Infinite electrostatic plate acceleration

Question

So, let's assume you have an infinite charged plate of some charge density, and a like-charged test charge positioned near it. In this case, the test charge is accelerated away from the plate at a finite and uniform acceleration rate, determined by the charge density on the plate. As the test charge gets further from the plate, the force vectors from each of the point charges on the plate become more and more perpendicular to the plate, contributing more and more of their force to accelerating the test charge away, and less and less force to pushing the point charge to stay centered.

The acceleration is constant, out to an infinite distance, indicating that the test charge has an infinite potential energy, which is converted to kinetic energy at a consistent rate determined by the charge density of the surface.

Now the question: If I go from this unphysical infinite system, to a very large finite plate with a finite charge, and a test charge positioned just a tiny distance above the plate's center, how far away from the plate will my test charge get before the gradient transitions from linear (i.e. constant acceleration) to falling off as the square of the distance (i.e. ever diminishing acceleration)?

Obviously, this happens immediately; what I'd like to understand is the contour of the field as it happens, and at what point the square of the distance becomes dominant. How would I go about calculating this?

Seems like in the original system, the point charge is converting potential vectors that push it to center into potential vectors that push it upward at the same rate that it is converting upward potential into kinetic energy. This implies to me that there would be a critical angle between the test charge and the edge of the finite plate where the new upward vector contribution drops below 50% of the potential energy consumed by acceleration, and you could then say that the square of the distance becomes dominant, and fringing effects take over. Seems like that angle would probably be the same, no matter the charge density on the plate, or the size of the plate, as long as the plate is circular. But, I don't know how to do the math to find that angle.

Answer 1

On way to do this is make your not-so-infinite plate a charge distribution on a disc (aligned on $z$ for reasons):

$$\rho(r, \theta, \phi) = \sigma (1-\Theta(r-R))\delta(\theta-\frac{\pi} 4)$$

Now you can just say the answer is $R$, because it is the only length-scale available in the problem.

For a more quantitative answer, do a multipole expansion (https://en.wikipedia.org/wiki/Multipole_expansion) of the potential.

Since there is symmetry w.r.t $\phi$, you will only have $l=0$ terms, and because the charge is one sign, only even $n$ are needed.