The linear charge density on a dielectric rin | vedclass

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Question

(D) The charge on an infinitesimal element of the arc which subtends an angle $d	heta$ at the center of the ring is given by:$dQ = \lambda R d	heta

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(D) The charge on an infinitesimal element of the arc which subtends an angle $d	heta$ at the center of the ring is given by:$dQ = \lambda R d	heta = \lambda_0 \cos(	heta/2) R d	heta$The potential $dV$ at the centre of the ring due to this charge $dQ$ is:$dV = \frac{1}{4 \pi \epsilon_0} \frac{dQ}{R} = \frac{\lambda_0 \cos(	heta/2) R d	heta}{4 \pi \epsilon_0 R} = \frac{\lambda_0}{4 \pi \epsilon_0} \cos(	heta/2) d	heta$To find the total potential $V$,we integrate over the entire ring from $	heta = 0$ to $	heta = 2\pi$:$V = \int_0^{2\pi} dV = \frac{\lambda_0}{4 \pi \epsilon_0} \int_0^{2\pi} \cos(	heta/2) d	heta$$V = \frac{\lambda_0}{4 \pi \epsilon_0} \left[ 2 \sin(	heta/2) ight]_0^{2\pi}$$V = \frac{\lambda_0}{4 \pi \epsilon_0} \left[ 2 \sin(\pi) - 2 \sin(0) ight] = \frac{\lambda_0}{4 \pi \epsilon_0} [0 - 0] = 0 	ext{ V}$

The linear charge density on a dielectric ring of radius $R$ varies with $\theta$ as $\lambda = \lambda_0 \cos(\theta/2)$,where $\lambda_0$ is a constant. Find the potential at the centre $O$ of the ring.

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