Method of image charges (part I)

The electric field of the dipole results $\vec E(x,y,z) = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{[x^2 + y^2 + (z + d)^2]^{3/2}} \left( \begin{array}{c} x \\ y \\ z + d \end{array} \right) + \frac{-q}{[x^2 + y^2 + (z - d)^2]^{3/2}} \left( \begin{array}{c} x \\ y \\ z - d \end{array} \right) \right]$ In the case $z = 0$ we can simplify the equation to $\vec E(x,y,z = 0) = \frac{q}{2 \pi \varepsilon_0} \frac{d}{[x^2 + y^2 + d^2]^{3/2}} \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right)$ Note that the electric field is perpendicular to the surface $z = 0$ , so that this is an equipotential surface. Thus, the electric field of the dipole corresponds to the field of the actual charge configuration, since the equipotential surface is the same for both. The only difference is that in the metal $\vec E = 0$ must apply, so that our solution applies only to the half-space $z \leq 0$ .

We can use Gauss's law to calculate the charge density $\sigma(x,y)$ . We integrate the electric field over the surface of a cube $dV = dxdydz$ centered around the point $\vec r = (x, y, 0)$ . The charge enclosed by the infinitesimal cube is $dQ = \sigma (x, y) dx dy$ . $\begin{aligned} & & \oint_{dV} \vec E \cdot d\vec A &= \frac{dQ}{\varepsilon_0} \\ \Rightarrow & & - E_z(x,y,0) dx dy &= \frac{\sigma(x,y)}{\varepsilon_0} dx dy \\ \Rightarrow & & \sigma(x,y) &= - \varepsilon_0 E_z(x,y,0) \\ & & &= -\frac{q}{2 \pi} \frac{d}{[x^2 + y^2 + d^2]^{3/2}} \end{aligned}$ For $x = y = 0$ we get the final result $\sigma(0,0) = - \frac{q}{2 \pi d^2} = - \frac{10^{-3}}{2 \pi \cdot 0.2^2} \,\frac{\text{C}}{\text{m}^2} \approx - 4 \cdot 10^{-3} \,\frac{\text{C}}{\text{m}^2}$