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In the physics tutorial topic on current electricity it defines the positive terminal as the high potential terminal and the negative as the low potential. When talking about positive test charges this analogy makes sense, the charge moves from the +ve to the -ve where it loses potential till it has to be "topped" up by the cell.

What confuses me when they begin talking electrons. Using this analogy would mean that the electrons move from the -ve to the +ve but this means it would work against the electric field so the potential energy increases till it reaches the +ve terminal.

My question is how does this analogy, if it does, change to accommodate electrons? Does the electric field reverse so the -ve terminal is the high potential and the +ve the low potential?

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Line 4, don't you mean 'losing potential' instead of 'losing charge'? :) Just clarifying. – mikhailcazi Dec 28 '13 at 14:12
    
yes sorry. My mistake ill correct it – user36405 Dec 29 '13 at 13:40

It doesn't change. Whatever force acts on a positive test charge; the exact opposite acts on a negative charge (like an electron).

Any positive charge will attract the electrons, while any negative charge will repel them. Electrons therefore go from low potential states to higher potential states. This doesn't mean they're going from lower potential energy to higher potential energy! You may be knowing that electric potential energy is: $U = K\frac{qQ}{r}$. Since electrons have negative charges, putting in their value into the equation for potential energy, and changing the distance factor will show you that the closer the electrons are to the positive $Q$ charge, the lower their potential energy.

The electric field in a wire is from the positive terminal of the battery to its negative terminal. Electric field, if you remember, is the force experienced by a charge:

$$\vec{E}= \frac{\vec{F}}{q} \implies \vec{F} = q\vec{E}$$

When the charge is positive, the force will be in the direction of the electric field (note the vector sign). When it is negative, it will be in the opposite direction:
For an electron: $$\vec{F} = (-e)\vec{E}$$

which, as is visible, will accelerate electrons in the opposite direction as that of the current.

$P.S:$ Current in a conductor is due to electrons only. Protons don't move very freely, since they're bound to nuclei.

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As you said electrons move from the -ve terminal to the +ve. I was under the impression that moving against the electric field would mean work would have to be done meaning that the potential energy of the electrons increases thus meaning they would have a higher potential at the positive terminal then the negative. – user36405 Dec 29 '13 at 13:43

As you said electrons move from the -ve terminal to the +ve. I was under the impression that moving against the electric field would mean work would have to be done meaning that the potential energy of the electrons increases thus meaning they would have a higher potential at the positive terminal then the negative.

Think back to the basic principle that states that opposites attract and like signs repel, where a proton would attract an electron but two protons repel, etc. Now remember that the convention (arbitrary) is that electric field lines go out of positive and into negative charges, such that if you have a proton and an electron (or, say, a positively charged plate and a negatively charged plate), the field goes out of the proton and into the electron (always at right angles, etc.). This field direction matches the projected motion of a proton simply by convention (just like by convention current is seen as the flow of "positive" charges when in reality electrons are what move -- it's sort of arbitrary). Thinking back to like charges repelling, if a proton is moving away from a proton and towards an electron, if you were to put an electron into that field, it would move against the field lines (because it moves in the opposite direction as the proton but along the same line) simply because, again, like charges repel and opposite charges attract.

So, all this to say: when an electron moves against the field line, this is its natural motion from high potential to low potential (think of it as an object in a gravitational field falling such that it goes from high potential energy mgh_1 to low potential energy mgh_2); if you were to move the electron away from a positive charge and towards a negative charge (in the direction of the field), you'd have to perform work. The electric field lines are always "out of positive" and "into negative" no matter if you're looking at a negatively or positively charged test charge.

(+) - - - - - > - - - - - (-)

To sum it all up, just remember that work is only performed by you if the charged particle/object is being moved AGAINST its natural motion (by which I mean, against the direction in which it wants to move). Otherwise, it is the electric field performing work (imagine you throwing a ball up--you're doing work; and gravity making the ball fall back down--its natural motion/tendency is the gravitational field doing the work).

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