Two equal charges are located at a distance of $2d$ from each other. Determine where the electric field strength is the greatest in the plane perpendicular to the segment connecting the two charges, passing through the midpoint of this segment. Determine the maximum value of the electric field strength!
I. Solution. Due to the cylindrical symmetry of the problem, it is sufficient to perform the calculations in any plane containing the two charges.

Let the charges be $Q_{1}$ and $Q_{2}\left(Q_{1}=Q_{2}\right)$, the electric fields they create be $E_{1}$ and $E_{2}$ (on the perpendicular bisector of the segment connecting the charges, $\left|E_{1}\right|=\left|E_{2}\right|$), and let $r$ denote the distance from the midpoint $F$ measured along the perpendicular bisector.

![](https://cdn.mathpix.com/cropped/2024_05_02_39beb72a5cbdd4bbbeaag-1.jpg?height=412&width=480&top_left_y=355&top_left_x=817)

The magnitude of the electric field created by each charge on the perpendicular bisector is:

$$
\left|E_{1}\right|=\left|E_{2}\right|=\frac{k Q}{r^{2}+d^{2}}
$$

The magnitude of the resultant electric field from vector addition is:

$$
\begin{gathered}
|E(r)|=\left|E_{1}\right| \frac{2 r}{\sqrt{r^{2}+d^{2}}} \\
|E(r)|=\frac{2 k Q r}{\left(r^{2}+d^{2}\right)^{3 / 2}}
\end{gathered}
$$

The direction is as shown in the diagram.

Since $E(r)$ is in the same direction for all $r>0$ and $E(r)$ approaches zero as $r$ approaches infinity, if the function (3) has only one extremum, it must be a maximum.

The first derivative is:

$$
\frac{\mathrm{d}|E|}{\mathrm{d} r}=\frac{2 k Q}{\left(r^{2}+d^{2}\right)^{5 / 2}}\left(d^{2}-2 r^{2}\right)
$$

In the domain $r>0$, the derivative is zero only at $r=d \sqrt{2} / 2$, where the derivative function also changes sign. This is therefore the location of the maximum.

Returning to the three-dimensional problem, the maximum value of the electric field - on the plane perpendicular to the segment connecting the two charges and passing through the midpoint - is taken on a circle centered at $F$ with radius $r=d \sqrt{2} / 2$. The maximum value of the electric field here is:

$$
E_{\max }=\frac{4 \sqrt{3}}{9} \frac{Q}{d^{2}}
$$

II. Solution. In most electrostatic problems, it is advantageous to work with potentials, since they add up scalarly according to the principle of superposition, thus avoiding the more complex vector addition. The electric field can be obtained by taking the gradient of the potential (for a straight-line examination, this is the first derivative).

Using the notation from the first solution:

$$
U_{1}=U_{2}=-\frac{k Q}{\sqrt{r^{2}+d^{2}}} \quad(6) \quad U=U_{1}+U_{2}=-\frac{2 k Q}{\sqrt{r^{2}+d^{2}}}
$$

from which

$$
E(r)=\frac{\mathrm{d} U}{\mathrm{~d} r}=\frac{2 k Q r}{\left(r^{2}+d^{2}\right)^{3 / 2}}
$$

Finding the location of the maximum (a mathematical problem) from here is identical to what was stated in the first solution.