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