JEE Main 2022 · 30 Jul · Shift 2 · Physics Q10

← JEE Main 2022 · 30 Jul · Shift 2

PhysicsSingle correct+4 / -1Question 10 of 90

A charge q is uniformly distributed along the length L of a rod. It is then bent in the shape of a semicircle. The magnitude of electric field at the centre of semicircle will be :

\dfrac q{2\epsilon_0L^2}

\dfrac q{2\pi\epsilon_0L^2}

\dfrac q{2\pi\epsilon_0L}

\dfrac q{4\pi^2\epsilon_0L}

Answer and worked solution

Official answer: A

Semicircle of radius $R=\frac L\pi$ with $\lambda=\frac qL$ : $E=\dfrac{2k\lambda}R=\dfrac{2}{4\pi\epsilon_0}\cdot\dfrac qL\cdot\dfrac\pi L=\dfrac q{2\epsilon_0L^2}$ .

(A)

Topics:Electric field

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JEE Main 2022 · 30 Jul · Shift 2, Physics question 10