$x$-$y$ plane is a conducting plate stretching to infinity. A charged line is placed from $(0,0,0)$ to $(0,0,L)$. It has charge density function of ${\lambda=\lambda_L\frac{z^2}{L^2}}$.

Calculate the surface charge density $\sigma(X,Y)$ on the conducting plane.

Charged Line

My workings: I know there will be another virtual image with the opposite charge stretching from $(0,0,0)$ to $(0,0,-L)$. I then tried to calculate the electric field $\vec{E}(X,Y,0)$ on the plane, and I think there will only be $z$-component. I then came up with $E_z(X,Y,0)=\frac{1}{4\pi \epsilon_0}\int_0^L{\lambda_L\frac{z^2}{L^2}\frac{-z}{\sqrt{(X-x)^2+(Y-y)^2+(-z)^2}^3}}dz$. Then I got stuck, can someone help? Maybe I got wrong already?

  • $\begingroup$ Use substitution $(X-x)^2+(Y-y)^2=r^2$ and $z=r\tan(\theta)$ (assuming your expression for E is correct) $\endgroup$
    – KP99
    Oct 27 '21 at 11:52

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