$z$-component of electric field due to a static square loop

Question

I am having trouble finding the $z$-component of an electric field discussed in problem 2.4 of Griffith’s introduction to electrodynamics.

Suppose we have a square loop of side length $a$ carrying a linear charge density of $\lambda$. To find the electric field a distance $z$ above the center of the square loop, it is sufficient to find the electric field at $z$ due to one side of the square, take the component of that electric field in the $z$ direction, and multiply by $4$. This $z$ component is found by multiplying the electric field $\mathbf{E}$ by $\cos\theta$, where $\theta$ is the angle made by the separation vector $\mathbf{r}$ from the square loop to $z$ intersecting the line from the center of the square loop to $z.$ Now, according to Griffiths, $\cos\theta = \frac{z}{\sqrt{z^2 + (a/2)^2}}$. But I do not see how this is true. This expression just seems to be the cosine of the angle between $\mathbf{r}$ and $z$ when $\mathbf{r}$ is coming from the midpoint of the side of the loop. But as you vary the source of $\mathbf{r}$ across the side of the loop, to account for the contribution of the whole side of the loop to $\mathbf{E}$, that angle will change. If so then $\cos\theta$ should be $\frac{z}{\sqrt{z^2 + x^2 + (a/2)^2}}$ to reflect this. But apparently this cannot be right. Where is the error in my reasoning?

Answer 1

The electric field at a point P located at a distance $z$ above the center of a square, due to only one side of the loop is given by,

\begin{equation} \mathbf{E}(P)=\frac{\lambda}{4\pi\epsilon_0}\int_{-a/2}^{a/2}\frac{dx}{r^2}\mathbf{\hat{r}} \end{equation}

At this point, you can use a symmetry argument to say that the total field at P due to the total loop has only a $z$ component (the $x$ and $y$ components vanish due to symmetry). So you can focus only on the $z$ component of the field generated by a side and multiply by 4, as you said in your question. By symmetry, $\mathbf{E}(P)=E(P)\mathbf{\hat{z}}$, so

\begin{equation} E(P)=\frac{\lambda}{4\pi\epsilon_0}\int_{-a/2}^{a/2}\frac{dx}{r^2}\mathbf{\hat{r}}.\mathbf{\hat{z}} \end{equation}

The dot product is given by,

\begin{equation} \mathbf{\hat{r}}.\mathbf{\hat{z}}=\cos{\theta}=\frac{z}{r}=\frac{z}{\sqrt{z^2+x^2+a^2/4}} \end{equation}

Here is your formula for the cosine. But you still need to integrate over the length of one side to get the field, and then multiply by 4 to have the total field.

The Griffiths approach is to use the result of the example 2.2 by changing $z$ by $\sqrt{z^2+a^2/4}$. It represents the distance from M, the middle of the side, to P. Due to $x$ symmetry, the field lies in the plane containing M and P. This electric field has a component along $z$ which is equal to,

\begin{equation} E_z(P)=\frac{1}{4\pi\epsilon_0}\frac{a\lambda}{\sqrt{z^2+a^2/4}\sqrt{z^2+a^2/2}}\cos{\theta} \end{equation}

\begin{equation} \cos{\theta}=\frac{z}{\sqrt{z^2+a^2/4}} \end{equation}

It gives you the desired final result. So there are two approaches, one where you can use your formula to integrate over the length of the side. Griffiths uses a previous result and symmetry considerations to avoid integration.

Answer 2

So after reading Kyle Kanos’ comment, I am wondering if the answer is something like this:

If $\mathbf{E}$ is the electric field at $z$ due to the whole side of the square loop and we imagine the components of $\mathbf{E}$ laid out in space, there will be no $x$-component due to symmetry. So $\mathbf{E}$ is a vector from the midpoint of the side of the square up to $z$. In that case it would actually be inappropriate to let cos $\theta$ vary like I did in my question; the one and only angle we need to consider to find $\mathbf{|E|}$cos$\theta$ is the angle made by a ray from the midpoint of the side of the square intersecting a ray from the middle of the square loop to $z$, which is what Griffiths does.

Answer 3

Your reasoning about the variation of θ along the side is correct, but due to symmetry and the integral approach, the effective cosθ is taken at the midpoint for simplification.

The integral approach inherently accounts for the varying angle along the length of the side, leading to the simplified final expression for the electric field.

If you follow the steps above, you will see how the standard approach leads to the same result given by Griffiths, confirming that the cosθ factor you derived is indeed the correct one for the midpoint approximation used in the integral.