The electric field at the center of a semi-ci | vedclass

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Question

(C) Consider a small element of length $dl = a d	heta$ on the semi-circle at an angle $	heta$ with the vertical axis. The charge on this element is

Vedclass · Accepted Answer

$\frac{\lambda}{2\pi \epsilon_0 a}$

Vedclass · Answer

(C) Consider a small element of length $dl = a d	heta$ on the semi-circle at an angle $	heta$ with the vertical axis. The charge on this element is $dq = \lambda dl = \lambda a d	heta$.The electric field due to this element at the center is $dE = \frac{1}{4\pi \epsilon_0} \frac{dq}{a^2} = \frac{\lambda d	heta}{4\pi \epsilon_0 a}$.Due to symmetry,the horizontal components of the electric field cancel out,and the vertical components add up.The net electric field is $E = \int_{-\pi/2}^{\pi/2} dE \cos 	heta = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4\pi \epsilon_0 a} \cos 	heta d	heta$.$E = \frac{\lambda}{4\pi \epsilon_0 a} [\sin 	heta]_{-\pi/2}^{\pi/2} = \frac{\lambda}{4\pi \epsilon_0 a} (1 - (-1)) = \frac{2\lambda}{4\pi \epsilon_0 a} = \frac{\lambda}{2\pi \epsilon_0 a}$.

The electric field at the center of a semi-circle of radius $a$ and linear charge density $\lambda$ is given by:

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