The electric field at the centre O of a semic | vedclass

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Question

(A) Consider a small element of length $dl = a d	heta$ at an angle $	heta$ with the axis of symmetry $PO$. The charge on this element is $dq = \lamb

Vedclass · Accepted Answer

$\frac{\lambda}{2\pi \varepsilon_0 a}$

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(A) Consider a small element of length $dl = a d	heta$ at an angle $	heta$ with the axis of symmetry $PO$. The charge on this element is $dq = \lambda dl = \lambda a d	heta$.The electric field $dE$ at the centre $O$ due to this element is $dE = \frac{1}{4\pi \varepsilon_0} \frac{dq}{a^2} = \frac{1}{4\pi \varepsilon_0} \frac{\lambda a d	heta}{a^2} = \frac{\lambda d	heta}{4\pi \varepsilon_0 a}$.By symmetry,the components of the electric field perpendicular to the axis of symmetry $PO$ cancel out,while the components along $PO$ add up.The component of $dE$ along $PO$ is $dE_{\parallel} = dE \cos 	heta = \frac{\lambda}{4\pi \varepsilon_0 a} \cos 	heta d	heta$.Integrating from $	heta = -\pi/2$ to $\pi/2$:$E = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4\pi \varepsilon_0 a} \cos 	heta d	heta = \frac{\lambda}{4\pi \varepsilon_0 a} [\sin 	heta]_{-\pi/2}^{\pi/2} = \frac{\lambda}{4\pi \varepsilon_0 a} (1 - (-1)) = \frac{2\lambda}{4\pi \varepsilon_0 a} = \frac{\lambda}{2\pi \varepsilon_0 a}$.

The electric field at the centre $O$ of a semicircle of radius $a$ having a linear charge density $\lambda$ is given by:

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