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$a$ ત્રિજ્યા અને રેખીય વિદ્યુતભાર ઘનતા $\lambda$ વાળા એક અર્ધ વર્તૂળના કેન્દ્ર $O$ આગળ વિદ્યુતક્ષેત્ર શેના દ્વારા આપી શકાય છે?
$\frac{\lambda }{{2\pi {\varepsilon _0}a}}$
$\frac{\lambda }{{2\pi {\varepsilon _0}{a^2}}}$
$\frac{\lambda }{{4{\pi ^2}{\varepsilon _0}a}}$
$\frac{{{\lambda ^2}}}{{2\pi {\varepsilon _0}a}}$
Solution
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 $dl$ ( $a d \theta$ ) along $PO$
$\mathrm{d} E=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\mathrm{d} q}{a^{2}} \cos \theta$
$=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda \mathrm{d} l}{a^{2}} \cos \theta$
$=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda(a \mathrm{d} \theta)}{a^{2}} \cos \theta$
Net electric field at $\mathrm{O}$
$E=\int_{-\pi / 2}^{\pi / 2} \mathrm{d} E=2 \int_{0}^{\pi / 2} \frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda \cos \theta \mathrm{d} \theta}{a^{2}}$
$2 \cdot \frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda}{a}[\sin \theta]_{0}^{\pi / 2}=2 \cdot \frac{1}{4 \pi \mathcal{E}_{0}} \cdot \frac{\lambda}{a} \cdot 1=\frac{\lambda}{2 \pi \mathcal{E}_{0} a}$
