Question

A thin glass rod forms a semicircle of radius r= 5.00 cm. Charge is uniformly distributed along the rod, with +q = 4.50 pC in the upper half and -q = -4.50 pC in the lower half. What are the: 

(a) Magnitude and: 

(b) Direction (relative to the positive direction of the x axis) of the electric field E at P, the center of the semicircle?

Answer 1

Let the linear density of charge, λ = q/0.5πR

The charge on the element of arc length, dq = λdS

Electric field, dE at the center due to the charge, dq isgiven by,

dE = kdq/R² directed radially outward.

dE_y = dE cos θ =  kdq cosθ/R²

= kλdScos θ/R²

But dS = Rdθ

dE_y = kλRdθ cosθ/R²

dE_y = (kλ/R).cosθ dθ

Integrating from θ = 0 to θ = π/2

E_y = (kλ/R).[sin θ] for θ = 0 to θ = π/2

E_y = kλ/R

Using, λ = 2q/πR

E_y+=  2kq/πR²

Similarly for the lower quadrant.

But the x components cancel out. The y components addup.

So, Total El field,    E = 2E₊ = 4kq/πR²

E = 4 x 9 x10⁹ x 4.5 x 10^-12 / 3.14 x 25 x 10^-4

E = 20.63 N/C  Vertically downward..