# Electric field outside of a current carrying wire

In summary: So in summary, even though there is no net electric charge on the surface of a wire, there is still an electric field outside of it.
Hello,

Basically my problem is trying to understand the electric field outside of a current carrying wire. My first thought is that it is zero, but now I am not sure. The boundary condition of the electric field says that the parallel component inside and out must be the same, so this means the the electric field is non-zero just outside the wire. But if that were the case, then some current could go outside the wire right? Help me with my concept here, thanks.

## Homework Statement

The question is about finding the poyting vector inside and outside of a current carrying wire. I am stuck on the electric field outside the wire.

## The Attempt at a Solution

arpon
The boundary condition of the electric field says that the parallel component inside and out must be the same, so this means the the electric field is non-zero just outside the wire

That would be true if the parallel component of E-field just inside the surface were non-zero
Taking a highly symmetrical example as a cylindrical uniform wire, it's easy to see that that parallel component is zero due to symmetry, regardless of whether the wire is electrically neutral or not. If you force the electrons to flow in a weird way, there might be some chances that E-field also exists outside the wire. But I don't think it's practical or significant.
Just my 2cents

Well, there must be a component of the E field parallel to the wire inside the wire right? Otherwise there would be no current. Then the boundary condition at the edge of the wire says that the E field outside the wire also has a parallel component.

Yep, you're right. I was hasty back then.
After giving it a thought, my best guess is there is always E-field outside the conductor.

The parallel component at the surface remains the same:

$$E_{p-out}=E_{p-in}$$

Therefore: $$\frac{J_{p-out}}{\rho _{out}}=\frac{J_{p-in}}{\rho _{in}}$$

Since $$\rho _{out} >> \rho _{in}$$ and $$J_{p-in}$$ is finite, we have $$J_{p-out} \approx 0$$.

The discontinuity in normal component at the surface:
Consider the case of steady state.
At steady state: $$div\vec{J}=0$$

Thus: $$J_{n-out}=J_{n-in}$$

Or: $$\frac{E_{n-out}}{\rho _{out}}=\frac{E_{n-in}}{\rho _{in}}$$ (*)

When $$\rho _{out} >> \rho _{in}$$ the current is almost retained inside the conductor, i.e. $$J_{n} \approx 0$$ or $$E_{n-in} \approx 0$$ as $$\rho _{in}$$ is finite.

However, as $$\rho _{out}$$ is relatively large, from equation (*), we can see that no conclusion can be drawn about the magnitude of $$E_{n-out}$$. In highly symmetrical cases such as cylindrical uniform neutral wire, by Gauss's law, we can easily see that $$E_{n-out}=0$$. But in general, $$E_{n-out}\neq 0$$, which leads to a fact that there is charges on the surface of the conductor.

your question is very much my question.
i expect that the electric field outside a current carrying(regardless of whether the current is steady or unsteady) must NOT be zero.
this is because... let's assume the contrary... i.e. let's assume that there is no electric field outside a current carrying wire.
now, the wire has some resistance.
take two distince points inside the wire say A & B.
now ohm's law says that there is a potential difference between A & B. call that potential differnce = V.
for simplicity, assume charge density at any given point does not change. this also implies steady current.
now V = line integral of electric field along ANY path joining A & B.( since in case of rho=const curl of E is zero)
if the path lies completely inside the wire then there is no trouble since we know that there is electric field inside the wire which is responsible for creating the potential difference.
BUT. if the path( or part of that path) lies outside the connecting wire( i don't think that there is any restriction in choosing such a path) then since we had assumed that there was no field outside the wire, the potential difference between A & B would be ZERO(or at least less than V for sure).

this would be a contradiction. the only possibility is that there should be electric field present outside a current carrying wire.)

Many theorists believe that there can’t be any Electric Field outside a current carrying wire and their main argument is based on the assumption that the wire is neutral electrically.

Although the wire has no net charge as a whole, the charge density on the surface is not constant (in fact it varies linearly along the wire)!

This distribution produces not only an internal Electric Field of constant magnitude (which drifts the elections), but also an external Field of some kind.

In the extreme case of a perfect conductor all Electric Fields vanish, so any closed-line integral of the electric intensity is again zero.

Last edited:
Gruxg
Papikoss said:
Many theorists believe that there can’t be any Electric Field outside a current carrying wire and their main argument is based on the assumption that the wire is neutral electrically.

Although the wire has no net charge as a whole, the charge density on the surface is not constant (in fact it varies linearly along the wire)!

This distribution produces not only an internal Electric Field of constant magnitude (which drifts the elections), but also an external Field of some kind.

In the extreme case of a perfect conductor all Electric Fields vanish, so any closed-line integral of the electric intensity is again zero.

exactly what i wanted to say!
to argue in favour of this consider a wire carrying CONSTANT CURRENT.
its easy to see that the charge density inside the wire at any point is zero as div(E)=k*div(J)
now div(E) is proportioanl to charge density and div(J) is zero as the current is steady as assumed.

so div(E) is zero and hence charge density everywhere inside the wire is zero.
this, however, cannot be said for the SURFACE of the wire as div(J) really does not exist there.

let's assume that the surface charge density at the surface of the wire is also zero.
now,we know that there is electric field present inside the wire(ohm's law).
what is the cause of this electric field?
there is no varying magnetic field present here as the currents are steady.
there are no charge densities present anywhere since we have assumed that surface charge density is zero and we have proved that charge density inside the wire was zero.

then what is the cause of the electric field? varying magnetic field-- no, charge densities --no. then what??
well, this means that the assumption that the surface charge density is zero at the surface is WRONG.

what say papikoss??

Gruxg
I think that your arguments are solid and can only lead to the conclusion that there must be some kind of surface charge distribution producing the Electric Field within and outside the wire.

In fact such a distribution exists indeed and, as I mentioned above, it varies linearly along the path of the wire. The role of the battery is to sustain those surface free charges.

This is a subject not much covered by modern textbooks but you can find some very interesting information if you google these:

1) Surface Charges on Conductors Carrying Steady Currents by B. R. Russell

2) The Electric Field Outside a Stationary Resistive Wire Carrying a Constant Current by A. K. T. Assis, W. A. Rodrigues Jr., and A. J. Mania.

3) Static Surface Charge on Current-Carrying Conductors by R. Cade and D. Soto Bello (haven't read that yet)

Gruxg
thank you so much papikoss. that was great help. :)

Last edited by a moderator:

## 1. What is an electric field?

An electric field is a force field created by electrically charged particles. It is a fundamental concept in physics that describes the influence of electrically charged objects on each other.

## 2. How is an electric field created outside of a current-carrying wire?

An electric field is created outside of a current-carrying wire due to the movement of electric charges, such as electrons, within the wire. As the charges move, they create a force that extends beyond the wire and forms an electric field.

## 3. How strong is the electric field outside of a current-carrying wire?

The strength of the electric field outside of a current-carrying wire depends on the distance from the wire and the amount of current flowing through it. The closer you are to the wire, the stronger the electric field will be. Additionally, the greater the current, the stronger the electric field will be.

## 4. What factors affect the direction of the electric field outside of a current-carrying wire?

The direction of the electric field outside of a current-carrying wire is affected by the direction of the current flow, the distance from the wire, and the orientation of the wire. The electric field always points in the direction of the force that a positive test charge would experience if placed in the field.

## 5. How does the electric field outside of a current-carrying wire affect nearby objects?

The electric field outside of a current-carrying wire can cause nearby objects to experience a force. If the object has a charge, it will be influenced by the electric field and may be attracted or repelled by the wire. If the object is neutral, the electric field may induce a charge on the object, causing it to experience a force.

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