Fernando Garcia

Useful context: In section 3.2.1: The Classic Image Problem, we studied the problem of an infinite grounded plane with a charge $q$ held a distance $d$ above. We found (using the method of images) that

$$\begin{array}{}\text{(Eq. 3.9, page 124)}& V(x,y,z)=\frac{1}{4\pi {ϵ}_0}[\frac{q}{\sqrt{x^2+y^2+(z-d{)}^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d{)}^2}}]\end{array}$$ A bit further down, in section 3.2.2: Induced Surface Charge, we recalled Equation 2.49: $\sigma =-{ϵ}_0\frac{\partial V}{\partial n}$. Using this and Eq. 3.9, we see that the induced charge on the conductor (the infinite grounded plane) is: $$\begin{array}{}\text{(Eq. 3.10, page 125)}& \sigma (x,y)=\frac{-qd}{2\pi (x^2+y^2+d^2{)}^{3/2}}\end{array}$$

Here's an alternative derivation of Eq. 3.10

This approach (which generalizes to many other problems) does NOT rely on the method of images.

The total field is due in part to $q$, and in part to the induced surface charge. Write down the $z$ components of these fields - In terms of $q$ and the as-yet-unknown $\sigma (x,y)$ - just below the surface.

The sum must be zero, because this is inside a conductor. Use this argument to determine $\sigma$.