# Difference between W/Q and E/Q in Electricity

1. Apr 3, 2013

### CraigH

In the Electricity section of the AQA A Level Physics data sheet (link given bellow) it includes these 3 equations:
http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-PHYA5-W-A2-DFB.PDF

V=W/Q
Emf = E/Q
Emf = I(R+r)

I understand that Emf and Voltage both have the unit Volts, and Work Done and Energy both have the unit Joules, so V=W/Q and Emf = E/Q are basically the same equation.

And I also understand that it is convention to call the voltage across a battery V, and the voltage it would theoretically have across it if it had no internal resistance emf. That is why Emf = I(R+r). IR would be the voltage across the battery, (or whole circuit) and Ir would be the voltage across the internal resistance of the battery.

But why is V=W/Q and Emf = E/Q?

Thanks

2. Apr 3, 2013

### sophiecentaur

The first two equations are equivalent. The third equation involves a different set of variables so why are you surprised that it can still be correct?
V (potential difference) is defined as the ratio of the work done (or energy out) per unit of charge. (Joules per Coulomb)
R (resistance) is defined as the ratio V/I so the equation is right 'by definition'. ( that's Joule Seconds per Coulomb squared - if you like)

There is no disagreement and nothing is wrong here.

3. Apr 3, 2013

### CraigH

I shouldn't have really mentioned the Emf = I(R+r) equation. This doesn't confuse me, I just included it because it was grouped with the V=W/Q and Emf = E/Q equations in the data sheet.
What confuses me is the fact that they use work done for voltage and Energy for Emf. I know they are both the same thing but I thought it seemed strange to use two different symbols for the same thing when they are right next to each other on the same sheet.
Why did the exam board feel the need to tell me that voltage is defined using work done but Emf is defined using energy?
Is there some subtle difference or connotation implied when you use the word "work done" instead of "energy", and why is voltage defined with "work done" and emf defined with "energy"?

4. Apr 3, 2013

### sophiecentaur

The answer to that one is simple. Don't rely on the formula sheet. In my day, they didn't exist and it never did me no 'arm. Learn the formula you need and understand when and how to apply them.
How the two quantities are described is, I think, more a matter of fashion. They are both Joules per Coulomb. Don't let it bother you. You should never get too aerated about those sort of 'classification' issues; it just makes one needlessly anxious.

5. Apr 3, 2013

### mikeph

Good question... I remember being taught that the emf is the work done ON a charge, E = W/Q, which is equal to a coulomb of charge's energy increment when moving across the battery. So I don't know why you'd use one over the other.

The Voltage is defined as the potential difference across the outer terminals, which is usually defined as the energy expended per coulomb of charge as it goes around the circuit.

I have no idea why they use one and not the other. Never liked those exam sheets, endless letters and equations with no context whatsoever, no doubt the students learn very good memorisation skills in their physics classroom.

6. Apr 3, 2013

### MalachiK

In the Unit 1 exam you could very well be asked to write down the definition of p.d. or e.m.f. If that happens, think about this thread as you have to specify if the energy is being transferred in or out of the circuit to get the mark.

7. Apr 3, 2013

### CraigH

MikeyW and sophiecentaur, I also don't really like the formula sheet. If you understand the equations you don't need to memorise them, so yes the sheet isn't necessary. I'm just using it as a revision tool at the moment, I'm going through each equation, and using them to remind me of the theory taught around that equation. Some of the theory is never taught though, an equation is given and the student is expected to memorise it, so they can just plug the numbers in, and get a number out with no context. Stefans law is an example of one of these equations. I just presume that the maths behind it is too complicated for this level.

MalachiK, I think this may be the answer I was looking for. when talking about emf no work is actually being done it is just a potential energy. But with a voltage you are often talking about a voltage drop, where the reason for that voltage drop is that electrical potential energy has been converted into some other energy.

To be honest I'm not actually doing A levels, I'm helping my younger brother revise for his A levels. I'm an undergrad engineer and I should really know this stuff, I just forgot ;)

Thanks everyone!

Last edited: Apr 3, 2013
8. Apr 4, 2013

### technician

I think it is a simple way to differentiate between energy SUPPLIED from the source of energy (the EMF)
And the energy DISSIPATED in the resistors .
EMF and POTENTIAL DIFFERENCE are forms of energy but there is a subtle difference.
You need to know the definition of EMF and POTENTIAL DIFFERENCE for your exams. Check your text book definitions.
Get them mixed up and you will lose marks.
I agree with what Malachik says

9. Apr 4, 2013

### MalachiK

The definitions that your brother will want for the exam is that....

p.d. is the energy transferred out of the circuit by a component per unit of charge that passes through it

e.m.f. is the energy transferred into the circuit by a power supply per unit of charge that passes though it

In fact it's worth getting the rhythm of those types of definitions learnt by heart as they mostly follow the same pattern. Stress, strain, density etc. all want a blah blah blah per unit blah blah type response.

10. Apr 4, 2013

### sophiecentaur

I have seen many mark schemes. To allow easy marking criteria, they give full marks for a correct formula with the variables given a reasonable description. They don't get at all philosophical or pernickety.

11. Apr 4, 2013

### MalachiK

Err, I wasn't aware that I was being either. But the AQA mark schemes are quite keen on having the energy going in the correct direction.

12. Apr 4, 2013

### CraigH

After speaking with my brother he has told me that MalachiK's definitions seem very familiar, and he recognises them as being very similar to what his teacher taught him.

13. Apr 4, 2013

### technician

Agree totally with MalachiK. Definitions are there for a reason and it is part of physics vocabulary and grammar to get them correct. Just like in language. Sloppy use of the words causes confusion and leads to lack of understanding.
EMF is not Potential difference in the same way that speed is not velocity, how often have you seen these misused?

14. Apr 4, 2013

### sophiecentaur

No, of course not. My point was that very few (if any) questions are actually asked in those terms. The questions tend to lead you by the nose and ask for simple descriptions and the application of the appropriate formula (of course, you don't even need to remember it). The mark schemes will usually require key steps and key words on the way through solving a problem and I do not remember anyone being penalised for the way they put things. They need to provide questions and mark schemes that allow the marker to get through a script in a very short time.
I am not saying that I approve of the system - I don't even approve of the whole structure, these days. There just is not time to get to any appreciable depth of sophistication between starting in September and your first possible Module Exams in January - so they have to make the early modules very superficial - like learning the combinations of (totally not-understood) attributes of fundamental particles before students have an idea of the meaning of momentum or an electron volt. They may as well be learning the names of a Premier League Football team for all the real Science they can learn. The lovely Mr. Gove may be a heartless fascist but he has certainly put his finger correctly on the re-sit culture as a serious weak point.

But Potential Difference is difference in potential ( as the name suggests) and Electrical Potential is defined totally in terms of work and charge. The distinction between emf and 'measured PD' is just a practical one to acknowledge that a real electrical power source will include a loss mechanism. An ideal voltmeter will not distinguish between the two. There is no parallel with velocity and speed where the quantities are fundamentally different - vector vs scalar.

15. Apr 4, 2013

### MalachiK

Spot on!

This may very well be the most perfectly righteous thing I've ever read. I couldn't agree with you more. But every time I suggest to my colleagues that time spent rote learning Feynman diagrams would be better spent working on mechanics / thermal physics / gas laws / anything classical that the students have a hope of gaining an authentic understanding of and which anyway involves practising mathematics.... I get looked at as though I'm funny in the head. Apparently the gas laws aren't cool any more.

And it's not even as though they make them learn the real Feynman diagrams. The AQA spec has the anti-fermion lines pointing the wrong way because (and I've asked the chief examiner this very question) it simplifies things! I ask you! If it's all just mindless rote learning what the hell difference does it make which way they learn to draw the bloody arrows? We might as well mindlessly learn things the right way around if we're going to learn anything.

16. Apr 5, 2013

### technician

The distinction between emf and 'measured PD' is just a practical one to acknowledge that a real electrical power source will include a loss mechanism. An ideal voltmeter will not distinguish between the two. There is no parallel with velocity and speed where the quantities are fundamentally different - vector vs scalar.

I think the distinction between emf and potential difference is more than just a practical one.
emf relates to the source of energy (chemicals in a cell) and potential difference relates to the dissipation of energy (resistors). The difference is subtle and both are measured in the same units.
I dont think it would be correct to refer to the emf across a resistor or the potential difference of a cell (I know there is a terminal potential difference!!!)

The difference between speed and velocity is well known and it is essential that the terms are used correctly in physics.
Current is represented by arrows and + and - signs are used when referring to 'voltages'.....there is something of a 'vector' nature here. (hesitate to mention AC !!!!)
How these terms are used in chit chat and expressing opinions is another matter.

I dont suppose it makes no difference to nothing in the long run.
Dumb everything down. Dont learn nothing important. Dont bother checking text books and exam questions.

Last edited: Apr 5, 2013
17. Apr 5, 2013

### sophiecentaur

OK. You have an ideal voltmeter and one chance at a measurement across 'a pair' of terminals. What will it measure?
"Important" can be subjective and a personal assessment of yours. PD is PD, and involves Potential - which is the basis for all of it and which is perfectly well defined and doesn't depend on the path taken in a conservative field. Potential is a scalar. AC Phasors are a mathematical tool and not relevant to this.
See post #15 for some more comments on the A level system. All my responses here have been in the form of pragmatic advice to an AS student. Sufficient unto the day is the evil thereof.

18. Apr 5, 2013

### technician

Post #15 is exactly what you state: comments (opinions) about A level teaching and there are as many opinions as there are opinion providers. (students want advice and want their confidence building, not academic opinions)
Sections 4.2 to 5.3 in the AQA text book make it perfectly clear what is meant by Potential difference and emf.
A common question goes something like this:
Explain what is meant by an emf of 12V....2 marks are awarded for this; 1 for recognising that Voltage is a measure of energy and 1 for recognising whether energy is being supplied or dissipated.
(sometimes emf is replaced by potential difference in the question)
There is a subtle difference in the link between current and emf and current and potential difference.
It is essential that students recognise the physics behind this.

19. Apr 5, 2013

### vanhees71

Well, physics at school is a dangerous topic, and I'm not sure whether this is the right subforum for it, but also I couldn't agree more with sopiecentaur in. I'm from Germany, and also in our highschools we have an awful coverage of science and (even worse unfortanately!) math!

You are perfectly right in saying that one first has to get the basic foundations on classical physics (classical mechanics, electromagnetism, thermodynamics and statistics) right, before one can hope to give some insight into quantum physics. On the other hand, of course, you also have to keep in mind to provide the excitement of science to highschool pupils, and unfortunately classical physics doesn't draw their attential usually. Usually, they want to know about cosmology, particle physics (the Higgs boson of course first!), and so on.

I'm not a highschool teacher. So, I cannot say, how to solve this trouble, but I think you can use the "hype topics", which you can explain on a more qualitative level first, to motivate the students to learn also about the "old physics", in order to provide the necessary foundations to really understand "modern physics".

Then there are some age-old nogos in teaching physics at schools, which are hard to get rid of, especially when talking to teachers. Some of those are:

(1) The photoelectric effect demonstrates the quantization of the electromagnetic field and the existence of photons. This is wrong! The photoelectric effect demonstrates the quantum features of electrons only. The Einstein formula is derived from the treatment of the interaction of the quantized bound electron with a classical electromagnetic wave in time-dependent perturbation theory.

(2) Using Bohr's model for atoms to introduce quantum theory of particles. This is not only quantitatively wrong (except by accident for the energy levels of the hydrogen atom) but even provides a wrong qualitative picture of the atom. A hydrogen atom is not a tiny disk but a sphere, and the idea of classical orbits of electrons around the nucleus is inapplicable as we know from the correct "new quantum theory" (new is also a bit misleading; after all it's nearly 90 years old :-)).

(3) "Relativistic mass". This notion is superfluous and misleading. It's not plain wrong, but nowadays the mass of an object is its rest mass (or equivalent the "invariant mass"), while energy is named energy and not mass times $c^2$. The correct relationship between mass and momentum of a (free) particle is $E=\sqrt{p^2 c^2+m^2 c^4}$, and that's the temporal component of a four-vector while the invariant mass is a scalar. Of course, most probably you cannot teach the covariant tensor formalism at high school, but keeping this in mind helps.

(4) Somewhat more ontopic in this thread: The one and only correct electromotive force in the integral form of Faraday's law is given by
$$\mathscr{E}=\int_{\partial A} \mathrm{d} \vec{r} (\vec{E}+\vec{v} \times \vec{B}).$$
Here, $\partial A$ is the boundary curve of an arbitrary surface, oriented in relation to the surface's normal vectors according to the right-hand rule and in the most general case time dependent, $\vec{E}$ and $\vec{B}$ are the electric and magnetic components of the electromagnetic field wrt. to the reference frame, where the calculation is done, and $\vec{v}=\vec{v}(t,\vec{r})$ is the velocity of the line element at time $t$ and location $\vec{r}$ as measured in this very reference frame. Then and only then you can write Faraday's Law in the integral form as
$$\frac{\mathrm{d} \Phi}{\mathrm{d} t} = -\mathscr{E}$$
with the magnetic flux defined, again in the same reference frame, by
$$\Phi=\int_A \mathrm{d} \vec{A} \cdot \vec{B}.$$
Closely related with this often wrongly stated integral form of Faraday's law are phenomena like the homopolar generator, which are easily solved when referring to the above given correct integral form.

20. Apr 5, 2013

### CraigH

As a student who finished A level physics not long ago I can say that it isn't that bad. Its not all memorizing facts, they do teach important science techniques the value of thinking about problems "from first principles", however some of the maths needed to fully understand certain topics just cannot be taught in the 2 years you spend studying that subject.

This is actually what I am studying at the moment! I do think that this is too advanced for an A level student though. When I started undergrad engineering it took me ages to understand why integrals and derivative's suddenly started to appear in equations I've previously learnt, e.g
$W=F*d*cosθ$ becomes
$W=\int \underline{F}.\underline{x}$

It seems obvious now but through all of A level an integral was just an area under a graph, and it confused me when it was used anywhere but in pure core maths. However putting calculus and vector calculus into A level physics would scare all of the students off, they would drop out and very few would even get in to undergrad physics and engineering.