Objectives:

• To be able to understand and define the electromotive force (e.m.f.) of a source, such as a cell or a power supply
• To be able to define the term potential difference (p.d.)
• To be able to define the volt
• To describe how a voltmeter may be used to determine the p.d. across a component
• To describe the difference between e.m.f. and p.d. in terms of energy transfer
• Select and use the equations;  $W = QV$  and   $W = \epsilon V$
• To be able to select and use the following equation for the energy transfer of electrons and other charged particles;  $eV = \frac{1}{2} mv^{2}$

What is Voltage?

You will be familiar with the term voltage from GCSE, but what is it and what does it actually mean?

When electrons pass through a component in a circuit, they usually give up some of their energy (and hence allow the component to work). The voltage is simply the term used to describe the amount of energy PER unit of charge. Before a component each unit of charge will have some energy , they will then give up (some) of this energy to the component and therefore the voltage will be lower after that component.

Potential difference (p.d.)

The amount of voltage that has been given to the component can be measured using a voltmeter, the voltmeter must be placed in parallel to a component. A voltmeter works by measuring the voltage at two points (before and after a component), one of these points is a reference point, it then measures the difference in voltage. Voltage is the energy per unit of charge, so the charged particles have some potential energy. A voltmeter measures the difference in potential energy from one side of a component to another – hence the term potential difference.

Understanding the definition of potential difference is vital for almost all future work on electricity (many students take this definition for granted and don’t learn it as the definition – this is ill advised!)

Potential difference is the energy per unit charge;     $V = \frac{W}{Q}$
where;
$V$  is the voltage, measured in volts,  $V$
$W$  is the work done, measured in joules,  $J$
$Q$  is the charge, measured in coulombs,  $C$

Voltage in an electrical system can be thought of as the same thing as pressure in a water system; the cell being the pump;

Electromotive force (e.m.f.)

Current, as we know, is the flow of charged particles (in circuits this is primarily electrons) but how do these particles obtain the energy to move in the first place? If they are flowing through a wire with a mean drift velocity they must have some kinetic energy,  $\frac{1}{2} m _{e} v^{2}$  , since energy must be conserved this energy must have been provided at some point.

The energy is of course provided by a cell, battery or power supply of some kind. When the electrons travel through a component this energy is converted into another form, in the case of a bulb they dissipate this energy in the form of light and heat.

Electromotive force, e.m.f., is the energy supplied per unit of charge. Think back to your studies on work done, some of the energy supplied to an electron is the force exerted on it to get it to accelerate and get it moving around a circuit. The unit of electromotive force is the volt, not joules or newtons (the reason for this is discussed later).

Electromotive force is often denoted with the greek letter epsilon, $\epsilon$

What is the difference between voltage, e.m.f. and p.d.?

All three terms describe an amount of energy per unit of charge, and are all measured in volts – voltage being the general term used.

The voltage across a power source is described as the e.m.f.

The voltage across a component is describe as the p.d.

The definition of p.d.,  $V = \frac{W}{Q}$ , can be applied to any of these terms.

You may see  $\epsilon = \frac{W}{Q}$  used as an equation for electromotive force. This is accepted, and just uses the symbol for e.m.f. instead of p.d..

The volt

Providing you have understood the above information, the definition of the volt just flows on from this. The volt is the unit given to voltage, potential difference and/or electromotive force. Therefore the volt must be equivalent to units given in the definition of potential difference;

$p.d. = \frac{Work done}{charge}$

$V = \frac{W}{Q}$

Replacing all the quantities with units gives;

$1 V = \frac{1 J}{1 C}$

$1 volt = \frac{1 joule}{1 coulomb}$

1 volt is the equivalent of 1 joules per 1 coulomb of charge.

The equation for the energy transfer of charged particles

If the definition of p.d. is used and rearranged for work done;

$V = \frac{W}{Q}$

$W = VQ$

This could also be seen as  $W = \epsilon Q$ , if for the work done by a power source.

There are times when it is appropriate to equal the energy given to a unit of charge with the energy the charged particles have at that time, If a unit of charge has been forced to move along a circuit, they will inevitably have a form of kinetic energy. Therefore we can write that the work done per unit charge can be equivalent to the kinetic energy of that unit of charge, i.e.

$W = VQ$
$W = \frac{1}{2} m _{e} v^{2}$

Therefore;

$VQ = \frac{1}{2} m _{e} v^{2}$

Since charged particles we are referring to have a charge of $e = 1.6 \times 10^{-19}$, we can alter the equation above to read;

$eV = \frac{1}{2} m _{e} v^{2}$