An infinite, thin, plane sheet of conducting material has a circular hole of radius a cut in it. A thin, flat disc of the same material and slightly smaller radius lies in the plane, filling the hole, but separated from the sheet by a very narrow insulating ring. The disc is maintained at a fixed potential V , while the infinite sheet is kept at zero potential. Using appropriate cylindrical coordinates, find an integral expression involv- ing essel fucions for the potenial at any point above the plane.

Φ (ρ ,ϕ , z ) = V a∫In (0 ∞)  J₁ (ka) J₀ (k ρ)e ∧ −k z dk

Explanation:

(a)

By recalling the Laplace equation, we know that the general solution to the Laplace equation in cylindrical coordinates is:

Φ (ρ ,ϕ , z )= ( A₀,₀+B₀,₀  l n ρ) (C₀,₀ + D₀,₀ ϕ) ( F₀, ₀+ G₀, ₀ z )

+ Σ  In(ν≠0)  (Aν , 0ρν + Bν , 0 ρ ∧ −ν) (Cν ,0 e∧i ν ϕ + Dν ,0 e∧−i νϕ) (Fν ,0 + Gν ,0 z )

+Σ  In(k ≠0)   (A₀, k J₀(k ρ) + B₀, k N₀(k ρ))(C₀,k+D₀, k ϕ)(F₀,k e k z+G₀,k e−k z)

+Σ In(ν≠0)  Σ In(k ≠0)  ( Aν , k J ν (k ρ) + Bν ,k N ν(k ρ)) (Cν ,k e ∧ i νϕ+ Dν , k e−i νϕ) (Fν ,k e k z+Gν, k e−k z)

Therefore,

Consequentially, the problem has azimuthal symmetry. Hence,  immediately dictating that ν = 0 and the function be single valued  in the azimuthal angle, leading to:

Φ (ρ ,ϕ , z ) = ( A₀+B₀ ln ρ) ( F₀+ G₀ z ) + ΣIn( k  ≠0)  ( Ak J ₀ (k ρ) + Bk N₀ (k ρ))( Fk e k z+Gk e−k z)

Therefore,

The solution must then be finite as z approaches infinity, telling us that Fk = 0 and G0 =0.

Also, the solution must be finite along the z axis (ρ = 0), telling us that Bk = 0 and B₀ = 0.

Resolving further, we now have our solution to be:

Φ(ρ ,ϕ , z )=Σk  A ∨k J₀ (k ρ) e ∧−k z

So, typically at this point, we would apply a boundary condition at a finite radius, but there is not one for  this problem. With no constraint on the radius, k therefore becomes a continuous spectrum and not a discrete set.

So, Φ (ρ ,ϕ , z ) = ∫In(0 ∞)  A (k ) J₀ (k ρ)e ∧−k z dk

Now, if we apply the last boundary condition,

We have:

Φ ( z= 0 )= V (ρ)

V (ρ)=∫ In (0 ∞)  A(k ) J₀ (k ρ) dk

We can then take the advantage of the orthogonality statement to solve this:

∫In(0 ∞ ) x J₀ (k x) J₀ (k ' x)dx = 1 / k  δ(k '−k )

Therefore,

A(k ) = k ∫ In(0 ∞)  V (ρ) ρ J₀ (k ρ)d ρ

If we now insert the actual boundary condition in this problem,

We have,

A(k ) = V k ∫In(0 a)  J₀ (k ρ)ρd ρ

A(k  )= V a J₁ (ka)

Hence, the final solution therefore is

Φ (ρ ,ϕ , z ) = V a∫In (0 ∞)  J₁ (ka) J₀ (k ρ)e ∧ −k z dk