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