7. $\left(10\right.$ points) An arc, with a central angle of $\alpha=60^{\circ}$, is cut from a circle with radius $R=40 \mathrm{~cm}$. A charge $q=5$ μC is uniformly distributed along the arc. Determine the electric field strength $E$ at the center of curvature of this arc.
Answer: 269 kV/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-12.jpg?height=232&width=576&top_left_y=2077&top_left_x=817)

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 is: $d q=\frac{q}{R \cdot \frac{\pi}{3}} \cdot d l$.

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

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

![](https://cdn.mathpix.com/cropped/2024_05_06_96efd573cfdf000c103ag-13.jpg?height=186&width=168&top_left_y=381&top_left_x=1024)

Summing over all such elements, we get:

$E=k \frac{3 q \cdot 2 R \sin 30^{\circ}}{\pi R^{3}}=6 k \frac{q \sin 30^{\circ}}{\pi R^{2}}=6 \cdot 9 \cdot 10^{9} \frac{5 \cdot 10^{-6} \sin 30^{\circ}}{\pi \cdot 0.4^{2}}=268574 \frac{\mathrm{V}}{\mathrm{m}}$.