Question

Electric field at centre O of semicircle of radius a having linear charge density lambda is given as

Answer 1

Let a small element be of length dl and charge = dq = λ dl , at which point the radius makes an angle θ with the vertical through the center(o) of geometry of the semi circle.

dE = K dq/a² , makes an angle θ with the vertical. The components along the y axis (vertical) add up. The limits are θ = -π/2 to π/2.

Answer 2

Answer:

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Explanation:

consider a small element of length dl and charge = dq = λ dl  , at which point the radius makes an angle θ with the vertical through the center of geometry of the semi circle.

K = 1/(4πε) 

dE = K dq/a²    ,   makes an angle θ with the vertical.

The components along the y axis (vertical) add up.   The limits are θ = -π/2 to π/2.

dl = a dθ

dE_y = (K λ /a) Cos θ dθ

 integrating,   E = (K λ / a)  [sinθ ]  from θ = -pi/2  to pi/2

E = 2 K λ/a

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