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(C) The length of the arc is given by $L = r	heta = r 	imes \frac{\pi}{3} = \frac{r\pi}{3}$.Since the linear charge density is $\lambda$,the total c

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$\frac{\lambda}{12\varepsilon_0}$

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(C) The length of the arc is given by $L = r	heta = r 	imes \frac{\pi}{3} = \frac{r\pi}{3}$.Since the linear charge density is $\lambda$,the total charge $q$ on the arc is $q = \lambda 	imes L = \lambda 	imes \frac{r\pi}{3}$.All points on the arc are at an equal distance $r$ from the centre.The electric potential $V$ at the centre due to a charge $q$ at distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$.Substituting the value of $q$,we get $V = \frac{1}{4\pi\varepsilon_0} 	imes \frac{(\lambda r\pi / 3)}{r}$.Simplifying this,$V = \frac{\lambda r\pi}{12\pi\varepsilon_0 r} = \frac{\lambda}{12\varepsilon_0}$.

An arc of radius $r$ carries a charge. The linear charge density is $\lambda$ and the arc subtends an angle of $\frac{\pi}{3}$ at the centre. What is the electric potential at the centre?

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