Recap……
- Types of electricity
- Current Electricity: Net flow of charges in a certain direction
- Static Electricity: No net flow of charges in a certain direction
- Matter can be classified into 3 types according to their electrical properties:
- Conductors – Materials which have mobile charge carriers, mainly electrons and ions which will drift to constitute an electric current under the effect of an applied electric field. Hence they can conduct electricity. Examples include metals and electrolyte solutions.
- Insulators – Materials which have no mobile charge carriers that can drift under the effect of an applied electric field. Hence they cannot conduct electricity. Examples include rubber, wood and plastic.
- Semiconductors – Materials which have intermediate electrical conductivity which vary substantially with temperature. Examples include Germanium, Silicon.
Show an understanding that electric current is the rate of flow of charged particles.
- All matter is made up of tiny particles called atoms, each consisting of a positively charged nucleus with negatively charged electrons moving around it.
- Charge is measured in units called coulombs (C). The charge on an electron is $-1.6 \times 10^{-19} \mathrm{C}$.
- Normally atoms have equal number of positive and negative charges, so that their overall charge is zero.
- For some atoms, it is relatively easy to remove an electron, leaving an atom with an unbalanced number of positive charges. This is called positive ion.
- Atoms in metals have one or more electrons which are not held tightly to the nucleus.
- These free (or mobile) electrons wander at random throughout the metal.
- But when a battery (or source) is connected across the ends of the metal, the free electrons drift towards the positive terminal of the battery (or source) producing an electric current.
- The size of the electric current is given by the rate of flow of charge and is measured in units called amperes with symbol A.
- A current of 3 amperes means that 3 coulombs pass a point in the circuit every second. In 5 seconds, a total charge of 15 coulombs will have passed the point.
Charge is quantised
Because electric charge is carried by particles, it must come in amounts which are multiples of $e$. So, for example, $3.2 \times 10^{-19} \mathrm{C}$ is possible, because this is $+2 e$, but $2.5 \times 10^{-19} \mathrm{C}$ is impossible, because this is not an integer multiple of $e$.
We say that charge is ‘quantised’; this means that it can only come in amounts which are integer multiples of the elementary charge. If you are studying chemistry, you will know that ions have charges of $\pm e, \pm 2 e$, etc. The only exception is in the case of the fundamental particles called quarks, which are the building blocks from which particles such as protons and neutrons are made. These have charges of $\pm \frac{1}{3} e$ or $\pm \frac{2}{3} e$. However, quarks always appear in twos or threes in such a way that their combined charge is zero or a multiple of $e$.
Figure 9.9 A current $I$ in a wire of cross-sectional area $A$. The charge carriers are mobile conduction electrons with mean drift velocity $v$.
An equation for current
Copper, silver and gold are good conductors of electric current. There are large numbers of conduction electrons in a copper wire – as many conduction electrons as there are atoms. The number of conduction electrons per unit volume (e.g. in $1 \mathrm{~m}^3$ of the metal) is called the number density and has the symbol $n$. For copper, the value of $n$ is about $10^{29} \mathrm{~m}^{-3}$.
Figure 9.9 shows a length of wire, with cross-sectional area $A$, along which there is a current $I$. How fast do the electrons have to travel? The following equation allows us to answer this question:
$
I=n A v q
$
Here, $v$ is called the mean drift velocity of the electrons and $q$ is the charge of each particle carrying the current. Since these are usually electrons, we can replace $q$ by $e$, where $e$ is the elementary charge. The equation then becomes:
$
I=n A v e
$
Deriving $I=$ nAve
Look at the wire shown in Figure 9.9. Its length is 1 . We imagine that all of the electrons shown travel at the same speed $v$ along the wire.
Now imagine that you are timing the electrons to determine their speed. You start timing when the first electron emerges from the right-hand end of the wire. You stop timing when the last of the electrons shown in the diagram emerges. (This is the electron shown at the lefthand end of the wire in the diagram.) Your timer shows that this electron has taken time $t$ to travel the distance $l$.
In the time $t$, all of the electrons in the length $l$ of wire have emerged from the wire. We can calculate how many electrons this is, and hence the charge that has flowed in time $t:$
$
\begin{aligned}
\text { number of electrons } & =\text { number density } \times \text { volume of wire } \\
& =n \times A \times l \\
\text { charge of electrons } & =\text { number } \times \text { electron charge } \\
& =n \times A \times l \times e
\end{aligned}
$
We can find the current $l$ because we know that this is the charge that flows in time $t$, and current = charge/time:
$
I=n \times A \times l \times e / t
$
Substituting $v$ for $1 / t$ gives
$
I=n \text { Ave }
$
The moving charge carriers that make up a current are not always electrons. They might, for example, be ions (positive or negative) whose charge $q$ is a multiple of $e$. Hence we can write a more general version of the equation as
$
I=n A v q
$
Worked example 3 shows how to use this equation to calculate a typical value of $v$.
WORKED EXAMPLE
3 Calculate the mean drift velocity of the electrons in a copper wire of cross-sectional area $5.0 \times 10^{-6} \mathrm{~m}^2$ carrying a current of $1.0 \mathrm{~A}$. The electron number density for copper is $8.5 \times 10^{28} \mathrm{~m}^{-3}$.
Step 1 Rearrange the equation $I=n A$ e to make $v$ the subject:
$
v=\frac{1}{n A e}
$
Step 2 Substitute values and calculate :
$
\begin{aligned}
v & =\frac{1.0}{8.5 \times 10^{216} \times 5.0 \times 10^{-6} \times 1.6 \times 10^{-19}} \\
& =1.47 \times 10^{-5} \mathrm{~ms}^{-1} \\
& =0.015 \mathrm{mms}^{-1}
\end{aligned}
$
Slow flow
It may surprise you to find that, as suggested by the result of Worked example 3, electrons in a copper wire drift at a fraction of a millimetre per second. To understand this result fully, we need to closely examine how electrons behave in a metal. The conduction electrons are free to move around inside the metal. When the wire is connected to a battery or an external power supply, each electron within the metal experiences an electrical force that causes it to move towards the positive end of the battery. The electrons randomly collide with the fixed but vibrating metal ions. Their journey along the metal is very haphazard. The actual velocity of an electron between collisions is of the order of magnitude $10^5 \mathrm{~m} \mathrm{~s}^{-1}$. but its haphazard journey causes it to have a drift velocity towards the positive end of the battery. Since there are billions of electrons, we use the term mean drift velocity $v$ of the electrons.
Figure 9.10 shows how the mean drift velocity of electrons varies in different situations. We can understand this using the equation:
$
v=\frac{I}{n A e}
$
- If the current increases, the drift velocity $v$ must increase. That is:
$
v \propto I
$
- If the wire is thinner, the electrons move more quickly for a given current. That is:
$v \propto \frac{1}{A}$
There are fewer electrons in a thinner piece of wire, so an individual electron must travel more quickly.
- In a material with a lower density of electrons (smaller $n$ ), the mean drift velocity must be greater for a given current.
Figure 9.10 The mean drift velocity of electrons depends on the current, the cross-sectional area and the electron density of the material.
It may help you to picture how the drift velocity of electrons changes by thinking about the flow of water in a river. For a high rate of flow, the water moves fast – this corresponds to a greater current $l$. If the course of the river narrows, it speeds up – this corresponds to a smaller crosssectional area $A$.
Metals have a high electron number density – typically of the order of $10^{28}$ or $10^{29} \mathrm{~m}^{-3}$. Semiconductors, such as silicon and germanium, have much lower values of $n$ perhaps $10^{23} \mathrm{~m}^{-3}$. In a semiconductor, electron mean drift velocities are typically a million times greater than those in metals for the same current. Electrical insulators, such as rubber and plastic, have very few conduction electrons per unit volume to act as charge carriers.
Electric current
- Electric current is the rate of flow of electric charge.
- Mathematically, $I=\frac{Q}{t}$ where I is the electric current (unit: ampere, symbol: A);
$Q$ is the electric charge (unit: coulomb, symbol: C);
$t$ is the time taken (unit: second, symbol: s)
Charge & Coulomb
- From the definition of electric current $I=\frac{Q}{t}$ we obtain, $Q=I t$.
- Electric charge flowing through a section of a circuit is the product of the electric current and the time that it flows.
- $Q=$ It, substituting in units we obtain the following :
- $1 \mathrm{C}=(1 \mathrm{~A})(1 \mathrm{~s})=1 \mathrm{As}$
- One coulomb is the quantity of electric charge that passes through a section of a circuit when a steady current of one ampere flows for one second.
Solve problems using the equation $Q=I t$
Example 1
Given that the electric current flowing through a circuit is $0.76 \mathrm{~mA}$, calculate the electric charae which passes each section of the circuit over a time of $60 \mathrm{~s}$.
Answer/Explanation
Solution:
$
\begin{aligned}
& {[Q=\text { It }]} \\
& Q=\left(0.76 \times 10^{-3}\right)(60)=0.0456=4.56 \times 10^{-2} \mathrm{C}
\end{aligned}
$
Example 2
Over a time of $8.0 \mathrm{~s}$, the electric current flowing through a circuit component is reduced uniformly from $60 \mathrm{~mA}$ to $20 \mathrm{~mA}$. Calculate the charge that flows during this time.
Answer/Explanation
Solution:
$
\begin{aligned}
\text { Charge } & =\text { area under current-time graph } \\
& =\frac{1}{2}(8.0)(60+20)\left(10^{-3}\right)=0.32 \mathrm{C}
\end{aligned}
$