A circular conductor with radius $R$ is electrically charged. On the line perpendicular to the plane of the circle and passing through its center, where will the magnitude of the electric field be maximal?
Divide the ring into $n$ equal parts ($n \gg 1$). If the total charge is $Q$, then a small segment will have a charge of $Q / n$. This charge creates an electric field $E_{n}$ at a distance $x$ from the plane of the ring along the axis (1. see the diagram), whose magnitude is

$$
E_{n}=\frac{(Q / n)}{R^{2}+x^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_13f01ab1613c75ba504ag-1.jpg?height=461&width=377&top_left_y=417&top_left_x=863)

We are only interested in the axial component of this electric field, because the components perpendicular to the axis cancel out due to symmetry, so the resulting electric field is axial. Since each small charge segment creates an axial electric field of the same magnitude, the magnitude of the resulting electric field is:

$$
E=n \cdot \frac{|x|}{\sqrt{R^{2}+x^{2}}} E_{n}=Q \frac{|x|}{\left(R^{2}+x^{2}\right)^{3 / 2}}
$$

$E$ as a function of $x$ is an even function. Let's examine it on the interval $(0, \infty)$ for $x > 0$:

$$
E=Q \cdot \frac{x}{\left(R^{2}+x^{2}\right)^{2 / 3}}, \quad \text { so } \quad \frac{\mathrm{d} E}{\mathrm{~d} x}=Q \frac{R^{2}-2 x^{2}}{\left(R^{2}+x^{2}\right)^{3 / 2}}
$$

On the interval $(0, \infty)$, the maximum can occur where $\frac{\mathrm{d} E}{\mathrm{~d} x}=0$, which is at

$$
x_{0}=R \sqrt{2}
$$

Since on the interval $(0, R \sqrt{2})$, $\frac{\mathrm{d} E}{\mathrm{~d} x}>0$, so $E$ is strictly increasing, and on the interval $(R \sqrt{2}, \infty)$, $\frac{\mathrm{d} E}{\mathrm{~d} x}<0$, so $E$ is strictly decreasing; the function indeed takes its maximum value at $x_{0}=R \sqrt{2}$ in the interval $(0, \infty)$.

Since $E$ is even, at $-x_{0}=-R \sqrt{2}$, $E$ takes the same maximum value as before, and is smaller everywhere else.

Farkas Ferenc (Szeged, Radnóti M. Gymnasium, IV. grade)