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Consider diagram A. In diagram A a car is at the top of a cliff. Gravity is pulling it down, but it does't move since its on a flat surface. Even though it would be at a lower potential at point B, it can't get to point B because the force of gravity is perpendicular to the direction it can move.

Now consider diagram B. In diagram B, an electron is at the negative terminal. The electrical field is pulling in towards the positive terminal, but it can't get there since the conductor is perpendicular to it. Even though it would be at a lower potential at point B, it can't get to point B because the field is perpendicular to the conductor.

Yet, somehow the in the second diagram/scenario, the current does flow. Why? Gravity and electromagnetism are similar and the situations above are similar yet there is movement in the first one but no movement in the second. Why?

  • $\begingroup$ You should draw mobile electrons all along the wire since a conductor if 'filled' with them and these mobile electrons interact thus your reasoning here, drawing an analogy between the two drawings, is incorrect. There isn't a lone electron, like the car, at the negative terminal under the influence of a 'downward' force. $\endgroup$ – Alfred Centauri Feb 9 '14 at 13:11

In the left image, the vehicle is prevented from moving downwards as the ground is exerting a force upwards equal to the force of gravity on the vehicle. However, in the case of the electric circuit, there is no such opposing force. The orientations of the conductors attached to the battery terminals doesn't have an effect on the flow of electrons (unless there is a poor connection) as they flow through the conductive material as opposed to trying to move in a direct line towards the region of lower potential.

As such, most people find it easier to use a "water" analogy when explaining electric circuits. In this case the electric potential can be considered as a pump or other source of potential, as seen in the figure below (the valve would be analogous to a resistance). As soon as a pipe or path (wire) is provided between the pump and an area of lower pressure (potential) then the water will flow, regardless of the shape or orientation of the pipe. This same principle applies to the flow of electrons in the system you have described.

Water analogy for electric circuits

Hopefully this answers your question adequately, I come from an engineering background, so my answer isn't as rooted in electromagnetic theory as others may be.


In a stationary situation a conductor will have an electric field of zero, since all mobile charges/electrons will have moved to its surface.

In situation B the electrons will still be located on the surface of the conductor. But the electrons near the poles of the battery will be pulled towards or pushed away from it which causes a local dislocation of charges. These dislocations will cause a change in the local charge concentrations, which will induce an electric field and will push or pull nearby electrons. This way the potential difference between the poles of the battery can travel through the entire wire.

So to summarize, you can look at electrons in a wire as a train, so if you push on one side (and pull on the other) the entire train will start moving.


Here's the flaw in your diagram B: You've filled space outside battery with uniform $\vec F$ pointing downward. What's the source of it? To match with Gravity, you need a horizontal charged surface outside battery.

Second, you're thinking about Electrostatic Force with battery. Battery actually generates emf (Electromotive Force) to drive electrons. emf isn't really a force (emf is potential difference between battery terminals which can be measured in wire, not in space like that of in Gravity case), but direction of force (which isn't Electrostatic Force) caused by emf is along the length of wire (not along a direction in space). Here's the relationship:

$$\mathrm{E.M.F.}=\int_\text{Circuit} \frac{\vec{F}}{q}\cdot\mathrm{d}\vec{l}$$

Electric field caused by emf isn't conservative, that's why we get positive work done even in circuit. That wouldn't be the case if it was like electrostatic field generated by point charge.

  • $\begingroup$ Actually, emf is present only within the battery itself. In the rest of the circuit, the charge carriers are driven by almost electrostatic field due to charges distributed on the surface of the wires and other metallic parts. The battery maintains this charge distribution and its electromotive intensity acts directly on the charge carriers only in its interior. $\endgroup$ – Ján Lalinský Feb 9 '14 at 9:36
  • $\begingroup$ @JánLalinský I wanted to say you need to measure voltage in wire, not in space. When it comes to electrostatic field, it's not possible because it's conservative. See my last paragraph. The field here is vortex... $\endgroup$ – Schrödinger's Cat Feb 9 '14 at 10:11
  • $\begingroup$ Sure, voltage is measured for two points of the circuit. This voltage is an integral of the macroscopic electrostatic field. The electric field vorticity is negligible in this case - there is no oscillating current, everything is stationary. $\endgroup$ – Ján Lalinský Feb 9 '14 at 10:52
  • $\begingroup$ Vorticity may be negligible, but nature of the field is what I am targeting. Over a closed loop, total work done is positive here. Means, this field isn't conservative. $\endgroup$ – Schrödinger's Cat Feb 9 '14 at 14:36
  • $\begingroup$ There is no substantial work involved, only the Joule heat is evolved in the wires. What you probably mean is the integral of total electromotive force around the circuit, which contains additional contribution due to the electromotive intensity inside the battery. This contribution to the integral is not due to macroscopic electric field inside the battery, since this contributes equal magnitude but opposite sign to the integral. $\endgroup$ – Ján Lalinský Feb 9 '14 at 15:44

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