At the centre of a half ring of radius $R=10 \mathrm{~cm}$ and linear charge density $4 \mathrm{n} \mathrm{C} \mathrm{m}^{-1}$, the potential is $x \pi V$. The value of $x$ is . . . . .
JEE MAIN 2024, Diffcult
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Potential at centre of half ring
$\mathrm{V}=\frac{\mathrm{KQ}}{\mathrm{R}}$
$\mathrm{V}=\frac{\mathrm{K} \lambda \pi \mathrm{R}}{\mathrm{R}}$
$\mathrm{V}=\mathrm{K} \lambda \pi \Rightarrow \mathrm{V}=9 \times 10^9 \times 4 \times 10^{-9} \pi$
$\mathrm{V}=36 \pi$
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