Clarification of electric potential difference

[keroberous]

I have some questions regarding the sign of potential difference. For example, given the literal definition as the difference in potential between two locations it should be possible for potential difference to be negative if the final potential is less than the initial, but while reading texts they always use positive values for potential difference, like "a proton moves through a potential difference of (some positive number)".

Or we defined potential difference to be ## \Delta V=-\varepsilon \Delta d ## which rearranges to ## \varepsilon =-\frac {\Delta V} {\Delta d} ## but I'm not sure what purpose the negative serves here because in all of the examples in the text they just use the equation for ## \varepsilon ## without the negative.

This part of potential difference is what I find the most confusing no matter how many times I read about it.

 

[sophiecentaur]

they just use the equation for εε \varepsilon without the negative.

Could you quote some reference for this change of sign in action? Is it in more than one text?

 

[Dale]

it should be possible for potential difference to be negative if the final potential is less than the initial,

Definitely, yes.

they always use positive values for potential difference, like "a proton moves through a potential difference of (some positive number)".

They just chose positive numbers for their examples. It doesn’t mean negative numbers are not possible

For the rest, a specific reference would be helpful, as @sophiecentaur suggested

 

[keroberous]

Here are two questions and I'm not sure how they are different.

 

[Dale]

For 6 they are ambiguous about the direction so you can assume any direction convention you like. May as well choose the one that makes it positive.

For 8 they are explicit about the direction so you just have to follow the math, which leads to a negative number in this case

 

[sungod]

In 8 the field points along the +ve x axis. Therefore, this is the direction of force on a positive particle. Since the force is in the direction of decreasing potential, E = -dV/dx, ie V falls, as x increases.

 

[Kenwstr]

I think that in most practical situations, it doesn't matter which direction you model the current to run as long as your are consistent with that. Neg to Pos is correct of course but you often see it modeled the other way around. The main thing that matters is the magnitude of the potential as it determines amps, watts and ohms relationships.