# W=- electric potential

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1. Dec 7, 2014

### sanhuy

if w = - change in epe

why does power = +epe / time if power is supposed to equal w / t shouldnt it be equal to -EPE / time

Please help me clarify this.

epe= electric potential energy

and also if current equals to Q/t
is there a negative current if there is a negative charge? isnt current supposed to only go from positive to negative terminal

2. Dec 7, 2014

### Simon Bridge

Welcome to PF;
It can be. The sign of the work is an arbitrary convention.

3. Dec 7, 2014

### sanhuy

thanks,
this still doesnt really make sense to me, so your saying that you do not need to add the negative sign in front of epe?

4. Dec 7, 2014

### DivergentSpectrum

Whoops misread the first part. nvm
I noticed you mentioned current. The "standard" current is kinda backwards from what current actually is, and it is defined by the current of a positive particle going in the opposite direction as the direction electrons are actually going.
But it sorta makes since considering current=dq/dt
if we defined current as if it was negative charges going from negative to positive, then we would have to say
current=-dq/dt

Basically, as long as youre consistent itll be fine.

Last edited: Dec 7, 2014
5. Dec 7, 2014

### Simon Bridge

No - I am saying that what sort of work counts as positive and what negative is a matter of definition.
Basically - if something loses energy, then it has done work, and if it gains it, then it has work done on it.
Which is positive and which negative depends on what you want to pay attention to.
You may be quite happy to say that work is negative when the thing loses energy, i.e. it does work - but we measure a car engine power as positive don't we(?), even though the engine is the thing doing the work?

So what you need to do is think about what the sign convention in each case means.

6. Dec 8, 2014

### sophiecentaur

I think you need to point out that, despite the fact that it is an arbitrary convention, it's a convention that everyone (pretty much) sticks to. Where there is a attraction involved - as with gravity, we (all choose to) say that the potential is negative - hence the expression 'potential well'. Work is 'got out' as the attracted object approaches the attractor. When considering Electrical Charge and Potential, we use a positive charge to work with and when it's attracted to a negative charge, we say the EPE is more and more negative as the test charge approaches the negative attracting charge - another potential well situation (consistent with gravity).
When struggling to get this right, the very last thing to do is to get bothered by electron flow and 'conventional current'. That can be treated as an entirely separate issue and, in no way, affects anything in the study of Electricity - except to annoy students!! No one got anything 'wrong' when they chose the sign to put on the first charged glass rod, in early experiments. Work with positive charges, 'in your head' and, treat the problem of how to actually get hold of a positive charge as a different issue. It is pretty much just as easy to put a positive charge onto an object as to put a negative charge there. The only exception is with electrons, themselves and they happen to be very available and negatively charged - in all thermionic valves and Cathode ray tubes. In wires, which way the electrons drift is virtually of no consequence at all.

7. Dec 8, 2014

### sanhuy

so I did some reading and i found a passage in my textbook

"W=–ΔPE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative ΔPE. There must be a minus sign in front of ΔPE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point." which makes sense.

so if i apply this principle to a proton that is losing potential, i get a negative power. is it possible for a + terminal to have a initial voltage of zero? sorry if this is really basic im doing a intro physics course and i never taken it in highschool and this electricity stuff is very hard to conceptualize.

edit final voltage = x > 0 lol.
and also what does it mean when we have negative power in electricity?

8. Dec 8, 2014

### sophiecentaur

@sanhury
To resolve this I think you need to realise that the sign of the charges is secondary to whether work needs to be put in or comes out of the process. If your calculations tell you that there is "negative power" into a system then that implies that work / energy is coming out of the system - in the case of a proton in a vacuum, the work will be accelerating it and transferring to kinetic energy. That is Negative Power into the electric energy situation and positive power being put into the motion of the proton's KE. Energy is conserved here because there is nowhere else for it to go. If the proton were travelling through a substance (A Hydrogen Ion in a plasma, for instance) it could be colliding with other particles and losing energy in the form of heat.
You wrote "is it possible for a + terminal to have a initial voltage of zero?". Voltage is Potential Difference and always needs to be referred to another point. A "+" terminal can be at +1000V or -1000V relative to Earth Potential but the battery can still only provide a potential difference of 6V (or whatever) - so the resultant PD to Earth, of the "-" terminal would be 994V or -1006V, in those cases.

This can be very confusing, I know but looking over over what you have read in reputable sources again and again and you will get it eventually. Just apply the rules consistently.

9. Dec 8, 2014

### DivergentSpectrum

The reason for reversing the sign is that all forces are inverse square proportional to distance.
The E field is given by q*r-2/(4*pi*E0)
where r=sqrt(x^2+y^2+z^2) and E0 is the permissivity of free space and q is charge
then the indefinite integral of the force with respect to r is
-q*r-1/(4*pi*E0)
notice that the sign of potential depends purely on the sign of q, but the negative sign aquired through integration sorta makes it look messy
so to ensure that the potential due to a positive charge is positive and the potential from a negative charge is negative, we say
V=q*r-1/(4*pi*E0)
notice that if the charge is positive, then potential nearby is positive, while if the charge is negative, the potential nearby is negative.
One of the biggest advantages to this is that high potentials can be visualized as "hills" and low potentials can be visualized as "valleys", and the Electric field lines always point downhill(positive to negative).
You could imagine a positive charge at a high potential "rolling down" to an area of low potential, (vice versa with negative charges), which is actually a pretty good mnemonic, because positive charges will always try to go to low potentials, while negative charges seek out high potentials.

Another interesting fact about potentials: if the potential is "flat" (constant), no motion will occur

Last edited: Dec 8, 2014
10. Dec 9, 2014

### sanhuy

hmm, so if a positive charge is moving towards the negative terminal with the electric field it would mean that it is losing PE therefore work = - EPE, and if a negative charge is moving against the electric field towards the positive terminal it would gain potential energy? therefore W = + Delta EPE

can someone confirm if this concept is right and i should use it for life.

11. Dec 9, 2014

### sophiecentaur

That's far too sweeping as it only applies where there is spherical symmetry. The thing that generates the field is independent of the force caused by that field. The Field is just the differential of the Potential.

12. Dec 9, 2014

### DivergentSpectrum

This may be true for non-classical fields, but assuming the number of field lines at a distance r from the epicenter is related to surface area of a sphere (4*pi*r^2) it is a perfectly good generalization.

to the op: remember also that electric potential is different from potential energy.
electric potential is V (volts)
while the potential energy of a particle with charge q is given by PE=q*V
So, if a negative particle is approaching the positive terminal, it will coorespond to an increase in volts, but deltaV*q=deltaPE and q is negative so potential energy is still lost.
Nothing ever gains potential energy unless there are some other factors such as an initial velocity pointing against the force or 2 forces working against eachother. Nature always prefers to decrease potential energy and increase kinetic energy.

Last edited: Dec 9, 2014
13. Dec 9, 2014

### sophiecentaur

You are assuming spherical symmetry which is a bit limiting for a general discussion about Electrical Potential, inverse square only applies in a very limited number of cases. How would you fit this with a parallel plate capacitor?
In any system, the potential energy cannot change without work being done. That's a given.

14. Dec 9, 2014

### DivergentSpectrum

true, but were talking about point charges here.
If it was any other case, we would just have to integrate force due to point charges over the area where charge exists. The general principle that positive charges seek out low potentials and negative charges seek out high potentials is true for any type of shape though, which i believe was the source of the ops problem

15. Dec 9, 2014

### sophiecentaur

From what I can see if this thread, and the diagram in particular, no one has specified point charges. The issue was to do with work done and potential change.

16. Dec 9, 2014

### sanhuy

so would this be a correct assumption.

17. Dec 10, 2014

### sophiecentaur

Which way is the force, due to the charge in the field, in each case? To move it 'with' the force involves negative work. To move it 'against' the force involves positive work.
Apply that rule to those pictures. Which ones are correct and which are not, according to the above rule?

18. Dec 10, 2014

### sanhuy

Both of them are correct right?
but if we say the ΔV = +2500v for both the pictures and use concervation of energy
we would get
KEo + PEo = KEf + PEf
0 + qVo = ½mv2 + qVf
-qΔV = ½mv2
and if you solve for velocity

V = √-(2(q)(2500V)/M))

which would not work for a positive charge but would for a negative

so im kind of unsure of these rules.

19. Dec 10, 2014

### sophiecentaur

Ask yourself is this consistent with the idea that when the charge moves against the attractive force you do positive work on it and when it moves with the force you do negative work. If you stick to that way of viewing it, can you go wrong?

20. Dec 10, 2014

### nasu

The potential energy decreases in both cases so the KE is positive in both cases. It does not matter what is the sign of the charge. The field is accelerating the charge in both cases. For electron the potential energy decreases when it goes against the field lines. So the work is positive and the KE increases.
The work (done by the field) is the negative of change in potential energy by definition. Your second drawing has the sign wrong.

21. Dec 10, 2014

### sophiecentaur

Forget about kinetic energy and acceleration. They are of secondary importance or negligible. What counts is the work done. If a charge is moved through a wire in a generator or in a resistor by a battery, the KE of the charges is virtually Zero. The only thing of importance here is the Work done on the charge in question. If it goes against the way the field is acting, then the work done is positive. If it goes with the field, the work is negative (i.e. work is got out. How is it possible that he is still getting this wrong? It's because of all the incidental stuff he's trying to bring into it. And KE is another bit of incidental stuff that is clouding the basic issue of work and potential.

22. Dec 10, 2014

### nasu

His drawing shows just a charge in uniform field. No wire, no generator, just the field and the charge. I was referring to that drawing and nothing else.
That charge gains KE in both cases, so this shows that the work is positive in both cases.
The electrons in a wire are a much more complicated case and maybe this should not be brought in the discussion yet.
But you never know what will make someone finally understand. :)

23. Dec 11, 2014

### sophiecentaur

Hang on.When Energy comes out of it, the potential change is Negative as negative work is done on the charge. Energy is conserved and KE (with a v squared in it) is always positive, so the PE change must be Negative when a charge moved under attraction. That is how potential is defined by convention. The + terminal of a 6V battery means that it takes +5 Joules to take a Coulomb of positive charge from the Negative terminal to the positive terminal.
There is too much in the way of personal hand waving and 'description' in this thread and not enough emphasis on the 'rules' that the conventions give you.
All of those pictures show situations where a charge is 'pulled' in a particular direction by the Volts on the plates. They all involve negative potential change.
It is only when you put Work into the system (with a string or a rod, for instance) that you can do work On the system and increase the Potential Energy in it by moving the charge by some external agent.

We learned the rules of Arithmetic about how signs work and the various sign conventions that we use in Maths. I cannot imagine a lengthy thread questioning why 'A minus times a minus gives a plus' and this is essentially what people are doing here. Perhaps the problem here is in not treating the Force and the Displacement as vectors because the result of a Dot Product ( W = F.D) would not be wrong.

24. Dec 12, 2014

### nasu

I am confused. You don't agree with the statement in blue?

25. Dec 12, 2014

### sophiecentaur

I may be misunderstanding what you are trying to say but, to my interpretation of those diagrams, the charge seems to be moving in the direction that the Field 'wants to take it' (sorry for anthropomorphising; I hate it usually). That means Negative Work needs to be done on the charge. If you are talking about work done on the charge by an external agent to move the charge (and that is the convention of Work and Potential) then the signs are as I have said.

I think you are getting too preoccupied by the signs of the charges and the Electric Potential at a position and not the Work situation. This is why I objected to bringing KE into this discussion because it is outside the question. We all know that PE + KE is constant in a conservative system so, if the KE would increase if you let the charge go, then the PE (W) change must be negative. If you use the signs correctly, you will always get the right answer for the sign of the Work.

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