7. (10 points) An arc, with a central angle of $\alpha=30^{\circ}$, is cut from a circle with radius $R=50$ cm. A charge $q=2$ μC is uniformly distributed along the arc. Determine the electric field strength $E$ at the center of curvature of this arc.
Answer: $71 \mathrm{kB} / \mathrm{m}$.

Solution. Consider a small element of the arc of length $d l$, on which charge $d q$ is located. It creates a field strength at the point of interest:

$d E=k \frac{d q}{R^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96efd573cfdf000c103ag-05.jpg?height=246&width=580&top_left_y=2127&top_left_x=818)

The projection of this field strength onto the horizontal direction is:

$d E \cos \beta=k \frac{d q}{R^{2}} \cos \beta$

The charge of this element $d q=\frac{q}{R \cdot \frac{\pi}{6}} \cdot d l$.

We obtain $d E \cos \beta=k \frac{\frac{q}{R \cdot \frac{\pi}{6}} \cdot d l}{R^{2}} \cos \beta=k \frac{6 q \cdot d l \cdot \cos \beta}{\pi R^{3}}$.

Note that $d l \cdot \cos \beta=d y$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96efd573cfdf000c103ag-06.jpg?height=177&width=166&top_left_y=525&top_left_x=1022)

Summing over all such elements, we get:

$E=k \frac{6 q \cdot 2 R \sin 15^{\circ}}{\pi R^{3}}=12 k \frac{q \sin 15^{\circ}}{\pi R^{2}}=12 \cdot 9 \cdot 10^{9} \frac{2 \cdot 10^{-6} \sin 15^{\circ}}{\pi \cdot 0.5^{2}}=71180 \frac{\mathrm{B}}{\mathrm{M}}$.