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 _________.
Answer (integer)
36
Solution
<p>$$\begin{aligned}
V & =\frac{K Q}{R} \\
& =\frac{9 \times 10^9 \times 4 \times 10^{-9} \pi R}{R} \\
& =36 \pi
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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