The electric field at the centre of a semi-circular arc of radius 'r'…

The correct answer is Option (4) → $\frac{\lambda}{2 \pi \varepsilon_0 r}$

Semi-circular are has a linear charge density λ, so the total charge on the arc -

$Q=λL=λπr$

and,

$dE=\frac{Kλrdθ}{r^2}=\frac{Kλdθ}{r}$

$∴dE_y=dE\sin θ=\frac{Kλdθ}{r}\sin θ$

Hence,

$\int dE_y=\int\limits_{-π/2}^{π/2}\frac{Kλ\sin θ}{r}dθ$

On evaluating the limits,

$E_y=0$

But, $E_x=\frac{Kλ}{r}\int\limits_{-π/2}^{π/2}\cos θ\,dθ$

$=\frac{Kλ}{r}\left[\sin θ\right]_{-π/2}^{π/2}$

$=\frac{Kλ}{r}\left[\sin\left(\frac{π}{2}\right)-\sin\left(-\frac{π}{2}\right)\right]$

$=\frac{2Kλ}{r}$