# How can we verify mathematically that electric fields inside a current carrying conductor have same magnitude?

So, I was revising my concepts of current and electricity, and while watching a related video on YouTube I came across this video:

A simpler representation of above circuit is shown below:

NOTE: Here, actually it is the cell that seems like two parallel plates... There is no gap in between; it is just the visuals.

For simplicity, consider a conductor of uniform cross-sectional area. Here, the narrator says that the field inside the current carrying conductor (which is generated due to charges on surfaces and the battery) is the same in magnitude everywhere (if the area is same), but they didn't give a clear explanation of the same...

Why do charges have to align in such a way that the magnitude of the field could be the same? Is it because otherwise it would violate the conservation of charge (charge entering = charge leaving)?

Or is it to maintain steady current?

I know, and my teachers also told me, that the field had the same magnitude. I am trying to find the answer but couldn't figure out how I can verify it mathematically.

• Which textbook of electromagnetism are you reading? Commented Jan 15 at 9:52

What you say is true for steady conditions, in conductors with uniform resistivity and section, for the average of the electric field on the sections of the conductors.

## Details

The answer, and the answers given by other members, should be clear if we consider the constitutive equation linking the electric field $$\mathbf{e}$$ and the current density $$\mathbf{j}$$ through the resistivity $$r$$ (or its inverse, the conductivity $$\sigma = \frac{1}{r})$$,

$$\mathbf{e} = r \mathbf{j} \qquad , \qquad \mathbf{j} = \sigma \mathbf{e}$$.

• perfect conductors have zero resistivity, $$r = 0$$, and thus $$\mathbf{e} = \mathbf{0}$$ inside the conductor for every value of the current;
• in non-perfect conductors wit non-zero electric current density $$\mathbf{j}$$, the electric field is not equal to zero, but it's equal to $$\mathbf{e} = r \mathbf{j}$$.

Steady conditions for conductors with constant section. Now, let's consider here the steady conditions, and take a volume of the conductor. Electric charges can leave the volume through its lateral surface, otherwise electric charge would accumulate on the surface of the conductor, or leave the conductor; the integral balance equation of electric charge reads,

$$0 = \displaystyle \int_{S_1} \mathbf{j} \cdot \mathbf{\hat{n}}_1 + \int_{S_2} \mathbf{j} \cdot \mathbf{\hat{n}}_2 = - \int_{S_1} \mathbf{j} \cdot \mathbf{\hat{t}}_1 + \int_{S_2} \mathbf{j} \cdot \mathbf{\hat{t}}_2 \qquad\rightarrow \qquad \int_{S_1} \mathbf{j} \cdot \mathbf{\hat{t}}_1 = \int_{S_2} \mathbf{j} \cdot \mathbf{\hat{t}}_2$$,

being $$S_1$$, $$S_2$$ two sections of the wire, with the respective normal unit-vector pointing outwards the volume, and $$\mathbf{\hat{t}}$$ is the unit vector "pointing always in the same direction when you move along the conductor", $$\mathbf{\hat{n}}_1 = -\mathbf{\hat{t}}_1$$, $$\mathbf{\hat{n}}_2 = \mathbf{\hat{t}}_2$$.

Using the constitutive law $$\mathbf{j} = \sigma \mathbf{e}$$, if the conductivity is uniform and equal on the two sections $$\sigma|_{S_1} = \sigma|_{S_2} = \sigma$$

$$\displaystyle \sigma \int_{S_1} \mathbf{e} \cdot \mathbf{\hat{t}}_1 = \sigma \int_{S_2} \mathbf{e} \cdot \mathbf{\hat{t}}_2$$

and thus the average flux of the electric field across each section of the conductor with the same resistivity has the same value. Far from any bend in the circuit, for symmetry considerations, the current density, and thus the electric field has the same direction as the axis of the conductor and thus,

$$e_1 A_1 = e_2 A_2 \quad \rightarrow (A_1 = A_2) \rightarrow \qquad e_1 = e_2$$.

Close to circuit bends, the electric field could be non uniform in space, and the relations hold only with the average quantities.

Circuit approximation. With the circuit approximation, lumping the conductor in a line with section as a property, we take the average value of a physical quantity on a section as the uniform value on that section, and both the current density and the electric field aligned with the unit vector tangent to the axis of the conductor,

$$\displaystyle \mathbf{e} = e \, \mathbf{\hat{t}} = \dfrac{1}{A} \int_A \mathbf{e} \cdot \mathbf{\hat{t}} dA \, \mathbf{\hat{t}}$$
$$\displaystyle \mathbf{j} = j \, \mathbf{\hat{t}} = \dfrac{1}{A} \int_A \mathbf{j} \cdot \mathbf{\hat{t}} dA \, \mathbf{\hat{t}}$$,

so that the relations obtained before holds $$e_1 = e_2$$.

Depending on how you look at it the answer is a little subtle. Recently there was a related question. In that answer I included some references.

As Jackson puts it

In general, the conductors of a current‐carrying circuit must have nonuniform surface charge densities on them (1) to maintain the potential around the circuit, (2) to provide the electric field in the space outside the conductors, and (3) to assure the confined flow of current. The surface charges and associated electric field can vary greatly, depending on the location and orientation of other parts of the circuit.

The short answer is that the wires have some conductivity.

$$J=\sigma E$$

and assuming the cross section and the resistivity or conductivity of the wires is the same throughout the wire, in the wire the electric field is parallel to the current and would be uniform if the conductivity is uniform.

A deeper look at the conductivity, you can think of the electrons as a gas moving at random throughout the conductor. However, when there is a an electric field applied, there is a net velocity of electrons called the drift velocity. So while there are individual electrons moving in all directions if you look at some small section of wire number of electrons entering the area of one end of the section, that equals the number of electrons leaving the cross sectional area of the section. $$J=qnv_{drift}$$

So why does does there need to be charge on the outside of the wire? It is as Jackson says, to have a potential outside of the conductor. If you thought of a very long conductor you could look at it as being enclosed by a cylinder at infinity and solve Lapalace's equation and you would find a surface charge is required on the surface for all the boundary conditions to be met.

While there is no radial component to the E-field inside the conductor, outside the conductor you do have have an electric field that is nearly perpendicular to the conductor because of the surface charge. Since current is flowing through the conductor that creates a radial steady state B field outside the conductor. This also helps explain how power flows, since the Poynting vector is proportional to $$E\times B$$.

If you start having a high frequency circuit, or look at a transient like shutting or opening a switch in the light bulb circuit it becomes more complicated because you can also radiate like an antenna, or it takes time for RC times constants or because information about what is happening in one part of the circuit doesn't reach the other part of the circuit because of the speed of light limit.

In classical long-wire-current-flow, we consider all current to be in the axial direction, implying no radial current (and thus no radial electric field, by Ohm's law). There will be no circumferential E-field if there is no time-dependent current, because the curl of E is zero (Maxwell's equations say).

Thus, a long wire can be considered identical to many identical parallel strands of wire, insulated from each other. Applying resistors-in-parallel reasoning will tell us that the current (thus the average E-field) is expected to be identical. In the limit of strand cross sectional area approaching zero, that implies constant-everywhere axial E-field.

There may be minor corrections if the wire cools through its outer surface but generates significant internal heat, because the resistivity can be temperature-dependent. There can be major corrections if AC and skin depth effects are considered.

Consider a circular loop of wire as a perfect conductor. Lets say that its full of negatively charged electrons. Now all the electrons repel each other so they all try to get as far away from their neighbour as possible, so the distribution of charge anywhere in the circular loop of wire is uniform, therefore it follows that there is no Electric field (E field gradient anywhere in the loop is zero).

Now if a battery in the loop is causing a current flow (charged electrons moving round the loop) there is an E field gradient but its the same anywhere in the loop because if its not the electrons will rearrange again to stay apart (like charges repel) and the field will be uniform. (E field gradient anywhere in the loop is a constant value depending on the battery voltage).

What gets interesting is if you start moving the loop through a magnetic field, depending on the geometry, if the magnetic field is changing so will the electric field in the loop. Then you have to find out about Faraday's law and Ampere's law to make sense of it.

PS the YouTube video - it doesn't look like a good video to me.

There is electric field inside a conductor because Ohm's law $$\text{J}=\sigma\text{E}$$, states that electric field is due to resistivity to flow of current, if there is no resistivity there is no electric field inside conductor. Electric field inside conductor is loss of energy.

This loss of charges account for continuity equation. In Faraday's law, potential across two points of conductor shows that in region of cross sectional area, electric field opposes current by producing opposite magnetic field. This means if we consider region as path for current then there is electric field there and any material through which current flow is a conductor.

• Please Review the edited question again.. i think you misunderstood....i hope it would be more clear now
– TPL
Commented Sep 5, 2022 at 15:16
• I guess that the OP is considering a conductor with some internal resistivity, i e. a non perfect conductor. Commented Sep 5, 2022 at 16:01
• This answer is incorrect. Commented Jan 15 at 9:47