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Electric field at centre O of semicircle of radius a having linear charge density lambda is given as
Consider a uniformly charged thin rod bent into a semicircle of radius R.Find the electric field generated at the origin of the coordinate system.Charge per unit length: ? = Q/pRCharge on slice: dq = ?Rd? (assumed positive)Electric field generated by slice: dE = kdq/r^2=k*lamda/r *dthetaComponents of d~E : dEx = dE cos ?, dEy = -dE sin ?Electric field from all slices added up:Ex =0Ey= -2klambda/rthanks and regardssunil kraskIItian faculty
Hello StudentConsider a small element of length dl and charge = dq =λ dl , at which point the radius makes an angleθ with the vertical through the center of geometry of the semi circle.K = 1/(4πε)dE = K dq/r² , makes an angle θ with the vertical.Ex =0The components along the y axis (vertical) add up. The limits are θ = -π/2 to π/2.dl = r dθdEy = (K λ /r) Cos θ dθintegrating, E = (K λ / r) [sinθ ] from θ = -pi/2 to pi/2E = 2 K λ/rI hope this answer will help you.
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