#### [SOLVED] Potential difference (PD) and electromotive force (EMF) in terms of electrons?

I am trying to understand potential difference and electromotive force in terms of electrons.

In the above 3 circuit diagrams the red shows (what I think to be) the distribution of electrons around the circuit. (on reflection I do not think there would be more electrons at B, F and L as indicated on the diagram but rather an even distribution of electrons between B and D, F and H and J, and L and P respectively).

I have several parts to by question:

1. Is emf simply the potential difference between A and B (just focusing on the first diagram for now) i.e. if you sum all the $\frac{e}{4\pi \varepsilon_0 r}$ from all the electrons for position A and then for B and subtract them.
2. On the same principle is the pd across the resistor the same as above expect done for C and D.
3. Do these diagrams seem to be write or have I got totally the wrong idea.

#### @DIYser 2015-11-08 21:30:03

1. Emf is the force at the terminals & within the circuit. It is proportional to the energy stored in the cell & it varies throughout a circuit depending on how the energy moves through the circuit components (higher resistance components dissipate more energy within them--which is what causes a proportional voltage drop (higher Vd with higher resistance components & lower Vd with lower resistance components)).

2. Unknown - please restate your question in definitive terms.

3. Your diagrams are fine. They represent one way of visualizing certain fundamentals, however, electrons actually move quite slowly along the path of a circuit. What is actually moving is energy from electron to electron--like a wave in a pond when a dropped pebble in the center of the pond causes a force that dissipates the water--the energy is what causes waves of water to eventually crash against the bank of the pond--the water molecules (and their electrons) nearest the pebble do not wash up on the bank.)

#### @phd physics 2015-09-20 11:52:01

The charge is the same on either side of the two points, but the energy that the charges have is different.

#### @Alfred Centauri 2014-08-08 21:13:52

The distinction between emf and potential difference is often glossed over and often misunderstood so this is an appropriate and interesting question.

Since both emf and potential difference are measured in volts, it is quite easy to use the terms interchangeably and, in many cases, there's no harm done but that fact is that emf and potential difference are distinctly different concepts.

Essentially, it is the emf that moves charge around a closed path - a circuit - while potential difference is due to the charge distribution along the circuit.

Consider the first circuit; there is a battery (or cell actually), the source of emf. If there is no circuit connected to the battery, the potential difference between the battery terminals is equal in magnitude to the emf.

From the Wikipedia article "Electromotive force":

In the case of an open circuit, the electric charge that has been separated by the mechanism generating the emf creates an electric field opposing the separation mechanism. For example, the chemical reaction in a voltaic cell stops when the opposing electric field at each electrode is strong enough to arrest the reactions.

The electric charge that has been separated creates an electric potential difference that can be measured with a voltmeter between the terminals of the device. The magnitude of the emf for the battery (or other source) is the value of this 'open circuit' voltage.

If an external circuit (path for charge to flow) is connected to the emf source, electrons from the more negative terminal will flow through the circuit to the more positive terminal.

There is no better demonstration of the distinction between emf and potential difference than to place a short circuit across the terminals of the battery (this must be done carefully and quickly) to determine the short-circuit current.

Assuming an effectively ideal short circuit, the potential difference between the battery terminals will be effectively 0V yet there will be a large current through the short-circuit.

It is the emf that 'drives' this current round the closed path.

Now, once again consider the first circuit. Clearly, the emf is driving the electrons clockwise but it is also clear that, for there to be a current through the resistor, there must be a potential difference (Ohm's law). Thus, the electron density is slightly greater on the conductor connected to the more negative terminal of the resistor than on the conductor connected to the more positive terminal.

There's much more to emf than what I've outlined here and the Wikipedia article (and links and reference therein) I linked above gives much more detail.

#### @AlanZ2223 2014-08-08 16:16:12

You should not think of electrons throughout the circuit as one part having more electrons than the other. You should instead think of the electrons within the conductors in terms of their energy. For example when you connect have a simple circuit with a battery, a set of wires, and a load. When you close the circuit, the electrons at the negative side of the battery will feel attraction to the positive side as this side has a deficiency of electrons. The electrons however do not just speed through the circuit and get through to the other side. They act as sort of a clothesline. Since electrons repel one another when the first electrons move the electrons in front of them, those electrons get repelled and this starts a chain reaction throughout the whole circuit. Now whenever the electrons come across a resistor, for example they lose some of their energy and are not as active to get to the other side of the battery and this is measured as a voltage or potential drop.

#### @user43487 2014-08-08 16:33:10

I understand what you have said but surly by the defintion of a potential difference across a resistor it means that the charge distriubtion is different either side?

### [SOLVED] In a circuit with two identical resistors, how does the battery 'know' to only deposit half of the emf at one and half at the other (KVL)?

• 2018-10-20 08:57:09
• Cr0xx
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• Tags:   electricity