Re or Im of Analytic Functions  as Potentials

Clear["`*"]

Here is the "Complex" Potential:  We need the Im part of Φ.

Φ = (8/π^2) Sum[(-1)^
              m Exp[(2  m + 1) I π  z]/(2  m + 1)^2, {m, 0, 
            Infinity}]

Plot3D[
      Im[Φ] /.  z -> x +  I y, {x, 0, 1}, {y, 0, 1/2}, 
      AxesLabel ->  {"x", "y", "Φ"}, 
      PlotRange ->  {{0, 1}, {0, 1/2}, {0, 1}}]

Find the "Complex"  Electric field as a function of (x,y)

Ez  =  -D[Φ, z]

Ez  =  Ez  /.  z ->  x  +  I y

Find  σx,  the charge density on the surface y = 0.
Note:  σx = Ey  = -(∂_y) Im[Φ] = - Im[(∂_y) Φ] = - Im[i (∂_z) Φ] = Im[ i Ez] = Re[Ez]

σx  =  Re[ Ez]  /.  y ->  0

Find σy, 
            the charge density on the surface x = 0. 
Note :  σy=Ex=-(∂_x) Im[Φ]=-Im[(∂_x) Φ]=-Im[(∂_z) Φ]=Im[Ez]

σy  =  Im[ Ez]  /.  x ->  0

Plot[σx, {x, 0, 1}, PlotRange  ->  {0, 5}, 
      PlotStyle ->  {Thickness[0.01], Hue[0.95]}]

Qx  =  Re[2 NIntegrate[σx, {x, 0, 1/2}]]

Capacitance/L    C = (Subscript[ϵ, 0 ])Qx/V =  (Subscript[ϵ, 0 ])Qx = 1.485 (Subscript[ϵ, 0 ])

Plot[σy, {y, 0, 1}, 
      PlotStyle ->  {Thickness[0.01], Hue[0.5]}]

Qy  =  NIntegrate[σy, {y, 0, Infinity}]

Q =  Qx + 2 Qy