Charge q is uniformly distributed over a thin | vedclass

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Question

From figure, $\mathrm{d} \ell=\mathrm{Rd} 	heta$

Charge on $d \ell=\lambda \mathrm{Rd} 	heta$

where $\lambda=$ linear charge density.

Elect

Vedclass · Accepted Answer

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

Vedclass · Answer

From figure, $\mathrm{d} \ell=\mathrm{Rd} 	heta$

Charge on $d \ell=\lambda \mathrm{Rd} 	heta$

where $\lambda=$ linear charge density.

Electric field at centre due to $d \ell$

$\mathrm{dE}=\mathrm{k} \cdot \frac{\lambda \mathrm{Rd} 	heta}{\mathrm{R}^{2}}$

We need to consider only the component $dE$ $\cos 	heta,$ as the component $dE$ sin $	heta$ will cancel out.

$	herefore  $ Total field at centre $=2 \int_{0}^{\pi / 2} \mathrm{dE}\, \cos 	heta$

$=2 \int_{0}^{\pi / 2} \frac{\mathrm{k} \lambda \mathrm{R}\, \cos 	heta}{\mathrm{R}^{2}} \mathrm{d} 	heta=\frac{2 \mathrm{k} \lambda}{\mathrm{R}} \int_{0}^{\pi / 2} \cos\, 	heta \mathrm{d} 	heta$

$=\frac{9}{2 \pi^{2} \epsilon_{0} R^{2}} \quad\left(	ext { since } \lambda=\frac{q}{\pi R}ight)$

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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