Question

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

• A. $\frac{2\lambda }{{\epsilon }_{0}a}$
• B. $\frac{\lambda \pi }{{\epsilon }_{0}a}$
• C. $\frac{\lambda }{2\pi {\epsilon }_{0}a}$
• D. $\frac{\lambda }{\pi {\epsilon }_{0}a}$

Answer 1

Answer: (C)

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 $dl(=ad\theta )$ along PO.
$dE=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{a^2}\cos \theta$
$(\because dl=ad\theta )$
$=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dl}{a^2}\cos \theta$
$=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda (ad\theta )}{a^2}\cos \theta$ Net electric field at O
$E={\int }_{-\pi /2}^{\pi /2}dE=2{\int }_O^{\pi /2}\frac{1}{4\pi {\epsilon }_0}\frac{\lambda a\cos \theta d\theta }{a^2}$
$=2\cdot \frac{1}{4\pi {\epsilon }_O}\frac{\lambda }{a}[\sin \theta {]}_o^{\pi /2}$
$=2\cdot \frac{1}{4\pi {\epsilon }_O}\cdot \frac{\lambda }{a}\cdot 1=\frac{\lambda }{2\pi {\epsilon }_{o}a}$