# Calculating the surface charge density as a function of $X,Y$

$$x$$-$$y$$ 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.

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?

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