Dielectrics and Gauss' Law — Question 10

Section 25-7 · Physics pages 4150–4151

Question 10 A potential profile can expose an impossible charge-free dielectric model A plate gap has prescribed potential V (z) = V0[1 −(z/d)2] for 0 < z < d, with V0>0. First suppose there is no free bulk charge and the lower electrode supplies a nonzero positive constant displacement D0. z V d Axes and labels specify the model; the sketch is schematic. (a) Infer the required relative permittivity and determine whether it stays finite at z=0. (b) Instead impose uniform finite kappa. Find the free volume density and both electrode free densities, assuming no exterior field.
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Question 10 - Solution A potential profile can expose an impossible charge-free dielectric model z V d (a) Method and calculation Ez = −dV/dz = 2V0z/d2, Dz = D0. κ(z) = D0d2 2ϵ0V0z →∞ (z →0+). (b) Interpretation and further calculation Dz = 2ϵ0κV0z/d2 ⇒ρf = 2ϵ0κV0/d2. σf,0 = Dz(0) = 0, σf,d = −Dz(d) = −2ϵ0κV0/d, which cancels the integrated free bulk charge per area. Physical check Potential, material law and source distribution cannot be specified independently; Gauss law supplies the consistency constraint.

Question 10

A potential profile can expose an impossible charge-free dielectric model

A plate gap has prescribed potential V(z)=V0[1−(z/d)2]V(z)=V_0[1-(z/d)^2] for 0<z<d0<z<d, with V0>0. First suppose there is no free bulk charge and the lower electrode supplies a nonzero positive constant displacement D0.

Diagram shown in the original SVG worksheet.

  1. Infer the required relative permittivity and determine whether it stays finite at z=0.

  2. Instead impose uniform finite kappa. Find the free volume density and both electrode free densities, assuming no exterior field.

Question 10 - Solution

A potential profile can expose an impossible charge-free dielectric model

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Ez=−dV/dz=2V0z/d2,Dz=D0.E_z=-dV/dz=2V_0z/d^2,\quad D_z=D_0.κ(z)=D0d22ϵ0V0z→∞(z→0+).\kappa(z)=\frac{D_0d^2}{2\epsilon_0V_0z}\to\infty\quad(z\to0^+).

(b) Interpretation and further calculation

Dz=2ϵ0κV0z/d2⇒ρf=2ϵ0κV0/d2.D_z=2\epsilon_0\kappa V_0z/d^2\Rightarrow\rho_f=2\epsilon_0\kappa V_0/d^2.σf,0=Dz(0)=0,σf,d=−Dz(d)=−2ϵ0κV0/d,\sigma_{f,0}=D_z(0)=0,\quad\sigma_{f,d}=-D_z(d)=-2\epsilon_0\kappa V_0/d,which cancels the integrated free bulk charge per area.

Physical check

Potential, material law and source distribution cannot be specified independently; Gauss law supplies the consistency constraint.