How Does Current Flow in an Equipotential Conductor?

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ajaysabarish
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i have studied that a conductor is equipotential and charges don't flow through it but how does it flow through a wire?
 
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ajaysabarish said:
i have studied that a conductor is equipotential and charges don't flow through it but how does it flow through a wire?

It is because there is an external "source" that keeps one end of it at one potential, and the other end at another potential.

Note that the conductor does not have zero resistance for this to occur.

Zz
 
Also note that if the conductor is in electrostatic equilibrium, then there is no (net) electric field on the charged particles inside (hence a constant electric potential). This means that if the conductor is not in electrostatic equilibrium, then there is a (net) electric field on the charged particles inside, and thus, a potential difference, which causes current.
 
ZapperZ said:
It is because there is an external "source" that keeps one end of it at one potential, and the other end at another potential.

Note that the conductor does not have zero resistance for this to occur.

Zz
but potential difference across a wire(conductor) is taken as zero while writing kirchhoffs law?so there is no potential difference hence charges shouldn't flow through it
 
Prannu said:
Also note that if the conductor is in electrostatic equilibrium, then there is no (net) electric field on the charged particles inside (hence a constant electric potential). This means that if the conductor is not in electrostatic equilibrium, then there is a (net) electric field on the charged particles inside, and thus, a potential difference, which causes current.
but potential difference across a wire(conductor) is taken as zero while writing kirchhoffs law?so there is no potential difference hence charges shouldn't flow through it
 
Dale said:
I think you may be talking about a conductor in electrostatics.
iam talking about both,just comparing them.
 
ajaysabarish said:
but potential difference across a wire(conductor) is taken as zero while writing kirchhoffs law?so there is no potential difference hence charges shouldn't flow through it

Try doing Kirchoff law on such a circuit, i.e. with a zero-resistance conductor connecting the ends of a battery. You will have a SHORT!

Zz.
 
ZapperZ said:
Try doing Kirchoff law on such a circuit, i.e. with a zero-resistance conductor connecting the ends of a battery. You will have a SHORT!

Zz.
yes,so shouldn't it have a resistance,but why don't we take it during kirchhoffs law
 
ajaysabarish said:
yes,so shouldn't it have a resistance,but why don't we take it during kirchhoffs law

Because you are representing the resistance in a REAL conductor via the resistor element/s in the circuit. This way, the "line" that you use to connect one element to the next is nothing more than a schematic representation of the path connecting one element to the other!

Zz.
 
It is incorrect to say that electrons will not flow in an equipotential, or that current will not flow in a equipotential. It is more correct to say that at in an equipotential there is nothing driving the current. For example you connect one terminal of a voltage source to a set of electrical components with a connecting wire. No current will flow, because everything will be at the same potential.
Now connect (with a wire) the electrical components to the other terminal of the voltage source. (This provides a return path to the voltage source). Because the two terminals of the voltage source are at different potentials, this time current will flow through the electrical components. The voltage difference between the terminals of the voltage source drives the current. The voltage difference will be distributed through the electrical components in proportion to their resistances or reactances.
Note a return path may not be needed of one end that is not connected to the voltage source is connected to a ground wire. (Be careful, if working with harmful voltage, that you (your body and/or wet feet) do not supply the return path to the ground)