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I have heard from numerous sources (most notably "Surface charges on circuit wires and resistors play three roles" by J. D. Jackson) that there is surface charge on any steady current carrying conductor of uniform conductivity (i.e the conducting wires for a simple battery resistor circuit). This surface charge is required in order to ensure that the electric field within the conducting wires is such that we have a steady uniform current throughout the circuit. I understand how Ohms law in combination with Gauss's law permit this build up of surface charge during the initial transient period when $\nabla \cdot J \neq0$. What I dont understand is how this surface charge is maintained during steady state conditions when $\nabla \cdot J =0$

For suppose we have such a wire of uniform conductivity $\sigma$. Within the wire, the charge density must be zero since if we take the divergence of ohms law we get $$\vec{E}=1/\sigma \vec{J} \,\,\,\Rightarrow \,\,\,\nabla \cdot \vec{E}=1/\sigma\, \nabla \cdot \vec{J}=0$$ since $\nabla \cdot \vec{J}=0$ for a steady current. Then from Gauss's law, we know that the charge density must be zero everywhere within the wire. The above argument rests on the fact that $\sigma$ is uniform within the wire though. As we approach the surface of the wire though, the conductivity must either abruptly or continuously drop until is reaches the value of the conductivity of the surrounding air (virtually zero). Thus, if we apply ohms law near or at the surface of the wire, we must take into account that $\sigma$ is no longer a constant and so taking the divergence we get $$\vec{E}=1/\sigma \vec{J}\,\,\, \Rightarrow \,\,\, \nabla \cdot \vec{E}=1/\sigma\, \nabla \cdot \vec{J}+\vec{J}\cdot (\nabla \frac{1}{\sigma})=0$$ but now since $\nabla \cdot \vec{J}=0$ we get that $$\Rightarrow \nabla \cdot \vec{E}=\vec{J}\cdot \nabla \rho$$ where $\nabla \rho $ is the gradient of the resistivity ($\rho$ is simply a function that gives the value of resistivity at all points in space and is a constant function within the wire but rises continuously or abruptly near the surface of the wire until it reaches the value of the resistivity of air). My problem is that the current density $\vec{J}$ is always in the axial direction(even near or at the surface) whilst $\nabla \rho$ must always point radially outward near or at the surface (since this gradient by definition points in the direction of increasing $\rho$ and this direction is outwards towards the highly resistive surrounding air). So that means the dot product $\nabla \cdot \vec{E}=\vec{J}\cdot \nabla \rho$ should always equal zero at or near the surface and so by gauss's law, the surface charge density at the surface must also always equal zero. But if this is the case, then how can a surface charge density ever build up on the surface of a conducting wire?

Any help on this issue would be most appreciated because it has been driving me mad recently!

Edit:

My current thinking regarding this issue is that even a current carrying conducting wire that contains a bend should still have $\vec{J}\cdot \nabla \rho =0$ anywhere along the surface of the bend. This is because I would expect $\vec{J}$ to be directed along the curvature of the bend and hence be parallel to a tangent line on the surface of the bend. I would also expect $\nabla \rho$ to be normal to the surface of the bend. Hence $\nabla \cdot \vec{E}=\vec{J}\cdot \nabla \rho=0$ anywhere along the surface of the bend and so no surface charge should accumulate along the surface of a bend.

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    $\begingroup$ Have you considered bends in the wire? For an ideal straight wire with a constant current, there is indeed no net charge. $\endgroup$
    – Gilbert
    Mar 26, 2021 at 19:29
  • $\begingroup$ @Gilbert Yes I have, I would expect that even in a bend, the current density $\vec{J}$ would be tangent to the wire and hence follow the bend while $\nabla\rho$ would always point directly out of the wire and hence it would be normal to the tangent lines along the bend in the wire. Thus $\vec{J}\cdot \nabla \rho =0$ for all points along the surface of the bend? $\endgroup$ Sep 2, 2022 at 8:20
  • $\begingroup$ How does $\nabla\cdot\vec{E}=0$ hold? Since the electric field is distributed outside the conductor, $\nabla\cdot\vec{E}=0$ does not hold for the entire region including space and conductor. $\endgroup$
    – HEMMI
    Sep 2, 2022 at 11:46
  • $\begingroup$ @HEMMI We are examining steady currents so that $\nabla \vec{J}=0$. Thus $\nabla \cdot \vec{E}=\vec{J}\cdot \nabla \rho$ as shown in my question. Now $\vec{J}$ is tangent to the surface of the wire everywhere along the wire (including bends). Also $\nabla \rho$ is normal to the wire everywhere along its surface. Hence $\vec{J}\cdot \nabla \rho=0$ anywhere along the surface of the wire. Thus $\nabla \cdot E=0$. This is all explained in my question. I realize there is a non zero $E$ field outside. How can that be given what I've just illustrated though? $\endgroup$ Sep 2, 2022 at 11:53

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I disagree with your math. Because of the surface discontinuity, you have some jump conditions that you have neglected.

From the continuity equation for the charges, you get:

$$ \frac{\partial \varrho }{\partial t} + \overrightarrow{ \nabla } . \overrightarrow{j} =0$$

Allowing for surface charge and current densities, and using distributions to calculate the divergence, one gets:

$$ \overrightarrow{ \nabla }. \overrightarrow{j}=\{ \overrightarrow{ \nabla }. \overrightarrow{j}\}+ \overrightarrow{n}. \big( \overrightarrow{j^{+}}-\overrightarrow{j^{-}} \big) \delta _{S}$$

$\overrightarrow{j^{+}}$ is the current density outside the conductor (null).

$\overrightarrow{j^{-}}=\overrightarrow{j_{S}}$ is the surface current.

$\{ \overrightarrow{ \nabla }. \overrightarrow{j}\}$ is the usual divergence where there is no discontinuity. From the charge conservation equation, one gets:

$$ \Rightarrow ~~~~ \frac{\partial \varrho_{S} }{\partial t} + \overrightarrow{n}. \big( \overrightarrow{j^{+}}-\overrightarrow{j^{-}}\big) =0$$

Since $\overrightarrow{j_{S}}$ is tangential to the surface, the dot product is null.

So you get: $$ \frac{\partial \varrho_{S} }{\partial t}=0$$

The surface charge density, if any, is constant.

Applying the same technic to :$$ \overrightarrow{ \nabla }.(\sigma \overrightarrow{E} )=\overrightarrow{ \nabla }.( \overrightarrow{j} )$$

One gets: $$\overrightarrow{n}. \big(\sigma^{+} \overrightarrow{E^{+}}-\sigma^{-}\overrightarrow{E^{-}}\big)=\overrightarrow{n}. \big( \overrightarrow{ j^{+}}- \overrightarrow{j^{-}}\big)$$

Outside the conductor: $$\sigma^{+}=0$$ $$ j^{+}=\overrightarrow{0}$$

On the conductor's surface:

$$\overrightarrow{j^{-}}.\overrightarrow{n}=\overrightarrow{j_{S}}.\overrightarrow{n}=0$$

So we get: $\overrightarrow{n}.\overrightarrow{E^{-}}=0$

Combining this result with the second Maxwell's equation we get: $$ \overrightarrow{ \nabla }. \overrightarrow{E} = \frac{ \varrho }{ \varepsilon _{0}} ~~ \Rightarrow ~~\overrightarrow{n}. \overrightarrow{E^{+}}= \frac{ \varrho_{S}}{\varepsilon _{0}}$$

So, you cannot exclude the possibility of a surface charge distribution: $\varrho_{S}$. However the reason cannot be found in Maxwell's equations jump conditions. If there is a charge density, it is for other reasons. Jackson's paper you have provided gives three of them:

  1. To maintain the potential around the circuit.
  2. To provide the electric field in the space outside the conductors.
  3. to assure the confined flow of current.
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This question is a little difficult because it says that you are seeking a canonical answer. However the canonical answer is contained in Jackson’s 1996 “Surface charges on circuit wires and resistors play three roles” paper which you are already aware of. So the canonical answer is clear and you are already aware of it.

What you seem to actually want is an understanding of where your argument fails. You already know the canonical answer but have a clear and convincing argument to the contrary.

The issue is that in your analysis you make two key assumptions that apply only during the steady state and then you ask:

if this is the case, then how can a surface charge density ever build up on the surface of a conducting wire?

Indeed, given your two steady-state assumptions the surface charge density cannot build up because it is assumed to have already built up completely! Let’s examine the specific assumptions:

since ∇⋅𝐽⃗=0 for a steady current

This is the big problem. By the continuity equation $\nabla \cdot \vec J +\partial_t \rho=0$, where $\partial_t=\frac{\partial}{\partial t}$. During the time that the surface charge is being built up $\partial_t \rho \ne 0$ and therefore $\nabla \cdot \vec J \ne 0$. So this condition is explicitly violated during the charge build up.

You used $\nabla \cdot \vec J = 0$ in two places. One was to reason that the charge density was zero everywhere in the wire, so this condition is no longer true. Before the steady state, while the surface charges are still being built up, there may in fact be non-zero charge density inside the conductor. Second you used it to reason that $\nabla \cdot \vec E = \vec J \cdot \nabla (1/\sigma)$ (note I use $1/\sigma$ for the resistivity since I am using $\rho$ for the charge density). So this requirement also fails during the non-steady-state period of surface charge build up.

the current density $\vec J$ is always in the axial direction(even near or at the surface) whilst $\nabla \rho$ must always point radially outward near or at the surface

The fact that the current density is in the axial direction is not a general law of nature. The conductor is an isotropic medium, so there is no naturally preferred direction and current can go in any direction that the E field points. The fact that the current density is in the axial direction is derived from the condition that $\nabla \cdot \vec J = 0$ which, as discussed above, is not valid during the build-up of surface charges.

During the build up of the surface charges the current density will have a component that is normal to the surface. It is precisely this component that leads to the build up of charge on the surface of the conductor. If you assume that it is absent then you assume that the charge cannot be built up.

So, in summary, you are correct that given the steady-state assumptions it is not possible for there to be any build up of surface charge. During that time, since $\partial_t \rho \ne 0$, the steady-state assumptions are necessarily violated. In particular, there can be non-zero charge density within the conductor and the current density will have a component that is normal to the surface. These two facts are what allow the surface charge density to build up.


EDIT: Now, after the surface charges have built up and we are in the steady state you had one remaining concern. Specifically,

So that means the dot product $\nabla \cdot \vec E = \vec J \cdot \nabla \rho$ should always equal zero at or near the surface and so by gauss's law, the surface charge density at the surface must also always equal zero.

This is simply an erroneous implication on the surface of a conductor. Consider a isolated conductor placed in an external E field. Such a conductor, in the electrostatic state, has a surface charge distribution which exactly cancels out the external E field (i.e. it acts as a Faraday cage) so that inside there is no E field. In the electrostatic state, by definition, $\vec J=0$ and therefore $\vec J \cdot \nabla (1/\sigma) = 0$, even though the surface charge is non-zero. So, contrary to your statement, Gauss' law together with $\vec J \cdot \nabla (1/\sigma) = 0$ does not imply that the surface charge is zero.

The issue appears to be that $\nabla (1/\sigma)$ is infinite and $\vec J$ is zero. So their product is undefined. In this case it happens to be finite, but this equation is of no use for discovering that.

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  • $\begingroup$ Thanks for the response. I am aware that there is a brief initial period where the steady state assumption does not apply and hence $\nabla \cdot J\neq0$. However, suppose we wait until this transient period is over. Once the transient is over, we must have that $\nabla \cdot J=0$ (for large $t$). It is at this point that $\nabla \cdot \vec{E}=\vec{J}\cdot \nabla \rho$ . But for all points of the surface of any wire, the dot product $\vec{J}\cdot \nabla \rho$ seemingly vanishes. Hence we must have $\nabla \cdot \vec{E}=0$. So no surface charge. Clearly I am wrong though. My question ... $\endgroup$ Sep 2, 2022 at 14:10
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    $\begingroup$ Sorry, I was still writing my answer but got interrupted, so you commented on an incomplete answer. Please look now and see if this answers your question. $\endgroup$
    – Dale
    Sep 2, 2022 at 14:12
  • $\begingroup$ can briefly be summarized as follows : How can the dot product $\vec{J}\cdot \nabla \rho$ ever be non vanishing on the surface of a current carrying conductor? If this dot product can be shown to be non vanishing on the surface , then we may have surface charge. Otherwise in steady state conditions, surface charge seemingly cannot exist? $\endgroup$ Sep 2, 2022 at 14:13
  • $\begingroup$ I have edited my original question slightly to make it more clear that I am only interested in the surface charge during steady state and that i understand that charge can build up on the surface during the transient period when $\nabla \cdot J \neq 0$. I dont understand how this charge is maintained during steady state while $\vec{J} \cdot \nabla \rho =0$ at the surface $\endgroup$ Sep 2, 2022 at 14:40
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    $\begingroup$ @SalahTheGoat I have added an edit to explain that issue $\endgroup$
    – Dale
    Sep 2, 2022 at 15:26
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There is another way of looking at the problem that was pointed out in the 1941 paper The Electric Field Associated with a Steady Current in Long Cylindrical Conductor by Alexander Marcus. Like the Jackson paper it puts the conductor along the axis of a cylinder far away to have a return path for the current.

Since it is a steady state problem he treats it as "an interesting example of Laplace's equation" Using separation of variables in cylindrical coordinate and he assumes a linear solution in z. Then he looks at the boundary conditions. Following the paper (in Gaussian units).

$$V=(Az+B)(\alpha \log{r}+\beta)$$

Inside the wire $\alpha $ can be set to 0 and $\beta $ to 1.

Taking the derivative to find the electric field we see that the E-field is a constant and points along the wire and inside the wire there is no radial component to E.

$$ \frac{-dV}{dZ}= E_1 =-A_1$$

Where the subscript 1 and 2 are inside and outside the wire and we could also say if the resistance of the wire per unit length is constant that $-A_1=RI$ or $J=\sigma E$ as we expect.

Outside the wire, $$ V_2=A_2z(\alpha\log{r}+\beta)$$

Letting $\alpha$ =1 and evaluating the boundary conditions $\beta=-log(r_1)$ giving $A_2=-log(\frac{r_0}{r_1})$

So inside the wire $$V_1=-E_1z$$

outside the wire

$$V_2=-\frac{E_1z}{log\frac{r_0}{r_1}}log\frac{r}{r_1}$$

Since the potentials and the electric fields are found we can also find the surface charge using the gaussian pillbox.

$$(\frac{dV_1}{dr})_{r=r_0}-(\frac{dV_2}{dr})_{r=r_0}=4\pi\sigma$$

which gives

$$\sigma= \frac{E_1z}{4\pi log(\frac{r_0}{r_1})}$$

Looking at the problem this way, it agrees with Jacksons paper statement that the surface charge is necessary to establish the potentials outside the wire.

Also in steady state, if you can find the potentials inside and outside the wire you can find the surface charge density. There are some pedagogical papers in the American Journal of physics that do this numerically. In doing this you can also see how the charge is distributed at bends in the wire. If appropriate time steps are taken one can see the transient of how the charge moves around and relaxes when a switch is thrown.

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  • $\begingroup$ @ UVphoton: I think you should mention you are working in Gaussian units. $\endgroup$
    – Shaktyai
    Sep 9, 2022 at 9:50
  • $\begingroup$ @Shaktyai Ok, will edit $\endgroup$
    – UVphoton
    Sep 9, 2022 at 9:52
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The local formulation of the Ohm law

$$ \mathbf j = \sigma \mathbf E $$ is not always exactly correct. It is applicable inside a metallic conductor or other so-called linear medium, where other forces on the mobile charge are negligible when compared to electric force.

But when mobile charged particle is very near the surface of the wire, it experiences additional force that cannot be ignored: the force of constraint, from the body of the wire, pushing the particle back into the conducting medium (in direction perpendicular to the surface). This force prevents the charged particles forming the current from jumping out of the wire into the nonconducting medium. They would otherwise jump out because they all repel each other.

From the reductionist/microscopic point of view, this force of constraint is a result of zillions of microscopic attractive and repelling forces due to all the other charged particles in the system, and is present and nonzero almost everywhere, both inside and outside the wire.

However, in macroscopic EM theory, this force is relevant only when the charged particle is very near the boundary where characteristics of the medium change with position. In this case, the surface of the wire is such boundary: usually, the wire attracts the charge much more than the nonconducting medium outside the wire. This constraint force is not usually included in the macroscopic electric field $\mathbf E$, because it is not determined by macroscopic distribution of electric charge, but rather by the type of materials on the boundary. So the constraint force in macroscopic EM theory is an independent force that acts in addition to the force $q\mathbf E$ due to macroscopic electric field.

So the "fixed" Ohm law on the surface of the wire would be something like

$$ \mathbf j = \sigma' \mathbf E + \sigma' \mathbf C $$

where $\sigma'$ is effective conductivity on the surface of the wire (which may be different from that inside the wire) and $\mathbf C$ characterizes constraint force per unit charge, which keeps the charges in the negatively charged wire surface layer bound to the wire.

Let's think of a surface patch that is charged negatively. Thus electric field just above the patch in nonconducting medium points towards the conductor, and electric field acting on charges in the surface layer, while it may be somewhat smaller than field just above, will be in the same direction. So the negative charge in that surface layer will experience electric force that is pushing it out of the wire. Despite this, usually there is no negative charge flowing in that direction in there. Here, we can see the usual Ohm's law $\mathbf j = \sigma \mathbf E$ is not valid in that surface layer.

That is, unless there is field emission or thermal emission going on. For example, field emission means the electric field in the surface layer is so strong it overcomes the maximum possible value of the constraint force that the conductor can exert per unit charge, and negative charges start jumping out of the conductor. This however happens only when electric field in there is pointing towards the wire and is strong enough, which can be achieved by cranking up voltage somewhere or concentrating enough negative charges on an isolated conductor. Typically the field emission will start on the sharpest edges of the conductor, because electric field there is typically the strongest. However, even if the current in the usually nonconducting medium(vacuum) appears, it should not be expected that it obeys Ohm's law. Current in vacuum certainly doesn't.

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I think that unlike the answers stated above, there should be a different explanation. Consider a steadily flowing cathode ray. In a region around it, it will definitely create an electric field but unlike that in electrostatics, this is not due to static but dynamic charges. Similarly what I think is that in a particular cross section of the thick wire,there will be definitely a non uniform charge distribution in wire but not a static charge, but that due to flowing electrons. Thus there will be non uniform current density in the wire which shows that the calculations are wrong. However I think that the surface charge density is static one and one must take its contribution to electric field also.

As pointed by RW bird, the surface charge density and the current density distribution will also vary along the length without violating Kirchhoff 1st law

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I agree with @Shaktyai's answer. I hope what I have written here is consistent with Shaktyai's answer, but if contains any misunderstandings, that is my responsibility.

Assuming steady state is achieved, consider a simple connected capacitor and battery problem.enter image description here.

$\vec{E}$ and $\vec{J}$ on the internal wires and plates disappear. I think this capacitor + battery issue is a special case of SalahTheGoat's concerns. Since this is a special case, the theorem he or she wrote should apply to this problem, even though $\vec{J}=0$. However, it is known that charge accumulates on the conductor surfaces of capacitors. Therefore, it is incorrect to say "Surface charge on a current carrying conductor is impossible".

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You are missing an important part of the puzzle. For simplicity, let us consider a wire of uniform resistivity. For a steady current, the charges must be moving at constant speed and along the wire (macroscopically at least). This means that the net forces acting on them must be zero.

The first force to consider is the Lorentz force due to the field generated by the current. Assuming a uniform current density in the wire, we find that the magnetic field inside it is given by \begin{align} \mathbf{B}(r) &= \frac{\mu_0 I}{2 \pi r} \hat{\mathbf{\phi}} \\ &= \frac{\mu_0 r}{2} j \hat{\mathbf{\phi}}. \tag{1} \end{align} This generates a Lorentz force on the charges: \begin{align} \mathbf{F}_B &= \rho ~ \mathbf{v} \times \mathbf{B} \\ &= -\mathbf{j} \times \mathbf{B}\\ &= j~B_\phi~\mathbf{\hat{r}} \tag{2} \end{align} If there are no other forces acting on the charges, they will move outwards, accumulating on the surface of the wire. After a while, this charge accumulation will generate an electric field in the radial direction which will counteract the Lorentz force. At steady state the two forces cancel out, which means the electric field must be \begin{align} \mathbf{E} &= -\frac{\mathbf{F}_B}{\rho} \\ &= -\frac{\mu_0 r j^2}{2 \rho} \hat{\mathbf{r}}. \tag{2} \end{align}

From the divergence of this electric field you find that it induces a charge density inside the wire of \begin{align} \rho &= \frac{\nabla \cdot \mathbf{E}}{\epsilon_0}\\ &= -\frac{1}{\epsilon_0} \nabla \cdot (\frac{\mu_0 r j^2}{2 \rho} \hat{\mathbf{r}})\\ &= -\frac{\mu_0 j^2}{2 \epsilon_0} \left(\nabla \cdot (\frac{r}{\rho} \hat{\mathbf{r}}) \right) \\ &= -\frac{\mu_0}{2 \epsilon_0} \left(\frac{1}{r}\partial_r (\frac{r j^2}{\rho}) \right)\\ &= -\frac{\mu_0}{2 r \epsilon_0}\left(\frac{1}{\rho}\partial_r( r j^2) + r j^2 (\partial_r \frac{1}{\rho}) \right)\\ &= -\frac{\mu_0}{2 r \epsilon_0 \rho}\left(j^2 + 2 rj \partial_r j - \frac{r j^2 \rho_r}{\rho} \right) \tag{4} \end{align}

We can then solve this differential equation to obtain $\rho$... (to be done)

Since the wire must be charge neutral, we must have an opposite charge at the surface, so that for a given cross section the total charge is zero: \begin{align} 2\pi\int_0^R(\pi R^2) r \rho(r) dr + (2 \pi R) \sigma &= 0. \tag{5} \end{align}

Finally, we find the charge density at the surface of the wire: (to be done) \begin{align} %\sigma = \frac{\mu_0 R j^2}{2}. \tag{6} \end{align}

Notice that this surface charge is generated by the relocation of the charges due to magnetic field induced by the current. There is no need for a varying resistivity. Moreover, you might feel conflicted to see a finite charge density inside a conductor. This is a bound charge, that appears because we are not dealing with electrostatics, since there is a finite current. This is essentially the same mechanism of the Hall Effect, with the peculiarity that here the field generating it is caused by the current in the same wire.

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  • $\begingroup$ A current carrying wire is not charge neutral. In order to maintain a uniform longitudinal field inside the wire there must be a gradient in the charge/unit length. One half of the wire will have a net positive charge and the other half will be negative. I would expect that the radial distribution of this excess charge would depend on magnetic effects, but any small selected volume should conform to Gauss's law. $\endgroup$
    – R.W. Bird
    Dec 21, 2021 at 15:46
  • $\begingroup$ @R.W.Bird what do you mean half of the wire has a positive charge? Which half? $\endgroup$ Dec 22, 2021 at 9:31
  • $\begingroup$ The half which is connected to the positive terminal of the power supply $\endgroup$
    – R.W. Bird
    Dec 22, 2021 at 14:52
  • $\begingroup$ I think I don't know to which phenomenon you're referring to. Could you give me some references? $\endgroup$ Dec 23, 2021 at 14:19
  • $\begingroup$ Continuity of flow in a uniform wire requires that the current and longitudinal electric field must both be uniform. A uniform field requires a uniform gradient in the distribution of charge along the wire. That means there will be extra electrons in the half of the wire which is connected to the negative terminal of the power supply and a deficiency of electrons in the other half. Gauss's law requires that there be electric field lines leaving the outer surface of the wire in the positive half and reentering the negative half. $\endgroup$
    – R.W. Bird
    Dec 23, 2021 at 16:51
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Steady (d.c.) currents in extended mediums usually ignore the weak force on the charge carriers produced by themselves - this is a second order effect (similar to other second order effects like self-heating of the wire and its backlash to resistivity; one evidence for 2nd order effect is in (4), (5): your surface charge is proportional to the square(!) of the current density). This is stated in William R. Smythe "Static and Dynamic Electricity", 3rd ed., Section 6.04: "The distortion of the current distribution in extended conductors caused by magnetic interactions is negligible." However, if you apply a strong external field to the conductor, it will increase the resistance (magneto-resistance) due to the Hall-effect - but the magnetic field of a current density on itself is much smaller.

The electric field outside a current carrying conductor can be explained by simple Maxwell theory (no need to go into details of charge carrier transport or modifying Ohms law at the surface of conductors), just as in the 1941 paper The Electric Field Associated with a Steady Current in Long Cylindrical Conductor by Alexander Marcus: The electric potential must be continuous across the surface of the conductor (unless there are double layers of electric charge on it, for which we have no evidence). Since the inside-wire-potential varies linearly along the length of the wire, also the outside-wire-potential has to do so (immediately on the surface of the wire). On the other hand, the potential in free space satisfies the Laplace equation with a log(r)-dependence. This gives a 1/r-dependence of the radial component of the E-field. Inside the conductor there is no radial E-field (only E-field in z-direction along the wire). Thus, the E_radial-component jumps at the surface of the conductor, and with Gauss' pillbox this leads to charges on the surface. Note that Marcus assumed a perfectly conducting return cylinder at radius r1 > r0. This cylinder forces zero potential at r=r1, and it confines the field to the volume inside the cylinder (outside there are no fields). In the equations of Marcus, you can make r1 infinitely large. Then the r-dependence of the potential vanishes, and you end up with V2(r,z)=-E1*z ... and then you have again NO surface charge on the conductor! Hence, surface charges can be there, but they do not have to be there - depending on the boundary conditions in the surrounding air volume. You could not explain this by magnetic interactions inside the wire or by surface terms of Ohms law.

Note that Markus also assumes $V=0$ for ($r_0<r<r_1, z=0$), and therefore his battery is placed at z=infinity, where it also has a large electrode with $r<=z_1$. This is a very special geometry. Any other, more realistic geometry probably leads to much more complicated maths.

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