Relating potential difference of electrostatics with circuit

Question

Since absolute potential isn't defined for a point, we take a reference point which is $\infty$, and then define potential of a point $A$ to be the work done per unit charge to move that charge from $\infty$ to that point. Moreover potential difference between two points $A$ and $B$ is the work done per unit charge to move that charge from $A$ to $B$. Now comes the case of battery and circuit. I have a series of doubts on this.

There are two ends of a battery $+,-$. And we say that these two have a potential difference. But i can't understand how to relate it with the definition i just leaned above. Here there is no electric field and we are not moving any charge from $\infty$ to those points. So analogy of the above with these small two points inside a small battery doesn't seem to click in my mind. Furthermore let's join that battery with a circuit with no resistance. What do we mean by all points in the circuit have equal potential?How do we relate it with the definition of potential above?Does it mean work done to bring a per unit positive charge from $\infty$ to that point in the circuit? Even if it is then which charge?Also,what does positive charge mean?Does it mean protons?

I have spent lion's share of high school physics in solving text book problems related to current and static electricity but didn't get to understand the very basics of these things and how to relate those. We were just taught how to apply formula to solve problems. Please help clear out my basics, respected physics lovers.

Answer 1

There is an electric field made by a battery. Moving a charge to that point would indeed move it through a field and take work.

If the terminal is connected to a conductive wire, then the whole wire is at the same potential. That means it takes no work to move the charge along the wire to the terminal or back. However you move there (even going off the wire and around the block), the $\int F dx$ and $\int E dx$ are zero along the whole path.

Similarly, bringing a charge from infinity to any of the various points on the equipotential configuration will take the same amount of work, regardless of path. If the charge is $q=+1$, then that work will be voltage (assuming voltage was defined relative to infinity).

Finally, what moves around is usually electrons. Usually, nobody is moving positive charges anywhere, and nobody is ever moving any charges from infinity. But within copper, electrons leaving makes positive “holes”. So you can have a surface with net positive charge on it. Any piece of copper that had zero charge and then we put it in a field and electrons moved around... there is an exact equivalent charge distribution situation as if positive particles had moved somewhere instead. They invented that before knowing the negative charges are usually what moves.