# Relating potential difference of electrostatics with circuit

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.

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).
• Could you please explain why the whole wire is at same potential in your 2nd paragraph?I don't seem to understand it.And another question,usually we are shown that charges in the battery are denoted by some $+$s accompanied by same number of $-$s.Do these indicate $+1$ or $-1$ charges? Sep 2 at 8:19
• One concept is that for a good conductor, doesn’t take work to move charges in it. If one part was a higher potential, that means theres a net field gradient between those locations, so charges move to get rid of that. If i go from A to B and change potential, then i did work on a charge, took it through a net field $\int Edx$. Those fields in absence of any blockage would force the charges around until enough charge moves that such fields were gone. Or: Potential means took work to make it. Charges try to flow downhill as it were, toward minimizing potential differences. Lower energy state. Sep 2 at 8:32