Electrostatics

JEE Main level, numerical answer, medium

171 of 311

Question 171 of 311

At the centre of a half ring of radius R=10 cm\mathrm{R}=10 \mathrm{~cm} and linear charge density 4nCm−14 \mathrm{n} \mathrm{C} \mathrm{m}^{-1}, the potential is xπVx \pi \mathrm{V}. The value of xx is _______

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Answer: 36

Potential at centre of half ring
V=KQRV=KλπRRV=Kλπ⇒V=9×109×4×10−9πV=36π\begin{aligned} & V=\frac{K Q}{R} \\ & V=\frac{K \lambda \pi R}{R} \\ & V=K \lambda \pi \Rightarrow V=9 \times 10^9 \times 4 \times 10^{-9} \pi \\ & V=36 \pi\end{aligned}

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