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(B) The electric potential $V$ due to a charged arc at its centre is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$,where $Q$ is the total c

Vedclass · Accepted Answer

$\frac{\lambda}{2 \varepsilon_0}$

Vedclass · Answer

(B) The electric potential $V$ due to a charged arc at its centre is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$,where $Q$ is the total charge on the arc and $R$ is its radius.For a half ring of radius $R$,the total charge $Q = \lambda 	imes (\pi R)$.Thus,the potential at the centre due to the first half ring is $V_1 = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda (\pi R_1)}{R_1} = \frac{\lambda}{4 \varepsilon_0}$.Similarly,the potential at the centre due to the second half ring is $V_2 = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda (\pi R_2)}{R_2} = \frac{\lambda}{4 \varepsilon_0}$.The net potential at the centre is $V_{net} = V_1 + V_2 = \frac{\lambda}{4 \varepsilon_0} + \frac{\lambda}{4 \varepsilon_0} = \frac{\lambda}{2 \varepsilon_0}$.

The electric potential at the centre of two concentric half rings of radii $R_1$ and $R_2$,having the same linear charge density $\lambda$,is $(\varepsilon_0 = \text{permittivity of free space})$

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