Question

A rod of length l has a total charge q distributed uniformly along its length it is bent in the shape of semicirclr

Answer 1

The magnitude of the electric field at the centre of curvature of the semicircle is = Q / 2πεL^2 ⟨.πR=L)

Explanation:

Correct statement:

A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.?

Solution:

dq=QLRdθ  

Net electric field have only vertical component  =∫d.Ecosθ

= 1 / 4πε0 x Q / L∫π /2 − π/2 Rdθ / R^2 cosθ

= 1 / 4πε0 x Q / LR [sinθ]π / 2− π / 2

= 1 / 4πε0 Q/ LR [1 − 0 ( − 1 ) ]

= 2.Q / 4πε0LR

= Q / 2πεL^2 ⟨.πR=L) .

Thus the magnitude of the electric field at the centre of curvature of the semicircle is = Q / 2πεL^2 ⟨.πR=L)

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