A semi-circular arc of radius ' $a$ ' is charged uniformly and the charge per unit length is $\lambda$. The electric field at the centre of this arc is

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Aicharya · Accepted Answer

$\lambda=$ linear charge density; Charge on elementary portion $d x=\lambda d x$.

Electric field at $O, d E=\frac{\lambda d x}{4 \pi \varepsilon_{0} a^{2}}$
Horizontal electric field, i.e., perpendicular to $A O$, will be cancelled. Hence, net electric field = addition of all electrical fields in direction of $A O$ $=\Sigma d E \cos \theta$
$\Rightarrow E=\int \frac{\lambda d x}{4 \pi \varepsilon_{0} a^{2}} \cos \theta$
Also, $d \theta=\frac{d x}{a}$ or $d x=a d \theta$
$E=\int_{-\pi / 2}^{\pi / 2} \frac{\lambda \cos \theta d \theta}{4 \pi \varepsilon_{0} a}=\frac{\lambda}{4 \pi \varepsilon_{0} a}[\sin \theta]_{-\pi / 2}^{\pi / 2}$
$=\frac{\lambda}{4 \pi \varepsilon_{0} a}[1-(-1)]=\frac{\lambda}{2 \pi \varepsilon_{0} a}$

A semi-circular arc of radius ' $a$ ' is charged uniformly and the charge per unit length is $\lambda$. The…

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