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Q.
A semicircular arc of radius $a$ is charged uniformly and the charge per unit length is $\lambda$. The electric field at its centre is
Electric Charges and Fields
Solution:
Electric intensity at centre $O$, due to small element dl of charged ring
$d E=\frac{\lambda d l}{4 \pi \varepsilon_{0} a^{2}}=\frac{\lambda(a d \theta)}{4 \pi \varepsilon_{0} a^{2}}$
As is clear from figure, horizontal components of $d E$ will cancel out in pairs and vertical components will add.
$\therefore E=\int\limits_{0}^{\pi} d E\, \sin\, \theta=\int\limits_{0}^{\pi} \frac{\lambda}{4 \pi \varepsilon_{0} a} \sin \,\theta \,d \theta$
$=\frac{\lambda}{4 \pi \varepsilon_{0} a}[-\cos \,\theta]_{0}^{\pi}$
$E=\frac{\lambda}{4 \pi \varepsilon_{0} a}(-\cos \,\pi+\cos 0)=\frac{\lambda}{2 \pi \varepsilon_{0} a}$
