Answer to Question #113487 in Electricity and Magnetism for ALI

Question

The electric field produced by, a thin rod of length L that lies along the z-axis and carries uniformly distributed positive charge q, at a point P located on the perpendicular bisector of the rod (the positive y-axis) a distance y from its center is given by equation.


E= 1/(4πε_0 ) q/(y√(y^2+L^2/4))

Answer 1

Assume that the linear charge density is $\lambda=q/L.$ Split our wire into many pieces charged $dq=\lambda dx$ each. The x-component of the field will cancel, the y-component will add. The field will be

$$dE_y=\frac{kdq}{r^2}\frac{y}{r}=\frac{k\lambda y}{(x^2+y^2)^{3/2}}dx.$$

Integrate this from $-L/2$ to $+L/2$ :

$$E=\int^{L/2}_{-L/2} \frac{k\lambda y}{(x^2+y^2)^{3/2}}dx=k\lambda \frac{L}{y\sqrt{y^2+L^2/4}}=\\ \space\\ =\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{y\sqrt{y^2+L^2/4}}.$$