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(A) Consider a small element of charge $dq$ on the half ring at an angle $	heta$ with the vertical. The charge per unit length is $\lambda = \frac{q}

Vedclass · Accepted Answer

$\frac{q}{2{\pi ^2}{\varepsilon _0}{R^2}}$

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(A) Consider a small element of charge $dq$ on the half ring at an angle $	heta$ with the vertical. The charge per unit length is $\lambda = \frac{q}{\pi R}$.The electric field due to this element at the centre is $dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{R^2}$.Due to symmetry,the horizontal components of the electric field cancel out.The vertical component is $dE_y = dE \cos	heta = \frac{1}{4\pi\varepsilon_0} \frac{\lambda R d	heta}{R^2} \cos	heta$.Integrating from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$:$E = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4\pi\varepsilon_0 R} \cos	heta d	heta = \frac{\lambda}{4\pi\varepsilon_0 R} [\sin	heta]_{-\pi/2}^{\pi/2} = \frac{\lambda}{4\pi\varepsilon_0 R} (1 - (-1)) = \frac{2\lambda}{4\pi\varepsilon_0 R} = \frac{\lambda}{2\pi\varepsilon_0 R}$.Substituting $\lambda = \frac{q}{\pi R}$,we get $E = \frac{q/(\pi R)}{2\pi\varepsilon_0 R} = \frac{q}{2\pi^2\varepsilon_0 R^2}$.

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

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