A circular wire of radius R carries a total charge Q distributed uniformly over its circumference. A small length of the wire subtending angle θ at the centre is cut off. Find the electric field at the centre due to the remaining portion.

$\frac{Q}{4 \pi^2 \varepsilon_0 R^2} \sin \left(\frac{\theta}{2}\right)$

Electric field due to an arc at its centre is

$\frac{k \lambda}{R} 2 \sin \left(\frac{\theta}{2}\right), \text { Where } k=\frac{1}{4 \pi \varepsilon_0}$,

$\theta=$ angle subtended by the wire at the centre,
$\lambda=$ Linear density of charge.

Let E be the electric field due to remaining portion. Since intensity at the centre due to the circular wire is zero.

Applying principle of superposition.

$\frac{k \lambda}{R} 2 \sin \left(\frac{\theta}{2}\right) \hat{n}+\vec{E}=0$

$|\vec{E}|=\frac{1}{4 \pi \varepsilon_0 R} \cdot \frac{Q}{2 \pi R} . 2 \sin \left(\frac{\theta}{2}\right)$

$=\frac{Q}{4 \pi^2 \varepsilon_0 R^2} \sin \left(\frac{\theta}{2}\right)$