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I understand that an electric field is setup inside a conductor when it is connected to the terminals of a battery and that this field is caused by induced surface charges on the conductor. These surface charges are also responsible for 'bending' the field along the shape of the conductor.

What causes these induced surface charges - Is it just the strength of the electric field at the terminal of the battery ? If so, do we really need the conductor to have contact with the terminal to set up this field inside the conductor? In other words , why doesn't a bulb glow if the circuit is brought sufficiently close to the battery without actually touching it ?

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  • $\begingroup$ I guess most of the time there is no "pure" contact, since copper wires will have an oxide layer. That's why electrons will tunnel through it. $\endgroup$
    – jinawee
    Commented May 28, 2014 at 8:12
  • $\begingroup$ @jinawee maybe -- the oxide layer is dang thin and its sheet resistivity pretty low, so tunneling probably isn't necessary in general. $\endgroup$
    – Carl Witthoft
    Commented May 28, 2014 at 11:39
  • $\begingroup$ possible duplicate of Does a AAA battery have a dipole moment? $\endgroup$
    – rob ♦
    Commented Aug 1, 2014 at 20:12
  • $\begingroup$ Get a 2 foot fluorescent bulb and in the evening go stand near some high voltage lines. The bulb should glow. What is making it glow? The electric field of the power lines. $\endgroup$
    – Bill N
    Commented Jan 21, 2015 at 22:22

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If the battery has a very high voltage, then it is possible that the air will break down and charge will flow through the air from the battery to the light bulb. When that happens, the air becomes a conductor - in the same way that the charge in the clouds can flow to the ground when lightning strikes.

For normal, low, voltages this cannot happen unless the gap is very small (a fraction of a millimeter). In that case, you'll most likely touch them together anyway. However, you may notice a small spark when you open or close the circuit. That spark is a sign that electric charges are flowing through the air instead of through a conductor.

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What causes these induced surface charges - Is it just the strength of the electric field at the terminal of the battery ?

Electric field is important to maintain the surface charges, but it is not sufficient for the current to flow. For that, the conductor has to be so close to the voltage source that carriers of electricity can actually leave the terminal of the battery and go to the conductor. For that to occur without solid contact, there has to be strong electric field between the terminal and the conductor. This strong electric field occurs when the gap is very small, but leads to sparks as hdhondt says - and to deterioration of the metals. The most practical way to have transfer of electric charge without sparks is to make sure there is a solid contact.

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  • $\begingroup$ Why do 'carriers of electricity' have to leave the terminal of the battery and go to the conductor for the current to flow ? Any visible effect of electricity(eg like a bulb glowing ) is because of the electric field that is setup inside the conductor and not because of charge carriers flowing through the conductor. Given that the electric field that is setup inside the conductor is purely based on the surface charges and these surface charges are caused purely by the electric field at the terminals of the battery, where is the need for charge carriers to flow from battery into the conductor ? $\endgroup$
    – shunya
    Commented May 28, 2014 at 13:30
  • $\begingroup$ "Any visible effect of electricity(eg like a bulb glowing ) is because of the electric field that is setup inside the conductor and not because of charge carriers flowing through the conductor. " That is incorrect, the bulb glows because the current is present which dissipates macroscopic energy of EM field into heat motion of the filament. Electric field is not enough, that's why dielectrics in static electric field do not glow. $\endgroup$
    – Ján Lalinský
    Commented May 28, 2014 at 17:02
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    $\begingroup$ In metals, electric field and current density are either both present, or both vanish. If the conductor is not connected to a circuit with a source of voltage, in any static electric field of the source it will quickly assume static equilibrium, where all the surface charges distribute in such a way that inside the conductor charge density, current density and electric field vanish. This is studied in electrostatics. $\endgroup$
    – Ján Lalinský
    Commented May 28, 2014 at 17:03
  • $\begingroup$ One way to disturb this static situation is to connect the conductor to source of voltage, say AA battery. When connected to the conductor, battery transforms its chemical energy to electric work done on the current in the conductor, which eventually changes into light and heat. No connection, no current, no chemical transformation, no work, no light and no heat evolved in the conductor. $\endgroup$
    – Ján Lalinský
    Commented May 28, 2014 at 17:06
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    $\begingroup$ This electric field does not decay to zero, since the source of voltage maintains constant potential difference between two ends of the conductor - the contact points. This is only because the source of voltage is able to push charge carriers against macroscopic electrostatic forces of the terminals and conductor. This is something that just another piece of conductor could not do; the latter would just become a part of electrostatic system. $\endgroup$
    – Ján Lalinský
    Commented May 30, 2014 at 5:25
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The word "current" contains the meaning of motion, what moves in an electric circuit is the electrons. In conductors electrons get an average drift velocity and create a current. In resistances this velocity falls very low ( that is what resistance to a current means, too) .

What causes these induced surface charges - Is it just the strength of the electric field at the terminal of the battery ? If so, do we really need the conductor to have contact with the terminal to set up this field inside the conductor?

When a battery terminal touches a wire, the field transiently propagates in the bulk of the wire, not just the surface. When a circuit has a direct field imposed on it, as battery terminals , electrons flow out of the battery and push by their drift electrons to go into the battery at the other pole.

In other words , why doesn't a bulb glow if the circuit is brought sufficiently close to the battery without actually touching it ?

If air intervenes, and the field first meets air and then the first metal of the circuit, there will be a transient current induced by the field on the conductor which will stop as the moving electrons will not be replaced, the air has large resistance.

A circuit can have an Alternating Current, AC, at the beginning, the electric field changing sign sinusoidally. Then the transient current, even through air, will be changing direction/sign and an AC current will appear, proportional to the capacitance between the source and the beginnings of the circuit. This happens because the electrons drift back and forth within the metal parts of the circuit, and do not need to cross through air. This current will light a lamp, although the capacitance will be very small between ends of wires. Capacitors have large surfaces and thus can allow larger currents . Capacitors are an integral part of AC circuits , from radios to computers , to air conditionerss, .....

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