Proving Charge Induced on Conductor is Equal to Sum of Imaginary Charges

[harshant]

Imagine that there is a point charge in vicinity of an infinite grounded sheet of conductor of arbitrary shape and size such that the problem of finding the potential can be solved by using the method of images. Is their a way to prove that the total charge induced on this sheet is always equal to the sum of the imaginary charges? (there exists a way for proving this for finite conductors using the Gauss law) I am confused because in all the texts I referred to this conclusion was stated to be 'expected' or 'obvious', and I couldn't see why it would be so.

 

[kuruman]

Assume the grounded sheet is in the xy-plane and charge $q$ is at distance $d$ above it. The potential in region $z>0$ due to the charge and its image is azimuthally symmetric and expressed in cylindrical coordinates as
$$V(\rho,z)=\frac{q}{4\pi \epsilon_0}\left[\frac{1}{[\rho^2+(z-d)^2]^{1/2}}-\frac{1}{[\rho^2+(z+d)^2]^{1/2}}\right]$$
The surface charge density is given by
$$\sigma(\rho)=\epsilon_0~E_z(z=0)=- \epsilon_0 \left. \frac{\partial V(\rho,z)}{\partial z}\right|_{z=0}$$
$$\sigma(\rho)=-\frac{q}{4\pi \epsilon_0} \frac{2d}{(\rho^2+d^2)^{3/2}}$$
A ring of radius $\rho$ and width $d\rho$ bears charge $dq=\sigma(\rho)2\pi \rho~ d\rho$. The total charge on the grounded sheet is
$$Q_{total}=\int_0^{\infty} \sigma(\rho)2\pi \rho~ d\rho$$
Do the integral and convince yourself that it is equal to $-q$.