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JEE MAIN - Physics (2024 - 9th April Morning Shift - No. 22)

At the centre of a half ring of radius $$\mathrm{R}=10 \mathrm{~cm}$$ and linear charge density $$4 \mathrm{~nC} \mathrm{~m}^{-1}$$, the potential is $$x \pi \mathrm{V}$$. The value of $$x$$ is _________.
Risposta
36

Spiegazione

$$\begin{aligned} V & =\frac{K Q}{R} \\ & =\frac{9 \times 10^9 \times 4 \times 10^{-9} \pi R}{R} \\ & =36 \pi \end{aligned}$$

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