and points P are aligned with their midpoints.Rod c (of length L/2) forms a complete circle about point P. Rank the rods according to the magnitude of the electric field theycreate at points P, greatest first. The rank of the rods according to the magnitude of the electric field they create at points P is .${E}_{c}={E}_{a}={E}_{b}$

See the step by step solution

## Step 1: Understanding the concept of electric field

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 $\lambda =\frac{q}{L}$ (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)

## Step 2: Calculation of the rank of the rods according to magnitude of the electric field at point P

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 $\left({E}_{a}={E}_{b}=0\right)$ .

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

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

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

$\int dV=\int \frac{kdq}{R}$

$\begin{array}{c}{V}_{c}=\int \frac{kQ}{RL/2}dx\\ =\frac{2kQ}{RL}\underset{0}{\overset{2\pi R}{\int }}dx\\ =\frac{kQ}{RL}\left(2\pi R\right)\\ =\frac{2\pi kQ}{L}\end{array}$

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

$\begin{array}{c}{E}_{c}=\frac{-d}{dx}\left(\frac{2\pi kQ}{L}\right)\\ =0\end{array}$

Hence, the rank of the rods according to the magnitude of the electric fields is ${E}_{c}={E}_{a}={E}_{b}$. ### Want to see more solutions like these? 