Sunday, 24 September 2017

homework and exercises - Calculating the electric field of an infinite flat 2D sheet of charge



I was trying to calculate the electric field of an infinite flat sheet of charge. I considered the sheet to be the plane $z=0$ and the position where the electric field is calculated to be $(0,0,z_0)$, I know that the electric field from a line charge is with charge density $\lambda$ is $E(r)=\frac{\lambda}{2\pi r\epsilon_0}$. I ended up with this integral: $$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 \sqrt{x^2+z_0^2}} \left(\frac{-x}{\sqrt{x^2+z_0^2}}i+\frac{z_0}{\sqrt{x^2+z_0^2}}k\right) dx=\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} \left(-xi+z_0k\right) dx.$$The $z$-component gives the correct answer. $$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} z_0 dx=\frac{\sigma}{2\pi\epsilon_0}\arctan\left(\frac{x}{z_0}\right)\Big|_{-\infty}^{\infty}=\frac{\sigma}{2\epsilon_0}.$$ But when I wanted to verify that the $x$-component is zero, I encountered a divergent integral.$$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} x dx=\ln\left(x^2+z_0^2\right)\Big|_{-\infty}^{\infty}.$$ Why is that? Where am I getting wrong?





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