Electric field at centre O of semicircle of r | vedclass

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Question

Considering symmetric elements each of length $dl$ at $A$ and $B,$ we note that electric fields

perpendicular to $PO$ are cancelled and those along

Vedclass · Accepted Answer

$\frac{\lambda }{{2\pi {\varepsilon _0}a}}$

Vedclass · Answer

Considering symmetric elements each of length $dl$ at $A$ and $B,$ we note that electric fields

perpendicular to $PO$ are cancelled and those along $PO$ are added. The electric field due to an

element of length $dl$ ( $a d 	heta$ ) along $PO$

$\mathrm{d} E=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\mathrm{d} q}{a^{2}} \cos 	heta$

$=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda \mathrm{d} l}{a^{2}} \cos 	heta$

$=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda(a \mathrm{d} 	heta)}{a^{2}} \cos 	heta$

Net electric field at $\mathrm{O}$

$E=\int_{-\pi / 2}^{\pi / 2} \mathrm{d} E=2 \int_{0}^{\pi / 2} \frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda \cos 	heta \mathrm{d} 	heta}{a^{2}}$

$2 \cdot \frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda}{a}[\sin 	heta]_{0}^{\pi / 2}=2 \cdot \frac{1}{4 \pi \mathcal{E}_{0}} \cdot \frac{\lambda}{a} \cdot 1=\frac{\lambda}{2 \pi \mathcal{E}_{0} a}$

Electric field at centre $O$ of semicircle of radius $a$ having linear charge density $\lambda$ given is given by

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