Answer to Question #211519 in Electricity and Magnetism for Rocky Valmores

Question

A thin non-conducting rod of finite length L has a charge q spread uniformly along it. Show that the magnitude of the electric field at point P a distance y on the perpendicular bisector of the rod is given by: E = q 2π0y 1 (L2 + 4y 2) 1/2

Answer 1

q= net charge

Liner charge density

$λ= \frac{q}{L}$

L=length of rod

dq=λdx

Electric field due to dq as p

$dE = \frac{kdq}{γ^2} \\ γ^2 = x^2 + y^2 \\ E_x = \int dEsinθ = \int \frac{kdq}{γ^2} \times \frac{x}{γ} \\ = \int \frac{k \times λ \times dx \times x}{(x^2+y^2)^{3/2}} \\ = k λ \int^{L/2}_{-L/2} \frac{xdx}{(x^2+y^2)^{3/2}} \\ E_x=0 \\ E_y= \int dEcosθ = \int \frac{kdq}{γ^2} \times \frac{y}{γ} \\ = \int \frac{k \times λ \times dx \times y}{γ^3} \\ = kλ y \int^{L/2}_{-L/2} \frac{dx}{(x^2+y^2)^{3/2}}$