The electric potential at the centre of two concentric half rings of radii $R_1$ and $R_2$, having same linear charge density $\lambda$ is
JEE MAIN 2023, Diffcult
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Potential at centre
$V=\frac{\left(\lambda \cdot \pi R_2\right)}{4 \pi \varepsilon_0 R_2}+\frac{\left(\lambda \cdot \pi R_1\right)}{4 \pi \varepsilon_0 R_1}$
$=\frac{\lambda}{2 \varepsilon_0}$
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