From electrostatics, we know that all the charges reside on the surface of a conductor. We also know that there is no skin effect for DC flow, which implies uniform flow of current throughout the conductor.

Now my confusion is: since charges accumulate on the surface of the conductor, shouldn't current flow only along the surface of conductor (for both AC and DC currents)?

Could you please clarify?

  • \$\begingroup\$ From the normal circuit analysis perspective (both AC and DC), don't think that charges accumulate in the conductor. When one charge enters one side of the wire, it pushes another charge out at the far side of the wire. If the far side is open-circuit, then charges do not enter the wire (unless we start to consider that the wire is a transmission line or has capacitance to some other part of the circuit). \$\endgroup\$ – mkeith Feb 19 at 19:28

I think your confusion is that it is not true that all charges are on the surface of the conductor. If the conductor has a net positive or net negative charge, the net charge will be near the surface, but there will still remain many charged particles in the interior, they are just balanced and cancel each other out.

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In metals, such as copper, electrons are free to move. They are mobile. They stay attached to the wire, but they can move around within the wire. What causes DC current to flow is that a voltage potential is applied across the wire. Because of how the universe works, the voltage across the ends of the wire creates a uniform electric field (voltage gradient) along the length of the wire. The electrons, since they are charged particles and free to move, travel along the wire "feeling" the electric potential gradient. But the number of electrons in the wire remains constant. For every electron departing the wire at one end, another one is being inserted at the other end. That is how circuits work at a basic level. An electric field causes charges to move in a conductor.

If there is a field, but one end of the wire is open-circuit, so no current can flow, then the electrons kind of crowd one end of the wire or the other as you say. But that doesn't really happen when there is a closed circuit. Or doesn't happen to any extent that you need to worry about for practical purposes when there is a closed circuit.

I suppose the superconductor is an exception because charges can move in a super conductor even without an electric field.

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    \$\begingroup\$ There is no 'field gradient'. The E field is constant along the conductor, and it is caused by surface charges whose (very low) density depends on the geometry of the circuit. (At least, this is the classical picture). \$\endgroup\$ – Sredni Vashtar Feb 20 at 14:35
  • \$\begingroup\$ Thanks for your answers @Justin and mkeith. What I understand is that charges will accumulate on the surface of conductor only if net charge of that conductor is nonzero at any instant. Please point out if anything is wrong in my understanding. \$\endgroup\$ – Nafis Sadik Feb 20 at 15:38
  • \$\begingroup\$ In order to have an E field that follows the direction of the conductor and is constant inside it, you need surface charge. It does not have to be net charge in excess, the whole conductor will remain neutral, but you will end up with negative rings of charge somewhere and positive rings of charge somewhere else. The distribution of charge on the surface is what creates that small E field inside, and that small E field pushes the charges inside the conductor to give a current whose density obeys j = sigma E. \$\endgroup\$ – Sredni Vashtar Feb 20 at 17:18
  • \$\begingroup\$ @SredniVashtar potential gradient then. If you write a good answer I will be happy to delete mine. I am sure you are aware that for DC purposes, the full conductor cross section is utilized by the current. This is the crux of what the OP is asking. \$\endgroup\$ – mkeith Feb 20 at 21:15

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