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

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia