Classical electromagnetic field theory is a description of the
interactions between electrical charges and electrical currents. The
theory was well-developed by the end of the nineteenth century, and yet
remarkably is consistent with the theory of special relativity. (In fact
you will recall from PH321 that it was electromagnetic theory which told
us we needed relativity!) Only in the limits of very small distances and
low field strengths does the theory begin to break down, in which case
we have to turn quantum electrodynamics, a relativistic quantum field
theory.
In this part of PH322 you will begin your study of classical
electromagnetism. We will study fundamental electrostatics - electric
fields generated by non-moving electrical charges - as well as taking a
brief look at magnetic fields. You can consider this section of PH322 to
be merely the first part of a three-module series (PH322 - PH504 -
PH604) which will develop classical electromagnetic field theory much
more fully, and give you a good understanding of Maxwell’s equations.
These equations completely describe the interactions between electric
and magnetic fields, but require mastery of the vector calculus you will
learn in PH312, and so will not be discussed at all in PH322.
Electric Charge
Electrical charge is a fundamental property of matter. It is has a
number of characteristics:
There are two types of charge: positive and negative.
Electrical forces are experienced between electrical
charges.
Like charges experience a repulsive force, unlike charges feel an
attractive force.
Electrical charge is measured in Coulombs (C). In SI units, 1 C =
1 Ampere-second (As).
Electrical charge is quantised, meaning it comes in discrete
chunks. The fundamental or elemental charge, which is also the magnitude
of the charge of an electron, is \(1.6~\times~10^{-19}\) C. All electrical
charges are integer multiples of the fundamental charge (except for
quarks which don’t appear in isolation and so we will ignore!)
Charge is conserved.
Non-examinable Aside: We can distinguish between
local and global charge conservation. Global charge
conservation says that, if I have a box and the total charge inside is
\(Q\) at some point in time, then at
some future point, the total charge inside the box is also \(Q\). This doesn’t preclude charge
disappearing at one point in the box and instantaneously appearing at
another point. In contrast, local charge conservation means that charge
does not appear and disappear, it can only flow between different
points. In the classical theory of electromagnetism, charge is
locally conserved.
Electrostatic Attraction
and Repulsion
As described above, electrical charges experience attractive and
repulsive forces. However, we mostly don’t notice these electrical
forces. This is because, in most matter, the positive and negative
charges are almost exactly balanced. However, we can create slight
inbalances in many ways. The classic demonstration is to rub glass on
silk. A small number of electrons are transferred from the glass to the
silk. Electrons are negatively charged (with a charge of \(-e\) each) and so the glass gains a
positive charge and the silk a negative charge. The total charge is
unchanged.
Coulomb Force
The force between two charged particles, with charges \(q_1\) and \(q_2\) and separated by a distance \(r_{12}\), is given by Coulomb’s law. The
magnitude of the force, F, is: \[F =
\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}^2}\]
where \(\epsilon_0\) (‘epsilon
zero’) is the permittivity of free space, sometimes
known as the vacuum permittivity or the electric constant, \(\epsilon_0 \approx 8.85 \times
10^{-12}~\mathrm{C^2N^{-1}m^{-2}}\). The constant \(1/(4\pi\epsilon_0)\) is sometimes written
as \(k\), and called Coulomb’s
constant, \(k \approx 9 \times
10^9~\mathrm{Nm^2C^{-2}}\).
While this provides the magnitude of the force, force is a vector and
hence also has a direction.
The vector form of Coulomb’s law is: \[\vec{F}_{12}=
\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}^2}
\hat{r}_{12}\]
The expression \(\vec{F} _{12}\)
means the force exerted by charge 1 on charge 2. Try to memorise which
way around this is to avoid notational confusion. Notice that the right
hand side now contains a unit vector, \(\hat{r}_{12}\), which points from charge 1
to charge 2. As a unit vector (i.e. a vector of length 1) it has no
effect on the magnitude of the force, it just tells us the direction in
which the force acts.
We can simplify the equations if we write it in terms of Coulomb’s
constant, \(k = 1/(4\pi\epsilon_0)\):
\[\vec{F}_{12}= k\frac{q_1q_2}{r_{12}^2}
\hat{r}_{12}\]
The unit vector \(\hat{r}_{12}\) is
given by \[\vec{r}_{12}=
\frac{\vec{r}_{12}}{|r_{12}|}\]
where \(|r_{12}|\) is the length
(magnitude) of the vector \(r_{12}\).
Another way to write Coulomb’s Law is therefore: \[\vec{F}_{12}= k\frac{q_1q_2}{r_{12}^3}
\vec{r}_{12}\]
We can see that the direction also depends on the sign of
the charges. If \(q_1\) and \(q_2\) are the same sign (both negative or
both positive) then \(q_1q_2\) is
positive. So the force on charge 2 points along the direction of charge
1 to charge 2. That is, charge 2 feels a force repelling it from charge
1. On the other hand, if the signs are opposite then \(q_1q_2\) is negative and hence the force is
in the opposite direction to the unit vector, i.e. charge 2 feels a
force pulling it towards charge 1.
Also notice that it now matters whether we are calculating the force
on charge 1 due to charge 2, or the force on charge 2 due to charge 1.
While the magnitude of these two forces is the same, the
direction is opposite, i.e. \(\vec{F}_{12} = -\vec{F}_{21}\). That is
because the unit vectors \(\hat{r}_{12}\) and \(\hat{r}_{21}\) are not the same, they point
in opposite directions, \(\hat{r}_{12} =
-\hat{r}_{21}\). This makes sense - if we have two particles
which are attracted to each other, then the force on the left particle
points to the right, and the force on the right particle point to the
left.
See Tipler/Mosca Examples 21-2, 21-3
Superposition Principle
Our study of electrostatics is greatly simplified by the
superposition principle. It says that the combined effect of two
charges, in terms of the force they exert on a third charge, is simply
equal to the sum of the forces each of the two charges individually
exert. This makes calculating the effect of multiple charges simple, we
just calculate the force due to each charge individually, and add them
up. However, we must add them as vectors - we care about the
direction!
The total force on a charge \(q_0\)
from a collected of \(N\) point charges
of charge \(q_i\) is given by: \[\vec{F} = kq_0\sum_{i=1}^{N} \frac{q_i}{r_i^2}
\hat{r_i}\]
where \(\hat{r_i}\) is the unit
vector pointing from the \(i\)th charge
to \(q_0\), and \(r_i\) is the distance between \(q_0\) and the \(i\)th charge.
Note that this is a vector sum. We have to add the
components of the vectors to obtain another vector. We cannot simply add
the magnitudes of the of the forces. This is easily seen from a simple
example: if we have a force of 1 N pushing a charge to the right, and
another force of 1 N pushing the charge to the left, the total force is
not 2 N but 0 N (the forces are in opposite directions, so they
cancel).
Non-examinable Aside - In certain media, known as
non-linear media, the superposition principle does not apply for very
high electric fields caused by, for example, intense laser light. This
leads to a branch of physics called non-linear optics which has lots of
useful and interesting applications, and is an active research field,
but sadly isn’t a topic for PH322.
Consider two electric charges, \(q_1 =
1~\mathrm{nC}\) and \(q_2 =
4~\mathrm{nC}\), lying on a straight line at distances of 1 m and
6 m from the origin, respectively, as shown in Figure 1.
(a) What is the force experienced by a charge of \(q_0 = -2~\mathrm{nC}\) lying 4 m from the
origin?
The unit vector from \(q_1\) to
\(q_0\), \(\hat{r}_{1,0}\), is along the +ve direction
of x-axis, and so \(\hat{r}_{1,0} =
\hat{i}\).
The unit vector from \(q_2\) to
\(q_0\), \(\hat{r}_{2,0}\), is along the -ve direction
of x-axis, and so \(\hat{r}_{1,0} =
-\hat{i}\).
\[\begin{aligned}
\vec{F} &= \frac{kq_1q_0}{r^2_{1,0}}\hat{i} -
\frac{kq_2q_0}{r^2_{2,0}}\hat{i} \\
&= (9\times10^9~\mathrm{kgm^3s^{-2}C^{-2}})
\bigg(\frac{(1~\mathrm{nC})(-2~\mathrm{nC})}{(3~\mathrm{m}^2)} -
\frac{(4~\mathrm{nC})(-2~\mathrm{nC})}{(2~\mathrm{m})^2}\bigg)\hat{i} \\
&= 1.6 \times 10^{-8}\hat{i}~\mathrm{N}
\end{aligned}\] Note that the force is along the +ve direction.
(b) Where should \(q_0\) be placed
so that there is no net force? We can see that this position will
be somewhere inbetween \(q_1\) and
\(q_2\). \[\begin{aligned}
\vec{F} &= \frac{kq_1q_0}{r^2_{1,0}}\hat{i} -
\frac{kq_2q_0}{r^2_{2,0}}\hat{i} \\
0 &= \frac{(1~\mathrm{nC})(-2~\mathrm{nC})}{(r_1-r_0)^2} -
\frac{(4~\mathrm{nC})(-2~\mathrm{nC})}{(r_2-r_0)^2} \\
\frac{1}{[(1~\mathrm{m})-r_0]^2} &=
\frac{4}{[(6~\mathrm{m})-r_0]^2} \\ %
\end{aligned}\]
Solving this quadratic gives \(r_0 =
8/3~\mathrm{m}\). There is another mathematical solution at \(r_0 = -4~\mathrm{m}\) (i.e. another point
where \(|F_{20}| = |F_{10}|\)), but
there the two forces are in the same direction and so they don’t
cancel.
Another way to see the answer is to note that force goes with \(1/r^2\) but linearly with \(q\). Since \(q_2\) is four times greater than \(q_1\), \(q_1\) need to be \(\sqrt{4} = 2\) times closer to balance this
out. So, we have: \[\begin{aligned}
\frac{r_2 - r_0}{r_0 - r_1} &= 2 \\
r_0 &= 3/8~\mathrm{m}
%
\end{aligned}\]
Five charges are arranged as shown in Figure 2. What is the force on \(q_0\)?
The total force is given by: \[\vec{F}_T =
\vec{F}_{1,0} + \vec{F}_{2,0} + \vec{F}_{3,0} +
\vec{F}_{4,0}\]
However, notice that \(\hat{r}_{1,0} =
-\hat{r}_{4,0}\) and so \(\vec{F}_{1,0}
= -\vec{F}_{4,0}\). Similarly, \(\vec{F}_{2,0} = -\vec{F}_{3,0}\). And so:
\[\begin{aligned}
\vec{F}_T = -\vec{F}_{4,0} -\vec{F}_{3,0} + \vec{F}_{3,0} +
\vec{F}_{4,0} = 0
\end{aligned}\] And so the net force is zero. In fact this is
obvious from the symmetry of the problem.
Charge 1, \(q_1 = 3\) nC, lies
at point (3,4) m and charge, \(q_2 =
-2\) nC, lies at point (6,2) m. What is the force on charge 2 due
to charge 1?
First write down the vector pointing from charge 1 to charge 2: \[\begin{aligned}
\vec{r}_{12} = (6-3)\hat{i}~\mathrm{m} + (2-4)\hat{j}~\mathrm{m} =
(3\hat{i} -2\hat{j})~\mathrm{m}
\end{aligned}\]
The magnitude of this vector is: \[|\vec{r}_{12}| = \sqrt{(3~\mathrm{m})^2 +
(2~\mathrm{m})^2} = \sqrt{13} ~\mathrm{m}\]
And so the unit vector is: \[\begin{aligned}
|\hat{r}_{12}| &= \frac{ \vec{r}_{12}^2 } { |\vec{r}_{12}| }
~\mathrm{m} \\
&= \frac{ 3\hat{i} -2\hat{j} }{\sqrt{13}}
~\mathrm{m}
\end{aligned}\]
We can then calculate the force: \[\begin{aligned}
\vec{F}_{12} &= k\frac{q_1q_2}{|\vec{r}_{12}^2|}
\hat{r}_{12} \\
&=
(9\times10^9)\frac{(3\times10^{-9})(-2\times10^{-9})}{|\vec{r}_{12}|^2}
\hat{r}_{12}~\mathrm{N} \\
&= \frac{(-5.4 \times 10^{-8})}{(13)^{3/2}} (3\hat{i} - 2
\hat{j})~\mathrm{N} \\
&= (-3.5 \hat{i} + 2.3 \hat{j}) \times
10^{-9}~\mathrm{N}
\end{aligned}\]
