Physics · Electrostatics

JEE Main 2026 — 6 April, Morning Shift — Question 9

A thin half ring of radius $35\mathrm{cm}$ is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is $100\mathrm{V/m}$ , then the value of Q is ______ nC. ( $\epsilon_0 = 8.85\times10^{-12}\mathrm{C^2/Nm^2}$ and $\pi = 3.14$ )

Option A:
2.14
Correct
Option B:
2.44
Option C:
3.25
Option D:
0.7

Answer: A

Step-by-step solution

Electric field at centre of half ring: $E = \frac{2k\lambda}{R} \sin(\theta/2)$ with $\theta=\pi$ for half ring, so $E = \frac{2kQ}{\pi R^2}$ . Given $E=100$ , $R=0.35$ m, $k=9\times10^9$ . Solve: $Q = \frac{E \pi R^2}{2k} = \frac{100 \times 3.14 \times (0.35)^2}{2 \times 9\times10^9} = 2.14\times10^{-9}$ C = 2.14 nC.

Answer key and solution verified before publishing.

Practise Electrostatics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Field & motion of charge

A three coulomb charge moves from the point (0, -2, -5) to the point (5, 1, 2) in an electric field expressed as vecE = 2xhati + 3y^2hatj +…

Answer: A

Step-by-step solution

Practise Electrostatics

More Electrostatics questions from this paper