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

Option 1 -

$\frac{3 \sqrt[]{3} Q}{8 π^{2} ε_{0} R^{2}} (- \hat{i})$

Option 2 -

$\frac{3 \sqrt[]{3} Q}{8 π^{2} ε_{0} R^{2}} (\hat{i})$

Option 3 -

$\frac{3 \sqrt[]{3} Q}{8 π ε_{0} R^{2}} (\hat{i})$

Option 4 -

$\frac{3 \sqrt[]{3} Q}{16 π^{2} ε_{0} R^{2}} (\hat{i})$

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$λ = \frac{- Q}{R \times 2 π / 3}$

$\vec{E} = \frac{2 k λ}{R} s i n (\frac{θ}{2}) (- \hat{i})$

$\Rightarrow \vec{E} = \frac{2 k}{R} \times \frac{- 3 Q}{2 π R} s i n 60 ° (- \hat{i})$

$\Rightarrow \vec{E} = \frac{3 \sqrt[]{3} Q}{8 π^{2} ε_{0} R^{2}} (\hat{i})$

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

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Figure shows a rod AB, which is bent in a 120° circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric field E → at the centre of curvature O?

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