Consider this complex circuit :

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  1. Consider an ideal resistor with uniform cross section and made of isotropic material with constant resistivity.

  2. Consider an ideal battery giving a voltage difference of $V$ across terminal.

  3. Consider that the wires are superconductors with $0$ resistance.

  4. The voltage drop across the length of current flow is linear, that is $\frac{\partial V}{\partial L}$ is constant.

  5. This implies that Electrostatic force in the direction of current flow is constant throughout the resistor, since $\vec E = \frac{\partial V}{\partial L}$.

** Q: Is there a way to prove that the Electric Field ($\vec E$) is constant all along the length of the resistor in the direction of current flow, derive this mathematically using fundamental laws of electrostatics, or prove why the $\frac{\partial V}{\partial L}$ is linear across the length of the wire in the direction of the current flow?

How do the charges in the wire (at the edges of the resistor) exert a constant electrostatic force/electric field throughout the resistor?


  1. Most other texts give the analogy of a capacitor with 2 charged plates, but I feel those examples are wrong, since they assume plates of infinite surface area which is clearly not analogous to the above circuit.

  2. Don't tell $E = \frac{V}{L}$ is constant at all points assuming $V = IR$ (ohm's law), because its actually the other way. $V=IR$ is probably derived from $E = \frac{V}{L}$ being constant.

  • 3
    $\begingroup$ Does this answer your question? If Electric field is constant in a region, does it imply potential is also constant? $\endgroup$
    – Bob D
    Mar 9, 2021 at 13:02
  • $\begingroup$ @BobD no, that's a totally different question $\endgroup$ Mar 9, 2021 at 13:51
  • $\begingroup$ Can we assume that the current is stationary? I.e. $I(t) = konst$ for all $t$? $\endgroup$
    – Julia
    Mar 9, 2021 at 14:19
  • $\begingroup$ Am I misunderstanding your question, or are you just exaggerating for the sake of humor when you call this a complex circuit? $\endgroup$
    – Hearth
    Mar 10, 2021 at 3:51
  • $\begingroup$ @Hearth yes, correct $\endgroup$ Mar 10, 2021 at 4:19

3 Answers 3


Let's assume the following:

  • The terminals (i.e., the points where the wires make contact with the resistive material) are plates having the same shape as the cross-section of the resistive material.
  • The material obeys the microscopic version of Ohm's Law, $\vec{J} = \sigma \vec{E}$, and has uniform conductivity.
  • We are in steady state, i.e., all quantities are constant with respect to time.

Then it can be proven as follows. (My argument is adapted from Chapter 7 of Griffiths's Electrodynamics.)

  1. The potentials on the plates are $V_1$ and $V_2$.
  2. At the boundary of the resistive material, we must have $\vec{J} \cdot \hat{n} = 0$, where $\hat{n}$ is an outward-pointing normal. If we didn't, then either charge would be building up on the surface or escaping the resistive material.
  3. Given Ohm's Law, this implies that $\vec{E} \cdot \hat{n} = 0$ on the surface of the resistive material.
  4. Since $\sigma$ is constant, and $\vec{J} = \sigma \vec{E}$, the steady-state current condition $\vec{\nabla} \cdot \vec{J} = 0$ implies that $\vec{\nabla} \cdot \vec{E} = - \nabla^2 V = 0$.
  5. Thus, the potential inside the resistive material satisfies Laplace's equation $\nabla^2 V = 0$, with boundary conditions $V = V_1$ at one terminal, $V = V_2$ at the other terminal, and $\hat{n} \cdot \vec{\nabla} V = 0$ along the sides.
  6. A solution to this potential problem inside the volume of interest is that the potential has a linear gradient parallel to the axis of the resistive material. By uniqueness of solutions to Laplace's equation, this is the only possibility.

Note that strictly speaking, the condition for uniform conductivity (used in step 4) could be relaxed to allow for situations where $\vec{E} \cdot \vec{\nabla} \sigma = 0$, and the result would still hold. For example, one could imagine a cylindrical object where $\sigma$ is a function of the distance from the axis, but is still translationally invariant along the axis.

  • $\begingroup$ a linear voltage gradient requires a constant conductivity of the resistive material $\endgroup$
    – Uwe
    Mar 10, 2021 at 3:38
  • $\begingroup$ @Uwe: I initially couldn't figure out where it was required, so I edited my answer to remove the requirement. But on further reflection you're right. Re-editing now. $\endgroup$ Mar 10, 2021 at 13:03

As to your last question: There will be an excess of electrons on the negative end of the resistor, and a deficiency of electrons at the positive end. A uniform field inside requires a gradient in the charge density.

  • $\begingroup$ When you say "a uniform field inside requires a gradient in the charge density" where are you referring to? A uniform field requires zero charge density over the region with the uniform field. The field is the integral of charge. If there is net charge, (which is necessary for a charge gradient), the integral is not constant. $\endgroup$
    – Matt
    Mar 9, 2021 at 21:30
  • $\begingroup$ @Matt he mean't gradient in surface charge density. A constant electric field E across a wire or resistor indeed results from a linearly varying surface charge density in the direction of the current flow. This is actually a little known secret, Check the whitepaper by Chabay, "A unified treatment of electrostatics and circuits" or this pdf www1.astrophysik.uni-kiel.de/~hhaertel/PUB/Voltage-PdN.pdf $\endgroup$ Mar 10, 2021 at 14:40
  • $\begingroup$ Actually, it had not occurred to me that a surface charge density gradient would produce a uniform field within a conductor. I'm still thinking that some of the excess charge may be distributed radially within the conductor. $\endgroup$
    – R.W. Bird
    Mar 10, 2021 at 20:19

You said:

"Don't tell $E = V/L$ is constant at all points assuming $V = IR$ (ohm's law), because its actually the other way. $V=IR$ is probably derived from $E = V/L$ being constant."

But this is the right answer. We have $V = IR$ first, then we derive "$E = V/L$ is constant at all points (of the resistor)". Why does $V = IR$? Well it is from your assumption "Consider an ideal resistor with uniform cross section and made of isotropic material with constant resistivity." We recall that resistance is defined with $V=IR$ thus this is the answer.


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