Given: A thin semicircular ring of radius $r$ carries total positive charge $q$ uniformly distributed.

Linear charge density: $\lambda = \frac{q}{\pi r}$ since the length of the semicircle is $\pi r$.

Take an element subtending angle $d\theta$ at the center: $dq = \lambda \, r \, d\theta = \frac{q}{\pi} d\theta$

Field due to $dq$ at center: Distance from the element to the center is $r$, so $dE = \frac{1}{4\pi\varepsilon_0}\frac{dq}{r^2}$

Direction: for a positively charged upper semicircle, field at the center is directed away from the charge element, i.e. toward the lower side. By symmetry, horizontal components cancel and vertical components add downward.

Resolve vertical component: If $\theta$ is measured from the positive $x$-axis over the upper semicircle, then $dE_y = -dE\sin\theta$ Hence, $dE_y = -\frac{1}{4\pi\varepsilon_0}\frac{dq}{r^2}\sin\theta$

*Integrate from $\theta=0$ to $\pi$: $E_y = -\frac{1}{4\pi\varepsilon_0 r^2}\int_0^\pi \sin\theta \, dq$

Using $dq = \frac{q}{\pi}d\theta$, $E_y = -\frac{1}{4\pi\varepsilon_0 r^2}\cdot \frac{q}{\pi}\int_0^\pi \sin\theta \, d\theta$

Evaluate the integral: $\int_0^\pi \sin\theta\, d\theta = 2$

Therefore, $E_y = -\frac{1}{4\pi\varepsilon_0 r^2}\cdot \frac{q}{\pi}\cdot 2$ $E_y = -\frac{q}{2\pi^2\varepsilon_0 r^2}$

Vector form: $\vec E = -\frac{q}{2\pi^2\varepsilon_0 r^2}\,\hat{j}$

Option matching: This is Option C.

Comparison with stored answer: Stored correct answer is C, which matches the derived result.