At the centre of a half ring of radius R =10 cm and linear charge density 4 n C m -1 , the potential is x π V . The value of x is .

Physics · Electrostatics

JEE Main 2024 — 9 April, Shift 1 — Question 51

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

Answer: 36

Numerical answer — enter this value.

Step-by-step solution

Potential at centre of half ring

$\mathrm{V}=\frac{\mathrm{KQ}}{\mathrm{R}}$

$\mathrm{V}=\frac{\mathrm{K} \lambda \pi \mathrm{R}}{\mathrm{R}}$

$\mathrm{V}=\mathrm{K} \lambda \pi \Rightarrow \mathrm{V}=9 \times 10^{9} \times 4 \times 10^{-9} \pi$

$\mathrm{V}=36 \pi$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential