# Difference between emf and potential difference

• cupid.callin

#### cupid.callin

I read this topic in many books but i really can't figure out what really the difference is!

One of the book differentiated them as follows ...

emf is the work done by the battery force (Fb) in moving unit charge from one electrode to another.
emf is due to the battery force which is not electrostatic and due to some battery mechanism. therefore it do not decrease with time when battery is in use.
while potential difference is due to the charges on 2 electrodes. as battery is used, charges on the 2 electrodes decrease and therefore potential difference decrease with time,

When battery is in open circuit...

Fb = Fe (electrostatic force due to charges on electrodes)
Fb = qE
Fb (d) = qEd
Wb = qEd

Wb/q = V

emf = V

Now what i really need to know is that if after using battery even for 1 sec emf is not = V (except for ideal battery which obviously don't exist! WHY DO WE DEFINE EMF? WhAT IS THE NEED TO DEFINE IT!

i can't explain my exact question as even i have no idea what i want to ask ... please explain me anything about emf and V ... anything that do not raise more questions in my mind!

If it genuinely states what you have quoted I would abandon this book. EMF is not a measure of work or energy.

Both EMF and PD are measured in volts.
An EMF is an energy source. A PD is not.
An EMF can introduce energy to a system, a PD cannot.

I note from your other posts you understand the concept of an electric field around a charge.

Consider a single isolated charge. It creates (induces) a surrounding electric field that varies with distance from the charge.
The difference between the electric potential at two points, A and B, is measured in volts and called the Potential Difference.
However if you connected a resistor between points A and B, no current would flow.

Now consider you battery, with the same voltage across its terminals as the PD between A and B above.

If you connect the same resistor across these terminals a current will flow, as defined by Ohms law.

The battery is introducing energy into the system, ie it is acting as an energy source. The rate of energy introduction (flow) is given by the product of the current and the EMF

Watts = Volts x Amps = (Volts)2/ Resistance

Since for a given system (resistor) the current is fixed by the resistor we say that the EMF is a measure of the energy available to that system.

Hope this helps.

Hi Studiot!

I don't mind admitting I get very confused by emf.

Is this right?
Consider a single isolated charge. It creates (induces) a surrounding electric field that varies with distance from the charge.
The difference between the electric potential at two points, A and B, is measured in volts and called the Potential Difference.
However if you connected a resistor between points A and B, no current would flow.

If there's a conductor connecting A and B, won't a charge opposite to the isolated charge build up on the nearer end, and an equal charge opposing it on the further end?

So while it's building up, isn't there a current, which gradually diminishes because of the gradually increasing opposing emf of the isolated charge of AB itself?

Hello Tiny-Tim, that is an interesting proposition,

So while it's building up, isn't there a current, which gradually diminishes because of the gradually increasing opposing emf of the isolated charge of AB itself?

Such a current in my resistor (remember I specified a resistor?) would dissipate ohmic heating energy.

Where would this energy come from?

Would it diminish the charge supplying the field?

Could I reduce said charge to zero by introducing many resistors?

If it didn't diminish the charge, could I use the effect to establish a perpetual motion (heating) machine by introducing resistor after resistor?

Well I'm sort of offering the reasoning behind my original statement.

If there's a conductor connecting A and B, won't a charge opposite to the isolated charge build up on the nearer end, and an equal charge opposing it on the further end?

And doesn't this describe a capacitor?

anything that do not raise more questions in my mind

I am trying to bear this in mind and avoid making things too complicated. I am also trying to avoid Kirchoff and the KVL controversy.

An electric field cannot exist inside a conductor, so the introduction of a conductor 'shorts' the points A and B to the same potential. An electric field can only exist in a dielectric medium.

The main point is that

an EMF introduces energy into a circuit.

a PD dissipates energy in a circuit.

And doesn't this describe a capacitor?

No, a capacitor has a gap … this conductor is continuous from A to B.
An electric field cannot exist inside a conductor, so the introduction of a conductor 'shorts' the points A and B to the same potential. An electric field can only exist in a dielectric medium.

No, a field can exist inside a conductor … it takes time to short A and B to the same potential, and the field exists until the current has had time to build up an emf opposing it … from Scott Hughes' 2005 MIT lecture http://web.mit.edu/sahughes/www/8.022/lec05.pdf" …
There is no electric ﬁeld inside a conductor. Why? Suppose we bring a plus charge near a conductor. For a very short moment, there will be an electric ﬁeld inside the conductor. However, this ﬁeld will act on and move the electrons, which are free to move about. The electrons will move close to the plus charge, leaving net positive charge behind. The conductor’s charges will continue to move until the “external” E-ﬁeld is canceled out — at that point there is no longer an E-ﬁeld to move them, so they stay still.

A more accurate statement of this rule is “After a very short time, there is no electric ﬁeld inside a conductor”. How short a time is it? Recall that in cgs units, resistivity (which tells us how good/bad something conducts electricity) is measured in seconds. It turns out that the time it takes for the charges to rearrange themselves to cancel out the external E-ﬁeld is just about equal to this resistivity. For metals, this is a time that is something like 10−16 − 10−17 seconds. This is so short that we can hardly complain that the original statement isn’t precise enough!​

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Hi,

Thanks for the replies!

actually my book defines emf approximately as wikipedia:

In physics, electromotive force, or most commonly emf (seldom capitalized), or (occasionally) electromotance is "that which tends to cause current (actual electrons and ions) to flow."

More formally, emf is the external work expended per unit of charge to produce an electric potential difference across two open-circuited terminals.[2][3] The electric potential difference is created by separating positive and negative charges, thereby generating an electric field.[4][5] The created electrical potential difference drives current flow if a circuit is attached to the source of emf. When current flows, however, the voltage across the terminals of the source of emf is no longer the open-circuit value, due to voltage drops inside the device due to its internal resistance.

And well i agree with tiny-tim that a single charge will also cause current in wire, the time of current depending on the magnitude of charge ...
and i guess this might not work for the resistor ... but still ... Studiot said that VD can't provide current, which i suppose is wrong!

An EMF is an energy source. A PD is not.

emf don't provide energy to the circuit, it just charges the battery

And the charges accumulated provide PD which provide energy to the battery,,, i suppose!

an EMF introduces energy into a circuit.

a PD dissipates energy in a circuit.

Actually an emf creates energy that can be utilized by the PD to provide current and heating...

and conductor AB is not comparable to a capacitor, if you say that charge cannot provide current to resistor, how can it provide current to capacitor ,,, capacitor is (in a way) like an open circuit(unless we use a dielectric) of infinite resistance ... and the energy lost in charging a capacitor is more that making a current through a resistor...

@ tiny tim

can you explain me how cgs unit of resistivity comes out to be s?
you may use this relation: ρ = 2m/nte^2

t is the average time b/w two collisions of electron of material.

In physics, electromotive force, or most commonly emf (seldom capitalized), or (occasionally) electromotance is "that which tends to cause current (actual electrons and ions) to flow."
More formally, emf is the external work expended per unit of charge to produce an electric potential difference across two open-circuited terminals...The created electrical potential difference drives current flow if a circuit is attached to the source of emf. ...
Devices that can provide emf include voltaic cells, ..."A source of emf can be thought of as a kind of charge pump that acts to move positive charge from a point of low potential through its interior to a point of high potential. … By chemical, mechanical or other means, the source of emf performs work dW on that charge to move it to the high potential terminal. The emf ℰ of the source is defined as the work dW done per charge dq: ℰ = dW/dq."
So states Wikipedia..

What is stated is that emf can cause Potential difference, and the potential difference can induce an emf.

1. if you induce an emf into a conductor, the emf causes charge to flow and align away from each other(seperates +ve from -ve). Hence the term workdone is also used. The effect of the emf is Voltage or potential difference.
2. If there exist 2 conductors with different potentials such that their difference in potential is V, then when a third neutral conductor or resistor is placed to connect them, an emf is induced in the third conductor by the potential difference of the earlier 2 conductors. this emf causes the charges to flow.

Hope it enabled you to understand the relationship of the emf and Potential Difference and the difference in the usage of the term. There is no difference in emf and Potential difference, the discretion is in the usage of the term, depending on what we are trying to talk about.

But better still, take a look at this example.. this kinda states it clearly and also why emf is associated with sources and potential diffences with sources and sinks.
Hope this makes some sense if not all... take care...

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Since you are all adamant that a current flows in my test resistor would anyone like to explain why this does not contravene the law of conservation of energy.

If what you say is true, I can extract energy indefinitely (as heat) from the charge by repeating the test, each time causing an (albeit small) current to flow through the resistor.

More formally, emf is the external work expended per unit of charge to produce an electric potential difference across two open-circuited terminals...

Voltage is not a unit of energy and EMF is measured in volts, so how can you define it as work?

emf (seldom capitalized)

EMF is usually capitalised and even afforded its own symbol a copperplate script capital E.

2. If there exist 2 conductors with different potentials such that their difference in potential is V, then when a third neutral conductor or resistor is placed to connect them, an emf is induced in the third conductor by the potential difference of the earlier 2 conductors. this emf causes the charges to flow.

EMFs are not 'induced'

No, a capacitor has a gap … this conductor is continuous from A to B.

There is a gap between A and B. It is filled with a dielectric medium (free space) and a capacitance exists even in this condition.

You are suggesting that a resistor can exist with a potential difference between its ends and pass no current in a steady state condition. Your temporary current is a transient.

What is the set-up?

If we're bringing a resistor AB near a static charge, then away, then near, and so on, then each time we move the resistor near, we have to do a tiny amount of work (against the resistance for a tiny amount of time) to get it there … that works ends up as heat.

If we're bringing a resistor AB near a static charge, then away, then near, and so on, then each time we move the resistor near, we have to do a tiny amount of work (against the resistance

What work is this?

We could also leave the resistor where it is and switch the charge on and off (via electrostatic shielding).

College Physics - Google Books Result
Raymond A. Serway, Chris Vuille, Jerry S. Faughn - 2008 - Science - 1120 pages
Further, because the electric field inside a conductor is zero, no work is required to move a charge between two points inside the conductor. ...books.google.co.uk/books?isbn=0495386936...

My underlining

http://books.google.co.uk/books?id=...a conductor in an electrostatic field&f=false

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EMF is usually capitalised and even afforded its own symbol a copperplate script capital E.

EMFs are not 'induced'

I know i aint smart nd neither is my english, so please excuse me on my follies... It is not that I know more, so I don't accuse of ur anger originating from jealousy. .. ;)

Well jokes apart.. I just found these references about emf also being induced.. (emf is induce-able) .. So if you would like to know more of induced emfs then please read the following.. And anyways the definition of emf being not written in capitals is not mine but of wiki...
http://physics.bu.edu/~duffy/PY106/InducedEMF.html
http://en.wikipedia.org/wiki/Electromotive_force
http://demonstrations.wolfram.com/InducedEMFThroughAWire/
http://physics.bu.edu/~duffy/PY106/InducedEMF.html

A lot of applications, principles and development owes itself to the fact that emfs can be induced...

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You are suggesting that a resistor can exist with a potential difference between its ends and pass no current in a steady state condition. Your temporary current is a transient.

NO, I'm never suggested a steady state current. From my first post, I was suggesting …
If there's a conductor connecting A and B, won't a charge opposite to the isolated charge build up on the nearer end, and an equal charge opposing it on the further end?

So while it's building up, isn't there a current, which gradually diminishes because of the gradually increasing opposing emf of the isolated charge of AB itself?
What work is this?

The work done moving an increasingly bipolar material through an electric field
We could also leave the resistor where it is and switch the charge on and off (via electrostatic shielding).

That just makes identifying the work more complicated … doesn't placing shielding inevitably do work against the field?

(Did you have a particular shielding set-up in mind?)

What magnetic effects are we talking about here vish_al210?

please explain me anything about emf and V ... anything that do not raise more questions in my mind!
Simply put, the potential difference of any voltage source can also be termed as emf as it causes charge to flow. The potential difference across any other passive components cannot be termed as emf as it is not the cause of the current flow.

NO, I'm never suggested a steady state current. From my first post, I was suggesting …

What is this if not a transient current, which results in the test resistor having a potential difference across its end permanently so long as it is in place? I didn't say a steady state current I (and you) said a steady state voltage. I am asking how you can have a steady state voltage across a resistor without a current then flowing?

If there's a conductor connecting A and B, won't a charge opposite to the isolated charge build up on the nearer end, and an equal charge opposing it on the further end?

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Simply put, the potential difference of any voltage source can also be termed as emf as it causes charge to flow. The potential difference across any other passive components cannot be termed as emf as it is not the cause of the current flow.

Excellent rephrasing of what I said.

sorry if I appear to be rephrasing dude... I didn't notice... anyways.. this is what I had posted earlier as well... anyways it is not about who but about what..
So if it makes u feel any better,
""If anything I posted is simillar to what'Studiot' posted, please take it as me seconding his opinion...""

Well even in a transformer the voltage in the secondary is an induced emf...
take care...

E.M.F. snd P.D. are both measured in volts and both can be defined in terms of energy as in accordance with the equation:
Volts=Joules/Coulombs
The difference is that E.M.F. refers to the supply and the total voltage(voltage across whole external circuit plus voltage lost across internal resistance of supply) whereas pd refers to the voltage across specific points in the circuit.We can make formal definitions.
Examples:
1.If a supply has an E.M.F.equal to E then in passing a charge Q round a circuit connected to it ,it transfers an amount of elecrical energy equal to QE.
2.The emf of a supply is the ratio of the electrical power which it generates to the current which it delivers.

You are suggesting that a resistor can exist with a potential difference between its ends and pass no current in a steady state condition. Your temporary current is a transient.
NO, I'm never suggested a steady state current. From my first post, I was suggesting …
If there's a conductor connecting A and B, won't a charge opposite to the isolated charge build up on the nearer end, and an equal charge opposing it on the further end?

So while it's building up, isn't there a current, which gradually diminishes because of the gradually increasing opposing emf of the isolated charge of AB itself?
What is this if not a transient current, which results in the test resistor having a potential difference across its end permanently so long as it is in place?

(I assume by "transient" you mean "temporary"?)

Yes, it's a a temporary current, which results in a displacement of charges within the resistor.

But no, there's no potential difference once the movement has stopped … the p.d. from the applied field is cancelled by the p.d. of the field due to the separated charges …

what else can cancel it?

total p.d. zero?​
I didn't say a stedy state current I (and you) said a steady state voltage. I am asking how you can have a steady state voltage across a resistor without a current then flowing?

Nobody mentioned "steady" until you did in post #13. I certainly made it clear from the beginning that it wasn't steady, using the phrases "while it's building up" and "gradually diminishes".

Hello Tiny Tim, I think our discussion is getting off topic for this thread and would better be continued elsewhere. My fault as much as anything as I introduced the herring.

Thank you to others for pulling it back.

So a resistor near a charge do not show any change… but if we move it back or forth or turn charge on and off, it will show some instantaneous current and voltage??

And studiot, a capacitor with dielectric and a conductor are not same.
Dielectric is a material with polar molecules which just align when electric field is applied. They don’t conduct like conductors as there are no free electrons, just polar molecules.

And wiki defines EMF as work “per unit charge” not work. Check V and W/q dimensionally … they match!

And please tell me one thing … ““ Iin case of battery, EMF is the charging force or not? ””

And guys you are welcome to continue your discussion here!

Everyone can use some extra knowledge!

And please tell me one thing … ““ Iin case of battery, EMF is the charging force or not? ””

Well cupid,
The potential difference in the battery was developed (considering a rechargeable battery) due to the emf of the charging ckt. Now the voltage created in the battery generates an emf in the ckt that it is connected to. But the emf ceases to exist when the ckt is removed or opened. Then it is only potential difference of the battery.

And there is no current induced or acted when u hold a conductor/resistor/inductor to one end of the battery alone, the potential is raised, but no consumption of power takes place(atleast not the least bit significant..). the current flow happens only when the ckt is closed.

And yes... even if you connect that resistor//conductor//inductor to one end of an AC Voltage source,... still no current//power is consumed...

So you don't agree with tiny-tim that by moving a resistor we can produce current in a resistor ... because his explanation sounds very satisfactory to me!

I am not disagreeing with him or anyone else as the context he is proposing the theory in, is different... It is a supposition..
but if you connect one end of the ac mains to one end of an ammeter and the other end of the ammeter to one end of a resistor, leaving the other end of the resistor and the other end of the AC mains open, you would find that no current flowed through the ckt. and that no real power was consumed...
I am also perplexed by this.. but reality s___s...It never let's our million dollar plans realize...
Or else regardless of you leaving a switch open or in off condition, the power would still be consumed... and that would not be a good thing for ur bill and the energy conservationists...

"Reality is generally fluid...until we fall face first on it... then we reaise that it actually is rock hard..., alas realisation is a little too late always..." by me...

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Tiny Tim you are correct, many thanks for pointing me in the right direction.

Here is a scanned extract.

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Thanks, Studiot.

I still get really confused about emf

people keep saying that there's a difference between emf and pd, but I've never managed to grasp it.

I have been doing some calculations, based upon what you said, about moving a dipole from infinity to A_B in the field of an isolated charge.

I have no time tonight as it is past the witching hour here, but am quite clear about the difference between the terms. I was trying to be too clever introducing potential fields and tripped myself up.

The difference really has nothing to do with fields. I will post again.