StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q13Q

Expert-verifiedFound in: Page 652

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 .${E}_{c}={E}_{a}={E}_{b}$

**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)

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:

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

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 ${E}_{c}={E}_{a}={E}_{b}$.

94% of StudySmarter users get better grades.

Sign up for free