JEE Main 2026 · 6 Apr · Shift 1 · Physics Q38

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PhysicsSingle correct+4 / -1Question 38 of 75

A thin half ring of radius 35 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 V/m, then the value of $Q$ is ________ nC.

( $\epsilon_o=8.85\times10^{-12}$ C²/Nm² and $\pi=3.14$ )

2.14

2.44

3.25

0.7

Answer and worked solution

Official answer: A

Field of a half ring. $E=\dfrac{2k\lambda}r$ with $\lambda=\dfrac Q{\pi r}$ , so $E=\dfrac{Q}{2\pi^2\epsilon_0r^2}$ .

Solve.
$Q=\frac{E\pi r^2}{2k}=\frac{100\times3.14\times0.1225}{1.8\times10^{10}}\approx2.14\times10^{-9}\text{ C}.$

(A)

Topics:Electric field

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JEE Main 2026 · 6 Apr · Shift 1, Physics question 38