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

Sunil Kumar FP
10 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
Yash Chourasiya
4 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.
Ex =0
The components along the y axis (vertical) add up. The limits are θ = -π/2 to π/2.

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

E = 2 K λ/r