Clear["`*"]

An AC field I = (Subscript[I, 0]) (e^(-iωt)) is applied to the plates of a capacutor with circular plates (radius a, separation d). Assuming that the E field between the plates is along z and that the B field is azimuthal, it is straightforward to show from Maxwell's equations that the fields in the gap are (Subscript[E, z]) = A (Subscript[J, 0])(k ρ)  and (Subscript[B, ϕ]) = - (i/c)A (Subscript[J, 1])(k ρ), where A is a constant that can be determined by requiring that the charge on the plate is (Subscript[Q, 0]), which is related to (Subscript[I, 0]) by (Subscript[I, 0 ])= -i ω (Subscript[Q, 0]). The charge density on the plate is related to (Subscript[E, z]) by  σ = (Subscript[ϵ, 0]) (Subscript[E, z]) = (Subscript[ϵ, 0  ])A (Subscript[J, 0])(k ρ).  It follows that  (Subscript[Q, 0]) = 2π (Subscript[ϵ, 0]) (Subscript[E, 0]) (∫_0^a) ρ (Subscript[J, 0])(
            k ρ) dρ.

eqn  :=  
      (Subscript[Q, 0])  ==  
        2 π (Subscript[ϵ, 0]) A Integrate[ρ BesselJ[0, 
                k ρ], {ρ, 0, a}, 
            Assumptions  ->  k a > 0]

sol  :=  Solve[eqn, A] // First

A  =  A /.  sol

Lets look at E and B to second order

Esec  :=  
      Normal[Series[ A BesselJ[0, k ρ], {k, 0, 2}]]

E2  :=  Expand[Esec/Esec[[2]]]

E2  =  Esec[[2]] (1 +  Coefficient[E2, k^2] k^2)

Bsec  :=    Normal[
        Series[-(I/c) A BesselJ[1, k ρ], {k, 0, 
            4}]]

B2  :=  Expand[Bsec/Bsec[[2]]]

Bcof  :=  
      Bsec[[2]] /.  {(Subscript[ϵ, 0]) ->  1/((Subscript[μ, 0]) c^2), 
          (Subscript[Q, 0]) ->  I (Subscript[I, 0])/(c k)}

B2  =  Bcof (1 + Coefficient[B2, k^2] k^2)

Evaluate second-order electric energy

we2  := 
      Expand[((Subscript[ϵ, 0])/
              4) 2 π d Integrate[ ρ E2^2, {ρ, 0, 
              a}]]

We2  =  we2[[1]] Expand[we2/we2[[1]]]

wm2  :=  
      Expand[2 π d/(4 (Subscript[μ, 0])) 
          Integrate[ ρ B2^2, {ρ, 0, a}]]

Wm2 =  wm2[[1]] Expand[wm2/wm2[[1]]]

General case

The energy stored in the E field of the capacitor is

we  =  ((Subscript[ϵ, 0])/
            4) 2  π A^2 d Integrate[ρ (BesselJ[0, 
                  k ρ])^2, {ρ, 0, a}, 
          Assumptions ->  k a  >  0]

We  =  Normal[Series[we, {k, 0, 8}]]

We  =   We[[1]] Expand[We/We[[1]]]

We  /.  (Subscript[Q, 0])  ->  (Subscript[I, 0])/ω

This agrees with the expression in the text (omitting corrections beyond second order)

wb  =  (1/(4 (Subscript[μ, 0]))) (1/
            c^4) 2 π (A)^2 d Integrate[ρ (BesselJ[1, 
                  k ρ])^2, {ρ, 0, a}, 
          Assumptions ->  k a  >  0]

Wb =  Normal[Series[wb, {k, 0, 10}]]

Wb  =  Wb[[1]] Expand[Wb/Wb[[1]]]

Wb  /.  {(Subscript[Q, 0])  ->  
          (Subscript[I, 0])/(k c),  (Subscript[ϵ, 0])  ->  
          1/(c^2  (Subscript[μ, 0]))}

This also agrees with the expression in the text (omitting corrections beyond second order).

Compare (Subscript[k, res]) a = (Sqrt[8]) with 1st root of (Subscript[J, 0])((Subscript[k, res])a)

Clear[x]

FindRoot[BesselJ[0, x] == 0, {x, 2.4}]

Sqrt[8] // N

\(Plot[BesselJ[0, Sqrt[8]\ ρ], {ρ, 0, 1}, 
      PlotStyle -> {Thickness[0.01], RGBColor[0, 1, 0]}, 
      PlotLabel -> \ \*""(Subscript[E, z])(ρ) for a capacitor at resonance"", 
      TextStyle -> \ {FontFamily -> \ "Times", 
          FontSize -> 14}]\)

Export["C:\Documents and Settings\johnson\My \
Documents\Filexfer\Class06S\capdist.eps", %32]

Wavelength of input current  at resonance = 2π (a/(Sqrt[8])). For a = 1 cm and the frequency = c/λ

λcm  =  N[2 π /Sqrt[8]]

Frequency (GHz)

νGHz  =  3 10^(10)/(λcm 10^9)