Surface charge distribution on a conductor derived by method of images techniques Kindly refer to "The classic image problem" of "THE METHOD OF IMAGES" of Introduction to electrodynamics book by Griffiths (Chapter 3).
A point charge is placed at $(0,0,d)$ on the $z$ axis. The $XY$ plane is a infinite CONDUCTING plane which is grounded. The author calculated the potential by the method of images. Then the author also calculates the induced charge on the conducting plane. The charge distribution comes out to be
$$\sigma(x,y)=\frac{-qd}{2\pi(x^2+y^2+d^2)^{3/2}}.$$
This suggests that the accumulation of charge is highest at $x=y=0$ and it gradually decreases as the value of $x$ and $y$ increases.
My question is as follows.
The plane is a conducting plane. How can it support a non uniform distribution of charge? Any non uniformity in charge accumulation is supposed to be distributed evenly since the plane conducts.
 A: 
Any non uniformity in charge accumulation is supposed to be distributed evenly since the plane conducts.

Not true. The charge distribution will be such that there is no component of electric field parallel to the surface of the conductor - because that is what would generate a force on the charges, and cause them to be redistributed.
Instead, what happens here is that there is a force of attraction from the charge above the plane - this force pulls charges closer to the origin and leads to the calculated unequal charge distribution.
The method of images works because in this case, the image charge (same distance below the surface, and with opposite charge) ensures that the electric field lines are indeed normal to the surface everywhere. In reality, you don't have a charge below the surface - but you have the charge distribution on the surface which has the same role.
A: Because charge distribution is even when there's no external affluence of course when there's a charge on top of it would force distribution to be uneven
