Possible source of confusion -For work done carrying a unit charge from Point 1 to Point 2, ie the Potential difference, is it
Voltage, $(V_{21}\:)=V_2\:-V_1\:$
OR is it
Voltage, $(V_{12}\:)=V_2\:-V_1\:$
Secondly, when we talk about the potential difference between $A$ and $B$, do we mean work done carrying a unit charge from $A$ to $B$ or the other way round. Is it equal to $(V_{AB}\:)$ or $(V_{BA}\:)?$
if you answer these alone, I would be very grateful!
Longer story, aka the inspiration...
...The positive electrode has a potential difference $V_{+}\:$ $(V_{+}\:> 0)$ between itself and the electrolyte solution immediately adjacent to it marked A in the figure. The negative electrode develops a negative potential $–V_{-}\:$ $(V_{-}\:> 0)$ relative to the electrolyte adjacent to it, marked as B in the figure. When there is no current, the electrolyte has the same potential throughout, so that the potential difference between P and N is $V_+\:-\left(-\:V_-\:\right) = V_+\:+\left(\:V_-\:\right)$ . This difference is called the electromotive force (emf) of the cell and is denoted by ε.
Can any one help me trace the path of the positive charge in this case and give a better definition of $V_+$ and $V_-$ in this case along with 2 things. The fact that both are defined as positive annoys me, maybe it could be a typo.
- The path of the positive charge in those $V_-$ and $V_+$
- Mathematical representation of in the form format $V_-$ = $V_{NB}$ = $V_N$- $V_B$ or whatever is actually correct