Proving: Method Of Electrical Images

So first of all what is 'Method of electrical images'.

Suppose you place a charge $Q$ in front of a infinitely large plane conducting sheet which is earthed.The sheet gets an induced charge $-Q$.This would attract the original charge.

The force of attraction is given by $F=\frac{-1}{4\pi \varepsilon}\frac{Q^2}{(2x)^2}$

The force is as if the conducting sheet acts as a mirror and forms a mirror image of the original charge $Q$.

Well the interpretation is beautiful and so is the equation.So how do we prove this.
Initially I raised this question to my physics teacher at P.C Thomas Classes.Failing to get a satisfactory answer I sat upon this problem that night.And finally........



Proof:
First we need to derive a equation for the induced charge density $\sigma (r)$ on the surface.

Due to the symmetry of the electric field lines produced by the charge $Q$.The charge density $\sigma(r)$ will have circular symmetry about the center of the plane surface (i.e. the point of least distance from the charge).

 Since inside a conductor the field is zero and at the surface field is perpendicular we have $E_{induced} = -E_{Q}cosθ$.

By Gauss's law,
Field due to an infinitesimally small element on the surface on the conductor is given by $\frac{\sigma (r)}{2\varepsilon }$

i.e. $E_{induced} = \frac{\sigma (r)}{2\varepsilon }$.
$=>-E_{Q}cos \theta  = \frac{\sigma (r)_{}}{2\varepsilon }$
$=>-\frac{1}{4\pi \varepsilon}\frac{Q}{l^2}cos \theta  = \frac{\sigma (r)_{}}{2\varepsilon }$
$=>-\frac{1}{2 \pi}\frac{Q}{(x^2+r^2)}.\frac{x}{\sqrt{x^2+r^2}}  = \sigma (r)$   $\because cos\theta = \frac{x}{\sqrt{x^2+r^2}}$
$\therefore \sigma(r)= -\frac{1}{2 \pi}\frac{Qx}{(x^2+r^2)^{3/2}}$

With the help of the about equation it can be easily verified that induced charge is equal to $-Q$.

Now we want to find the force on $Q$ by the induced charge on the sheet.

 

For that we divide the conducting sheet into rings of infinitesimally small thickness.

Force due to a ring of radius $r$ is given by (There won't be any horizontal force due to symmetry of the ring)
$F_{ring}(r)=\int_{0}^{2\pi r}\frac{(\sigma (r)dr)dl.Q.cos\theta}{4\pi \varepsilon (r^2+x^2)}$
               $=\int_{0}^{2\pi r}\frac{(\sigma (r)dr)Q.cos\theta}{4\pi \varepsilon (r^2+x^2)}dl$
               $=\frac{(\sigma (r)dr)Q.cos\theta .r}{4\pi \varepsilon (r^2+x^2)}2\pi r$
$\therefore F_{ring}(r)=-\frac{(dr)Q^2\cos^2\theta.r}{4\pi  \varepsilon (r^2+x^2)}$  $\because \sigma (r)=-\frac{Q\cos\theta }{2\pi (r^2+x^2)}$

To find the total force we need to add up the contributions from all such rings.
$F_{total}=\int_{0}^{\infty }F_{ring}(r)$
          $=\int_{0}^{\infty }-\frac{Q^2\cos^2 \theta .r}{4\pi \varepsilon (r^2+x^2)^2}dr$

$\because r=x.tan\theta $ and $dr=x.sec^2\theta.d\theta$
we have,
$F_{total}=-\int_{0}^{\pi/2 }\frac{Q^2\cos^2 \theta .x^2\tan\theta \sec^2\theta}{4\pi \varepsilon x^4.sec^4\theta}d\theta$
           $=-\int_{0}^{\pi/2 }\frac{Q^2}{4\pi \varepsilon x^2}\sin\theta \cos^3\theta d\theta$
           $=-\frac{Q^2}{4\pi \varepsilon x^2}\left [ \frac{\cos^4\theta}{4} \right ]_{0}^{\pi/2}$
           $=-\frac{Q^2}{4\pi \varepsilon x^2}.\frac{1}{4}$
           $=-\frac{1}{4\pi \varepsilon}.\frac{Q^2}{(2x)^2}$

This proof assumes the reader to have some basic knowledge in classical electrodynamics.If you need any further clarification, please comment.