Why is the electric field of a naturally equilibrating, isotropic, cylindrical conductor not a function of z (height)?

Question

Why is the electric field of a naturally equilibrating, isotropic, cylindrical conductor not a function of z (height)?

139 Views Asked by user213745 At 26 Mar 2026 - 6:01

Some math is presented below to make the question more specific.

TL;DR: If you calculate the electric field of a solid, cylindrical conductor, you find that this field is only a function of radius and time (unless I made a mistake with the calculation, presented below for the curious). Why is there no $z$ dependence? What mistake(s), if any, lie in the solution below?

Also, as a follow-on question: how does one obtain/determine the initial conditions of the field, in order to perform a Fourier-Bessel series summation and gain a solution that is consistent with said initial conditions?

Background:

Maxwell's equations are a set of four differential equations that completely describe classical electromagnetic phenomena in S.I. Units.

Equations of $3\text{D}$ position and time: \begin{eqnarray} \textbf{Div} [E] &=& \dfrac{p}{e} \tag{i}\\ \textbf{Curl}[E] &=& -\dfrac{dB}{dt} \tag{ii}\\ \textbf{Div} [B] &=& 0 \tag{iii}\\ \textbf{Curl}[B] &=& uJ + ue\dfrac{dE}{dt} \tag{iv} \end{eqnarray} Where \begin{eqnarray} 1&.& E \,\,\text{is the electric field vector},\\ 2&.& B \,\,\text{is the magnetic field vector},\\ 3&.& J \,\,\text{is the free current density field vector},\\ 4&.& e \,\,\text{is the electric permeability of the material},\\ 5&.& u \,\,\text{is the magnetic permeability of the material},\\ 6&.& p \,\,\text{is the free charge density}. \end{eqnarray} Situation:

Given a simple, homogeneous, solid, cylindrical conductor (i.e. $e$ and $u$ are constants), the natural question of the distribution of current (what does $J$ look like?) arises.

Setting up the problem:

The conductor is aligned in the $z$-axis direction, and neither of the fields are driven (not forced to behave in a given way). Find $E$, while assuming that the conductor is simple.

We define: $$J = \sigma E$$ which is a material constraint suggesting that the current density is directly proportional to the electric field, and the proportionality is ensured by a constant $\sigma$, which is the metal's conductivity. This form of $J$ implies that our conductor is isotropic and linear.

MY attempt at a solution:

$$\textbf{Curl}[(ii)] = \textbf{Curl}[\textbf{Curl}[E]] = \textbf{Grad}[\textbf{Div}[E]] - \Delta[E] = -\dfrac{d}{dt}(\textbf{Curl}[B]) \tag{v}$$

But from $\text{(iv)}$,

$$\textbf{Curl}[B] = uJ + ue\dfrac{dE}{dt} = u\sigma E + ue\dfrac{dE}{dt} \tag{vi}$$

We can plug $\text{(vi)}$ into $\textbf{Curl}[B]$ in equation $\text{(v)}$, and get: $$Grad[Div[E]] - \Delta[E] = -\dfrac{d}{dt}\left(u\sigma E + ue\dfrac{dE}{dt}\right)$$ $$\Delta[E] - \dfrac{d}{dt}\left(u\sigma E + ue\dfrac{dE}{dt}\right) = \textbf{Grad}[\textbf{Div}[E]]\tag{(vii)}$$

Brief Discussion of Grad[Div[E]].

Taking the Divergence of (iv), we get:

Div[Curl[B]] = Div[uJ + ue*dE/dt]

0 = uDiv[J] + ue*d/dt(Div[E])

But J = sigma*E, so

usigmaDiv[E] = -ued/dt(Div[E])

From (i), Div[E] = p/e, so...

sigmap/e = -ed/dt(p/e)

(sigma/e)*p + dp/dt = 0

This is a basic 1st order linear ode, and the solution is:

p(r,t) = p0*e^(-t/tau), where tau = e/sigma, and p0 = p(r,0) is our arbitrary boundary condition.

Note: We don't actually know the spatial distribution of p, but we know that it decays exponentially, on the time-scale of tau. Since p(r,t) decays to zero in "almost no time at-all", the common approximation is:

Div[E] = p(r,t)/e -> 0 .. this works for engineering applications, but as I am seeking an analytical solution, I will not make this approximation, because then I lose the spatial distribution information of p(r,0).

Thus Grad[Div[E]] is not approximated as zero on the timescale of tau.

Back to the solution:

We make the assumption that vector E is oriented exclusively in the z-axis direction, and travels along the axis of the conductor. Also, we will investigate the form of E(r,t) in cylindrical coordinates (r,T,z), because this appeals to the geometry of the problem.

z-axis direction of E vector:

Laplacian[E] = 1/r*(r*d/dr(E_z))' + 1/(r^2)*d^2/dT^2(E_z) + d^2/dz^2(E_z)

and...

Grad[Div[E]] = Grad[d/dz(E_z)] = d/dr(d/dz(E_z)) (r) + 1/r*d/dT(d/dz(E_z)) (T) + d^2/dz^2(E_z) (z)

, where (r),(T),(z) are directional vectors.

Notice that this geometry imposes the following constraints:

d/dr(d/dz(E_z)) (r) = 0

1/r*d/dT(d/dz(E_z)) (T) = 0

Thus;

Laplacian[E] - d/dt(usigmaE + uedE/dt) = Grad[Div[E]] ... (vii)

1/r*(r*d/dr(E_z))' + 1/(r^2)d^2/dT^2(E_z) + d^2/dz^2(E_z) - d/dt(usigmaE + ue*dE/dt) = d^2/dz^2(E_z)

1/r*(r*d/dr(E_z))' + 1/(r^2)d^2/dT^2(E_z) - d/dt(usigmaE + ue*dE/dt) = 0 ...(viii)

Assumption: The Current density does not have a preferred mode i.e. it is rotationally invariant. This implies....

E(r,T,t) = E(r,t)

And the theta (T) second partial term of (viii) becomes zero, prompting:

1/r*(rd/dr(E_z))' - d/dt(usigmaE + ue*dE/dt) = 0 ...(ix)

Expanding (ix), we get:

d^2/dr^2(E_z) + 1/rd/dr(E_z) - usigmad/dt(E_z) - ue*d^2/dt^2(E_z) = 0

Employ separation of variables:

E(r,t) = R(r)*N(t),

NR" + 1/rNR' - usigmaRN' - ueR*N" = 0

R"/R + 1/rR'/R - usigmaN'/N - ue*N"/N = 0

R"(r)/R(r) + 1/rR'(r)/R(r) = usigmaN'(t)/N(t) + ue*N"(t)/N(t) = -lambda ...(x)

lambda is a constant parameter (the eigenvalue). Notice that the units of lambda are [1/m^2], implying that it is a square spatial frequency.

Rearranging the R(r) based equation from (x), we get:

r^2*R"(r) + rR'(r) + lambdar^2*R(r) = 0

which is the a parametric Bessel equation of order 0.

There are many ways to solve this equation, but the Frobenius method of searching for a solution using a convergent power series is, perhaps, the most straight-forward approach. Doing this,

we end up with a power-series representation of:

R(r) = J[0,Sqrt[lambda]*r].

I choose not to prove this here, but it's pretty straightforward.

Now, our current is confined to the conductor, meaning the current density is zero for r > L, where L is the radius of our cylindrical conductor. As such, we can establish a boundary condition. We choose to solve for E such that the current density at surface is zero. In practice, this makes sense because the skin of the conductor is thin enough to have an infinite effective resistance. The surface would need to have an infinite conductivity to transport current in an infinitesimally thin layer of the conductor.

Thus: R(L) = 0 ...our chosen boundary condition.

Since R(r) = J[0,Sqrt[lambda]*r], we need to find a specific eigenvalue of lambda such that R[L] = 0.

Since J[0,r] has an infinite set of roots (p_j) that all tend to infinity, we will suppose that one of those roots satisfies the boundary condition. Of course, we don't know which one.

So:

R[L] = J[0,p_j] = J[0,Sqrt[lambda]*L]

invoking the relationship: Sqrt[lambda] = p_j/L, so the eigenvalue lambda is:

lambda = (p_j/L)^2, and R[r] = J[0,p_j/L*r].

Thus (picking the second equation from (x)):

usigmaN'(t)/N(t) + ueN"(t)/N(t) = -lambda

usigmaN'(t)/N(t) + ueN"(t)/N(t) = -(p_j/L)^2

ueN"(t) + usigmaN'(t) + (p_j/L)^2*N(t) = 0

Which is a simple 2nd order linear ode of constant coefficients.

We then get the two independent solutions:

N(t) = e^(-t/(2*tau))*[K1*e^(t/tq) + K2*e^(-t/tq)]

Where:

tq = (2*ue)/Sqrt[(usigma)^2 - 4*(u*e)*p_j/L] [seconds]

N(t) is therefore dampened at twice the rate of the relaxation frequency, but is subject to the term tq, which may be a real exponential or generally complex sinusoid.

Thus, it follows that without driving/exciting the fields of the conductor in any way: