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(D) Given,linear charge density $\lambda = \frac{q}{\pi r}$.Consider a small element of the ring subtending an angle $d	heta$ at the center. The char

Vedclass · Accepted Answer

$-\frac{q}{2{\pi ^2}{\varepsilon _0}{r^2}}\,\hat{j}$

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(D) Given,linear charge density $\lambda = \frac{q}{\pi r}$.Consider a small element of the ring subtending an angle $d	heta$ at the center. The charge on this element is $dq = \lambda (r d	heta) = \frac{q}{\pi r} (r d	heta) = \frac{q}{\pi} d	heta$.The electric field due to this element at the center $O$ is $dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q d	heta}{\pi r^2}$.Due to symmetry,the horizontal components of the electric field cancel out,and the vertical components add up.The vertical component of the field is $dE_y = dE \cos	heta$ (directed along $-\hat{j}$).Integrating from $-\pi/2$ to $\pi/2$:$E = \int_{-\pi/2}^{\pi/2} \frac{1}{4\pi\varepsilon_0} \frac{q d	heta}{\pi r^2} \cos	heta (-\hat{j})$$E = -\frac{q}{4\pi^2\varepsilon_0 r^2} \hat{j} \int_{-\pi/2}^{\pi/2} \cos	heta d	heta$$E = -\frac{q}{4\pi^2\varepsilon_0 r^2} \hat{j} [\sin	heta]_{-\pi/2}^{\pi/2}$$E = -\frac{q}{4\pi^2\varepsilon_0 r^2} \hat{j} [1 - (-1)] = -\frac{q}{4\pi^2\varepsilon_0 r^2} \hat{j} (2) = -\frac{q}{2\pi^2\varepsilon_0 r^2} \hat{j}$.

$A$ thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net electric field $\vec{E}$ at the center $O$ is:

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