Students who have studied high school physics know that in classical mechanics and electrostatic theory, universal gravitation and the Coulomb force share similar properties: they both follow the inverse-square law with respect to distance. Now, I would like to pose a question from the perspective of gravity:
For an infinite plane plate with uniform density, is the gravitational field it generates uniform? In other words, is the gravitational force experienced by an object of equal mass at any point outside the plate the same?
A similar question can be asked for electrostatic force, simply by replacing gravity with the Coulomb force and mass with charge. By analogy with finite cases, one might conclude: The field must be non-uniform! Because the force is inversely proportional to the square of the distance, different distances should result in different forces. Is this truly the case?
The reality is often surprising, and this is a classic example illustrating the heaven-and-earth difference between the finite and the infinite: The field is actually uniform!!
Elementary Proof
Friends who have studied multivariable calculus can calculate this result using double integrals, which will reveal that the field is independent of distance. However, for the benefit of those who only have basic knowledge of calculus and physics, BoJone decided to introduce a relatively “elementary” proof method he devised. First, let us consider a right circular cone with radius r and height h. A total charge +Q is uniformly distributed along the circumference of its base.
Due to isotropy, we can quickly write the field strength at its vertex as: E' = \frac{kQh}{(r^2+h^2)^{3/2}}
In other words, we only take the vertical component of the force, as the horizontal components cancel each other out. The direction of this field strength is vertically upward.
An infinite plane plate with charge density \sigma can be viewed as being composed of infinitely many rings of varying radii and infinitesimal widths. Any point above it can form infinitely many right circular cones with these rings. The total field strength E should be the superposition of the field strengths of each cone. The field strength of each cone is approximately: \Delta E = k \frac{2\pi r \sigma h}{(r^2+h^2)^{3/2}} \Delta r
Therefore: E = \int_0^{+\infty} \frac{2k\pi\sigma hr}{(r^2+h^2)^{3/2}} dr = 2k\pi\sigma
The final result is a constant, which is the unexpected uniform electric field! (Can everyone experience the difference between finite and infinite now?)
Further Understanding
It is possible that some readers still do not believe or cannot understand the above result. BoJone attempts to “explain” (rather than calculate) through a diagram that this is indeed a uniform field. As shown in the figure above, our goal is to prove that the forces at point A and point B are the same. First, we project points A and B onto the plane (the figure only shows two dimensions, but it is actually three-dimensional). Then, centered at the projection points, we divide the plane into concentric rings. The radius is proportional to the distance. As shown in the figure, \frac{xz}{x'z'} = \frac{AC}{BD}.
From the figure, we can see that the field strength direction of circle xy on point A and circle x'y' on point B are both along \vec{zA}, and their magnitudes are equal (because although the distance to point A is smaller, the amount of charge in circle xy is also smaller. The Coulomb force is inversely proportional to the square of the length, while the charge is proportional to the square of the length; the two effects cancel out). That is to say, for every force component in any direction at point A, an equivalent force component can be found at point B (remember it is an infinite plane). Therefore, the field strengths at the two points are equal.
Capacitance of Sufficiently Large Parallel Plates
Based on the above calculation, if there are two infinite parallel plates with surface charge density \sigma and opposite signs, the field strength between the plates is 4k\pi\sigma.
When two finite parallel plates are large, 4k\pi\sigma remains a very good approximation for the field strength between them (especially near the center). Let the overlapping area be S and the distance be d. Since Q = \sigma S, the capacitance is: C = \frac{Q}{U} = \frac{Q}{Ed} = \frac{Q}{4k\pi\sigma d} = \frac{S}{4k\pi d}
This is the formula for calculating the capacitance of parallel plates provided in high school textbooks! Of course, it is assumed here that the space between the plates is a vacuum. If it is not a vacuum, a correction factor must be multiplied, i.e., C = \frac{\varepsilon S}{4k\pi d}.
The above are BoJone’s calculation results from 2011.02.22, inspired by the second volume of “The Feynman Lectures on Physics”.