At the centre of a half ring of radius R=10 | vedclass

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(B) The potential $V$ at the centre of a charged ring is given by $V = \frac{1}{4\pi \epsilon_0} \int \frac{dq}{R}$.For a half ring of radius $R$,the

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$36$

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(B) The potential $V$ at the centre of a charged ring is given by $V = \frac{1}{4\pi \epsilon_0} \int \frac{dq}{R}$.For a half ring of radius $R$,the total charge $Q = \lambda 	imes (\pi R)$.Since every point on the half ring is at the same distance $R$ from the centre,the potential is $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}$.Substituting $Q = \lambda \pi R$,we get $V = \frac{1}{4\pi \epsilon_0} \frac{\lambda \pi R}{R} = \frac{\lambda}{4 \epsilon_0}$.Given $\lambda = 4 	imes 10^{-9} \ C/m$ and $k = \frac{1}{4\pi \epsilon_0} = 9 	imes 10^9 \ N \ m^2/C^2$.$V = k \lambda \pi = (9 	imes 10^9) 	imes (4 	imes 10^{-9}) 	imes \pi$.$V = 36 \pi \ V$.Comparing this with $x \pi \ V$,we find $x = 36$.

At the centre of a half ring of radius $R=10 \ cm$ and linear charge density $\lambda = 4 \ nC \ m^{-1}$,the potential is $x \pi \ V$. The value of $x$ is . . . . .

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