1
$\begingroup$

So, I was revising my concepts of Current and electricity,and while watching a related video on youtube, I came accross this video....

enter image description here

A simpler representation of above circuit is shown below--

enter image description here

NOTE:

HERE , ACTUALLY IT IS THE CELL WHAT 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 AREAHere, The narrator says that the field inside the current carrying conductor (which is generated due to Charges on sufaces and the battery),is same in magnitude everywhere (if the area is same) , But didn't gave any clear explanation of the same...

Why 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 because to maintain steady current ?

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

I am expecting a reply that could clear my ideas.... and help me to verify the same!

(FOR SIMPLICITY , CONSIDER A CONDUCTOR OF UNIFORM CROSS-SECTIONAL AREA)

$\endgroup$
1
  • $\begingroup$ Which textbook of electromagnetism are you reading? $\endgroup$
    – my2cts
    Jan 15 at 9:52

3 Answers 3

0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Short answer

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 $e_1 = e_2$.

$\endgroup$
-1
$\begingroup$

There is no electric field inside a conductor because gauss law states that electric field is due to net charge and there is no charge inside conductor, so there is no electric field. But there is constant potential inside a conductor but zero potential difference, thus no flow of current.

On surface of a conductor, there is also zero potential difference so no current flow untill it connected to a points have different potential. A surface acts as a equipotential surface.

Now your question should how to verify constant electric potential inspite of no charge inside. Well an electric field is negative gradient of electric potential. So even zero electric field can have non-zero constant electric potential.

$\endgroup$
3
  • 1
    $\begingroup$ Please Review the edited question again.. i think you misunderstood....i hope it would be more clear now $\endgroup$
    – TPL
    Sep 5, 2022 at 15:16
  • 1
    $\begingroup$ I guess that the OP is considering a conductor with some internal resistivity, i e. a non perfect conductor. $\endgroup$
    – basics
    Sep 5, 2022 at 16:01
  • $\begingroup$ This answer is incorrect. $\endgroup$
    – my2cts
    Jan 15 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.