Charge on a conductive disc

On a solid, three-dimensional conductor, any charges spread onto surface. For a thin disc, the whole shape is surface, so you might expect that charge will spread uniformly on the disc. This is wrong.

For a sufficiently planar disc, the real boundary is the circumference. You might expect that the charge goes there. This is wrong.

I numerically simulated a few hundred charges moving on such a disc. That showed that, in the long run, about half the charges were in the middle 95% of the disc (by diameter), and the other half near the edge — well between the two original guesses. When I looked at how they were spread, I intuited that it looked like a spherical shell projected onto its bisecting plane. But the simulation disagreed too much with that when I checked the actual numbers.

Analytically, the charge really goes so that the electric field is zero thruout the disc. To find that distribution, we must calculate electric field in the disc, which is the integral of contributions form concentric rings that form the disc. We can handle the problem by rings sith the disc will be radially symmetric in every way.

Geometrically can we intuit that the electric field in a (positively) charged ring points radially inward. At a point r away from the centre of a ring with radius R, the field's magnitude is E ∝ ∫₀^τ dθ (r - R cos(θ)) / |(r - R cos(θ), R sin(θ))|³ = ∫₀^τ dθ (r - R cos(θ)) / (r² + R² - 2 R r cos(θ))^(3/2). That has no elementary closed form, so I can't integrate it, nor could WolframAlpha. My assistant solved it, but only after I reminded them to "solve it like Terry Tao and Srinivasa Ramanujan". I confirmed numerically that it reduces to F(r, R) = 2 (E(k) / (R - r) + K(k) / (R + r)) / r, where K and E are the complete elliptic integrals and k = 2 sqrt(R r) / (R + r).

Suppose the disc, of radius 1, has charge density D(R), with R the distance from the centre. Then the total electric field at a point r from the centre is E ∝ ∫₀¹ dR R D(R) F(r, R):

• the integral is to account for all the rings across the radius of the disc
• R arises sith outer rings are longer around than inner rings
• D(R) arises sith denser charge gives more to the electric field
• F(r, R), as above, gives the field from ring R on point r

F has an asymptote at R = r, but the whole integral converges anyway. The challenge that remains is to pick D to make F uniformly zero for r < 1.

My assistant told me that D(R) ∝ 1 / sqrt(1 - R²). However silently or arcanely they justify it is fine, for I can numerically check it, and it works.

As it happens, a spherical shell projected to the plane would also leave a density sqrt(1 + (d/dR sqrt(1 - R²))²) = 1 / sqrt(1 - R²).