A Charged Isolated Conductor — Question 10

Section 23-5 · Physics pages 3730–3731

Question 10 A simple bulk-relaxation model explains the approach to zero interior charge density For a homogeneous conducting medium, assume the local laws J = γE, ∂ρ/∂t + ∇· J = 0, and ∇· E = ρ/ϵ0, with constant conductivity γ > 0. Treat these as a quasistatic continuum model. t/τ ρ/ρ0 Axes and labels specify the model; the sketch is schematic. (a) Derive the decay equation and relaxation time for an initial bulk charge density ρ0. (b) For γ/ϵ0 = 4.00 s−1, find the time for the density magnitude to fall to one percent of its initial value. Explain where the displaced charge can go in an isolated finite conductor.
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Question 10 - Solution A simple bulk-relaxation model explains the approach to zero interior charge density t/τ ρ/ρ0 (a) Method and calculation ∂ρ ∂t = −γ∇· E = −γ ϵ0 ρ. ρ(t) = ρ0e−t/τ, τ = ϵ0/γ. (b) Interpretation and further calculation τ = .250 s, e−t/τ = .0100 ⇒t = τ ln 100 ≈1.15129 s. Qtotal = Qbulk + Qsurfaces = constant. In an isolated finite sample, redistribution toward surfaces can remove bulk excess charge without destroying total charge. Physical check The exponential follows from the supplied homogeneous constitutive model; it is not a universal transient description at every microscopic time scale.

Question 10

A simple bulk-relaxation model explains the approach to zero interior charge density

For a homogeneous conducting medium, assume the local laws 𝑱=γ𝑬\mathbf J=\gamma\mathbf E, ∂ρ/∂t+𝜵⋅𝑱=0\partial\rho/\partial t+\boldsymbol\nabla\cdot\mathbf J=0, and 𝜵⋅𝑬=ρ/ϵ0\boldsymbol\nabla\cdot\mathbf E=\rho/\epsilon_0, with constant conductivity γ>0\gamma>0. Treat these as a quasistatic continuum model.

Diagram shown in the original SVG worksheet.

  1. Derive the decay equation and relaxation time for an initial bulk charge density ρ0\rho_0.

  2. For γ/ϵ0=4.00\gamma/\epsilon_0=4.00 s−1^{-1}, find the time for the density magnitude to fall to one percent of its initial value. Explain where the displaced charge can go in an isolated finite conductor.

Question 10 - Solution

A simple bulk-relaxation model explains the approach to zero interior charge density

Diagram shown in the original SVG worksheet.

(a) Method and calculation

∂ρ∂t=−γ𝜵⋅𝑬=−γϵ0ρ.\frac{\partial\rho}{\partial t}=-\gamma\boldsymbol\nabla\cdot\mathbf E=-\frac\gamma{\epsilon_0}\rho.ρ(t)=ρ0e−t/τ,τ=ϵ0/γ.\rho(t)=\rho_0e^{-t/\tau},\quad\tau=\epsilon_0/\gamma.

(b) Interpretation and further calculation

τ=.250s,e−t/τ=.0100⇒t=τln⁡100≈1.15129s.\tau=.250\,\mathrm s,\quad e^{-t/\tau}=.0100\Rightarrow t=\tau\ln100\approx1.15129\,\mathrm s.Qtotal=Qbulk+Qsurfaces=constant.Q_{\mathrm{total}}=Q_{\mathrm{bulk}}+Q_{\mathrm{surfaces}}=\text{constant}.In an isolated finite sample, redistribution toward surfaces can remove bulk excess charge without destroying total charge.

Physical check

The exponential follows from the supplied homogeneous constitutive model; it is not a universal transient description at every microscopic time scale.