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

Swagata Saha , 11 Years ago

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2 Answers

Sunil Kumar FP

Last Activity: 11 Years ago

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/pR

Charge on slice: dq = ?Rd? (assumed positive)

Electric field generated by slice: dE = kdq/r^2=k*lamda/r *dtheta

Components of d~E : dEx = dE cos ?, dEy = -dE sin ?

Electric field from all slices added up:

Ex =0

Ey= -2klambda/r

thanks and regards

sunil kr

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Yash Chourasiya

Last Activity: 5 Years ago

Hello Student

Consider 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.
E_x =0
The components along the y axis (vertical) add up. The limits are θ = -π/2 to π/2.

dl = r dθ
dE_y= (K λ /r) Cos θ dθ
integrating, E = (K λ / r) [sinθ ] from θ = -pi/2 to pi/2

E = 2 K λ/r

I hope this answer will help you.