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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Electric field above a square ring

Example 2.2 (page 62) showed us how to calculate the electrif field a distance $z$ above the midpoint of a straight line segment of length $2L$ with uniform line charge $\lambda$:

$${\mathbf{E}}_{\text{Ex. 2.2}}=\frac{1}{4\pi {ϵ}_0}\frac{2\lambda L}{z\sqrt{z^2+L^2}}\hat{z}$$

We can recycle this result to quickly find the electric field above a square ring. A square ring is simply 4 of these straight segments in a square arrangement.

In this case, we want $L$ to be (so we recycle the formula):

$$L=\frac{a}{2}$$

Consider for a second just one of the sides. The distance from the rod's middle point to the test charge is going to be:

$$\sqrt{z^2+(a/2{)}^2}$$

So in the original formula (result from Example 2.2) we will have to change

$$\begin{array}{rl}L& \to \frac{a}{2}\\ z& \to \sqrt{z^2+(a/2{)}^2}\end{array}$$

AND! We have to add a factor of $\cos \left(\theta \right)$ so we only get the vertical component. Now, instead of $\cos \left(\theta \right)$, consider writing:

$$\begin{array}{rl}\cos \left(\theta \right)& =\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ & =\frac{z}{\sqrt{z^2+(a/2{)}^2}}\end{array}$$

Since we have 4 sides, we have that the Electric field due to the square ring at a distance $z$ above from the origin is

$$\begin{array}{rl}\mathbf{E}& =4|{\mathbf{E}}_{\text{Modified Ex. 2.2}}|\cos \left(\theta \right)\hat{\mathbf{z}}\\ & =4\frac{1}{4\pi {ϵ}_0}\frac{\lambda a}{\sqrt{z^2+\frac{a^2}{4}}\sqrt{z^2+\frac{a^2}{2}}}\frac{z}{\sqrt{z^2+(a/2{)}^2}}\hat{\mathbf{z}}\\ & =\frac{1}{\pi {ϵ}_0}\frac{\lambda az}{(z^2+\frac{a^2}{4})\sqrt{z^2+\frac{a^2}{2}}}\hat{\mathbf{z}}\end{array}$$

If you are interested in studying how this electric field looks like, consider the following two manipulable plots (both with $\lambda ={ϵ}_0$):

1. At consatnt $a$, variable $z$.

2. At constant $z$, variable $a$.

For a given $z$, there is an $a$ which maximizes the field at that point, and vice versa.

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