Leaky Capacitor (Homework 2-323)

2. Last quarter we considered a “leaky” spherical capacitor of inner conducting-shell radius r1 and outer conducting-shell radius r2. The dielectric between the two shells is only slightly conducting with conductivity σ. Last quarter we found the magnetic field between the shells when the capacitor discharges through its own dielectric by several complicated approaches. We can use gauge freedom to make this calculation easier.

a. Suppose the initial current between the shells is I0. Find the voltage at radius r within the dielectric. (You can choose the outer shell as grounded.)

b. Find all the components of the vector potential A (in Lorentz gauge) between the shells. Hint: In the Lorentz gauge you have the “Lorentz condition” Griffiths equation 10.12.

c. Find the resulting B field.

For a spherical leaking capacitor the infinitesimal resistance of a slice of the dielectric having radius $R1<r<R2$ is

$d R(r)=ρ*d r/S(r) =1/σ*d r/(4πr^2)$

The total resistance of the dielectric is

$R_{tot}=1/4πσ*∫_(R_1)^(R_2) d r/r^2 =1/(4πσ)*(1/R_1 -1/R_2)$

and for the dielectric that extends between outer shell $(R2)$ and intermediate radius $r$


The potential drop (from outer shell grounded is)

$V(r)=I_0 R(r)=I_0/4πσ (1/r-1/R_2)$

Since the capacitor discharges through internal resistance (of the dielectric)

$V(r,t)=V(r)*exp⁡(-t/(R_tot C))=I_0/(4πσ) (1/r-1/R_2 )*exp⁡(-t/(R_tot C))$

In the Lorentz gauge (equation 10.12) one has

$(∇A ⃗ )=-ϵμ*∂V/∂t$


$(∇A ⃗)(r,t)=ϵμ/(R_tot C)*V(r,t)=ϵμ/(R_tot C)*I_0/4πσ (1/r-1/R_2 )*exp⁡(-t/(R_tot C))$

In spherical coordinate the divergence of a vector is

$∇A ⃗=1/r^2 *(∂〖(r〗^2 A_r))/∂r+1/(r*sin⁡θ )*(∂(A_θ*sin⁡〖θ)〗)/∂θ+1/(r*sin⁡θ )*(∂A_ϕ)/∂ϕ$


Here I am lost. For a radial current I could not find the form of the magnetic vector potential A. I just can assume (which can be wrong) that (because of the spherical symmetry)

$∇A ⃗=1/r^2 *(∂〖(r〗^2 A_r))/∂r$

and therefore

$r^2 A_r=∫ r^2 ∇A ⃗*d r=1/(R_tot C)*I_0/4πσ*∫r^2 (1/r-1/R_2 )*d r=$

$=1/(R_tot C)*I_0/4πσ (r^2/2-r^3/(3R_2 ))$

and thus

$A_r=1/(R_tot C)*I_0/4πσ (1/2-r/(3R_2 )) exp⁡ (-t/(R_tot C))$

and $A_θ=A_ϕ=0$

So that finally

$B ⃗=∇×A ⃗=0$


(But again, my assumption could be wrong)