EST_JEE-MAIN 2021_(S2) - Competishun – Live JEE & NEET Courses and Exam Prep

JEE-MAIN 2021

27-08-2021 S2

Question

Figure shows a rod $A B$, which is bent in a $120^{\circ}$ circular arc of radius $R$. A charge ( $-Q$ ) is uniformly distributed over rod $A B$. What is the electric field $\vec{E}$ at the centre of curvature O ?

Select the correct option:

$\frac{3 \sqrt{3} Q}{8 \pi \varepsilon_0 R^2}(\hat{i})$

$\frac{3 \sqrt{3} Q}{8 \pi^2 \varepsilon_0 R^2}(\hat{i})$

$\frac{3 \sqrt{3} Q}{16 \pi^2 \varepsilon_0 R^2}(\hat{i})$

$\frac{3 \sqrt{3} Q}{8 \pi^2 \varepsilon_0 R^2}(-\hat{i})$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Easy

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