Thin, very long, uniformly charged insulating rod ends in a quarter circle of radius $R$. The electric charge per unit length is $\sigma$ (see the figure).

Calculate the electric field strength at the center of the quarter circle!

![](https://cdn.mathpix.com/cropped/2024_05_02_3c8393415897d0b10dc6g-1.jpg?height=112&width=371&top_left_y=318&top_left_x=866)
$\mathrm{Az}$ is the superposition of the fields created by the charged quarter-circle and the semi-infinite rod. We will prove that the sought electric field strength is the same as that of an infinitely long linear charge. We need to show that the electric field strength created by the quarter-circle arc at the given point is the same as that of an infinitely long semi-infinite rod. Divide the arc into elementary segments, and for each elementary arc, assign a corresponding elementary segment on the semi-infinite rod as shown in the diagram. Thus, the quarter-circle arc can be mapped uniquely onto the semi-infinite rod. Now we will show that the electric field strength created by an elementary arc and the corresponding elementary segment is the same. The electric field strength created by the elementary arc at point $O$ is: $\Delta E_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\sigma \Delta \varphi}{R}$. The electric field strength created by the corresponding elementary segment is: $\Delta E_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\sigma \Delta x}{l^{2}}$. Using the fact that $\Delta x=\frac{l \Delta \varphi}{\cos \varphi}$ and $\cos \varphi=\frac{R}{l}$, it follows that $\Delta E_{1}=\Delta E_{2}$. Since the direction of $\Delta E_{1}$ and $\Delta E_{2}$ is also the same, our statement is proven. The electric field strength of an infinitely long linear charge at a distance $R$ is:

$$
E=\frac{1}{2 \pi \varepsilon_{0}} \frac{\sigma}{R}
$$

Therefore, the sought electric field strength at point $O$ is this value, and its direction is perpendicular to the semi-infinite rod. It also follows from the solution that the electric field strength created by a semicircle at its center is the same. The electric field of an infinitely long linear charge is well-known, and its calculation method can be found in high school textbooks.

Note: Several students solved the problem using integration. Of course, we accepted this solution as well, but we would like to point out that if the problems in the journal can be solved using the knowledge of the curriculum, then we should prefer this solution. It is also clear from the above solution that sometimes the solution is simpler if we think through the problem first and do not choose the mechanical calculation.