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I'm asking this because it seems to me that when we talk about PD around a charge (i.e. potential difference between infinity and a point), we're talking about the work done per unit charge by an external agent. But when we're talking about the PD in a circuit, we're talking about the work done by the battery's electric field.

Say on earth, there is a positive point charge at point A, then the potential at any point around point A (e.g. lets take the point as point B) will be the work done per unit charge by an external agent in moving a positive test charge from infinity to point B (i.e. to move from a region of lower potential to a region of higher potential). This potential will be positive since work has to be done by the external agent in order to move the test charge (since positive and positive charges repel). This potential can also be thought of as the potential difference between infinity to point B. Therefore, the PD between infinity to B is positive since work has to be done by an external agent (i.e. work done against the electric field).

Alternatively, in a simple circuit with a battery labelled e.g. 6 volts. The 6V refers to the PD between the positive and negative terminals of the battery. Due to the PD, there will be an electric field in the circuit's wire. The 6V also refers to the work done per unit charge in moving a positive charge from the positive terminal to the negative terminal.

However whose work done is the 6V referring to? Is it referring to the battery's electric field in the wire? Or is it referring to an external agent? Or is it referring to the work done by the positive charge? If it is referring to the battery's electric field, then 6V would make sense. If it is referring to the external agent, then the work done should be negative since the external agent do negative work to move the positive charge from a region of higher potential to a region of lower potential, and hence the battery will have a potential difference of -6V. If it is referring to the positive charge, then the charge would have done 6 joules of work per unit charge and hence the voltage will also be 6V.

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  • $\begingroup$ In the field picture of a simple circuit, it's a little easier to see that energy is leaving the battery and entering the resistor. But I'm not sure if that makes the situation clearer or murkier for you. $\endgroup$ – rob Aug 31 '18 at 21:47
  • $\begingroup$ "However whose work done is the 6V referring to? Is it referring to the battery's electric field in the wire?" - I highly recommend that you draw some type of schematic of the circuit and add it to your question. Typically, the voltage drop across the wires connecting the various circuit elements of a circuit have insignificant voltage across, i.e., there is very little work done on a mobile charge moving through the length of the wires. The work done by the source in a simple circuit is typically done on the load, e.g., a light bulb or resistor. Again, please add more context. $\endgroup$ – Alfred Centauri Sep 1 '18 at 0:12
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There are two definitions which are used to define electric potential difference:

  • The electric potential at position $B$ relative to the electric potential at position $A$ is the work done by an external force in taking unit positive charge from position $A$ to position $B$.
  • The electric potential at position $B$ relative to the electric potential at position $A$ is minus the work done by the field in taking unit positive charge from position $A$ to position $B$.

These two definitions are equivalent because the external force is equal in magnitude but opposite in direction to the force on the charge due to the field ie $\vec F_{\rm external} + \vec F_{\rm field}=0$ and so the net work done on the unit positive change is zero which in turn means that the kinetic energy of the charge does not change.
With actually stating it to be so the system is the unit positive change and technically both forces, $\vec F_{\rm external}$ and $\vec F_{\rm field}$ are external forces.

In circuit theory if potential rather that potential difference is to be used it is customary to assign one node as the zero of potential and call that the ground or the earth.

Although related another concept is that of electric potential energy and the definition of a difference in electric potential energy relates to the work done by external forces or minus the work done by the electric field in taking the charges from arrangement of charges $C$ to arrangement of charges $D$.

A single charge on its own cannot have electric potential energy that is a property that an assembly of charges has.

Now to try and answer your question.

Assume the system to be a single (positive for simplicity) charge $Q$ which finds itself in an external electric field generated by something outside the system.
The single charge will experience a force due to the field $\vec F_{\rm field}$ and accelerate ie gain kinetic energy due to the work done by the external electric field.
That single charge interacts with the lattice which consists of charged ions and the gain in kinetic energy is balanced by the amount of heat (or other forms of energy eg light in an LED) generated when the kinetic energy is lost.
Where does that heat come from?
It comes from the electric field produced by who knows what.
Rather than talk about forces another way of describing the situation is in terms of electric potential and to say that there has been a decrease in the electric potential of the single charge.
In this case one could say that the work done on the charge by the electric field is $-Q\Delta V$ where $\Delta V$ is the change in the electric potential (the electric potential difference).
Note that in this context you it would be wrong to say that the single charge has lost electric potential energy.

Now consider a charged capacitor.
The capacitor has a stpre of electric potential energy in its electric field and at some time previously something had to do work to separate charges and increase the electric potential energy of the capacitor.
The capacitor also would have a potential difference across its plates.

Consider a system which consists of a charged capacitor whose terminals are connected to resistor.
Positive charge flow from the positive terminal of the capacitor though the resistor and to the negative terminal of the capacitor.
In doing so the electric potential energy of the system is decreased and an equal amount of heat is generated in the resistor.
The amount of charge stored on the capacitor would decrease.
In simple terms you can think of a moving positive charge in the resistor being subjected to the electric field produced by the charges on the capacitor with that electric field doing work on the moving positive charge.
The moving positive charge moves from a position of relative high electric potential to a position of relative low potential and has work done on it in doing so.

Finally consider a "strange" capacitor which has an electrochemical reaction occurring between its plates (terminals).
That electrochemical reaction can move positive charges from the negative terminal to the positive terminal and maintains a constant potential difference across the terminals.
This is a cell in which the chemical energy of the system (battery) is converted into electric potential energy.
Now connect a resistor across the terminals of the cell.
The electric field outside the cell is from the positive terminal to the negative terminal.
Under the influence of this external electric field positive charges will move from the positive terminal through the resistor and arrive at the negative terminal of the cell.
That external electric field does work on the moving positive charge with the result that heat is generated in the resistor.
That moving positive charge has gone from a higher potential to a lower potential and as a result the system (cell & resistor$ has lost some electric potential energy.
However with the cell the moving positive charge moves from the negative terminal to the positive terminal increasing its potential and hence resulting in an increase the electric potential energy of the system.

Overall there is a balance and the electric potential energy of the system stays the same.
After completing a complete circuit the potential of the moving positive charge is unchanged.
Chemical energy has been converted to an equal amount of heat.

Whose work done is the 6V referring to?

In moving a charge $+Q$ from the positive terminal of the cell to the negative terminal of the cell the external electric field generated by the cell does $6\, Q$ units of work on the moving charge, the potential of the moving charge drops by $6\, V$ and $6\,Q$ of heat is generated in the resistor.

In moving the charge $+Q$ from the negative terminal of the cell to the positive terminal of the cell in the opposite direction to the direction of the electric field inside the cell the electrochemical process does $6\,Q$ units of work on the charge and raises its potential by $6\, V$.

If the moving positive charge goes round a complete circuit then its potential does not change, the electrical potential energy of the system does not change, forces acting on the moving charge do work which result in a change of $6\, Q$ units of chemical energy being converted into $6\,Q$ units of heat.

Take your pick.

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  • $\begingroup$ Thank you very much for the detailed answer; I made an edit as I think there is a typo. $\endgroup$ – Bøbby Leung Sep 3 '18 at 0:03
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In a circuit with a battery, the battery does its work by converting chemical energy to the potential energy of the charges at the terminal of the battery.

Then, this potential energy of the charges is converted to their kinetic energy and heat or some other forms of energy, depending on the type of a load, as the charges move round the circuit. So, we can say that the charges spend their potential energy or that the charges perform some work as they flow through the circuit.

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" it seems to me that when we talk about PD around a charge (i.e. potential difference between infinity and a point), we're talking about the work done per unit charge by an external agent"

You're absolutely right; most textbooks define potential and potential difference due to a static charge in this way. Such definitions are (usually!) correct, but, in my view, rather clumsy. There is no need to bring in an external agency. The definition that I prefer is that the pd between A and B (the amount by which the potential of A is greater than that of B) is the work done by the field per unit charge on a test charge going from A to B.

The amount of work per unit charge in this definition is indeed equal to the amount of work per unit charge that would have to be supplied by an external agency in moving a test charge from B to A (assuming no change in the KE of the test charge!), but why bring an external agency into the definition of an electrical quantity arising from an electric field?

[It is, of course, perfectly possible for a test charge to go from A to B in an electric field without any external agency other than the field. An example would be the acceleration of electrons in an electron gun. Electrical PE is lost and kinetic energy gained.]

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