Lecture 2

What is charge?

Electric charge is a fundamental property of some types of particle.
There is no way that one electron, for example, can have a different charge to another electron – the charge is an unchanging property.
Experimental evidence for this came from Robert Millikan and Harvey Fletcher in 1909. (For more details of this ingenious experiment see exercise 23.79 on p 807 of Y&F).
In our best modern theory of particles, known as quantum field theory, the fact that all electrons have the same charge is baked deeply into the mathematics, and appears quite ‘natural’.
But more surprising is that the charge of an electron is exactly minus the charge of the proton (so far as anyone can tell – to an accuracy of one part in $10^{20}$ or so¹).
Charge must also be conserved. This is built into classical electromagnetism, based on evidence dating all the way back to Benjamin Franklin’s astonishingly dangerous experiments with a kite² in the 1750s, based on which he imagined charge as being a kind of fluid that could be moved around but not created or destroyed.
Conservation of charge is deeply linked to an underlying symmetry of the physics, known as a gauge symmetry. This insight is one of many by Emmy Noether, but is not examinable in this course.

The SI unit of charge is Coulombs, named after the same physicist who demonstrated Coulomb’s law discussed last time. A Coulomb is a pretty big charge, or to put it another way, the charge $e$ on an electron is very small.

\[e \simeq 1.60 \times 10^{-19}\,{\rm C}\,.\]

Looking at the Coulomb force law from last time,

\[\vec{F}_{1,2} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{\vert \vec{r}_{1,2} \vert^2} \hat{\vec{r}}_{1,2}\]

we can deduce that $\epsilon_0$ must be measured in units of ${\rm C^2\, N^{-1}\, m^{-2}}$, so that the left hand side comes out in Newtons. Its value is derived from experiment to be

\[\epsilon_0 \simeq 8.85 \times 10^{-12}\, {\rm C^2\, N^{-1}\, m^{-2}}\,.\]

If we look at it in terms of the electric field, we had from last time

\[\vec{F} = \vec{E}(\vec{r}) q.\]

which means that the SI units associated with $\vec{E}$ must be ${\rm N\,C^{-1}}$. For reasons that will become clearer later in the course, this can also be written as volts per metre, i.e. ${\rm V\,m^{-1}}$. It’s the same unit, just expressed in a different way.

Force and motion in an electric field

Once we know the electric field — even if it’s due to some very complicated or even unknown charge distribution — it’s pretty easy to calculate forces, using the equation above.

To be clear,

An uncharged particle ($q=0$) feels no force at all;
A positively charged particle ($q>0$) feels a force along the fieldlines;
A negatively charged particle ($q<0$) feels a force backwards along the fieldlines.
This does not mean that the particle moves along electric fieldlines. Just like in any mechanics problem, velocity does not have to be in the same direction as acceleration.

What happens if we have something which is not a point particle, but some kind of composite object (i.e. made out of lots of particles)? An atomic nucleus, for example, is made out of multiple protons, say $N$ of them each with a charge of $+e$.

Provided the electric field does not change significantly across the nucleus, each one feels a force of $\vec{F}_{\rm proton} = \vec{E} e$
The net force the nucleus feels is equal to the vector sum of all the individual forces, $\vec{F}_{\rm total} = \vec{E} N e$.
$Ne$ is just the total charge of the nucleus, $q$.
So, even for a nucleus – or any object made out of lots of particles – the basic rule $\vec{F} = \vec{E} q$ still applies, where $q$ is the total charge.

That said, if $\vec{E}$ had varied between the different particles making up the nucleus, this argument would not hold. It relies on the $\vec{E}$ field taking the same value on each particle, so that we can sum up the total force.

But hang on a second – how can the electric field ‘not change significantly’ across something like a nucleus, when it has all of these positive charges in it? They all surely contribute to the electric field and make it vary a lot! The answer is that when calculating the net force we do not allow any of the charges in the nucleus to contribute to the field. Here’s why…

Why doesn’t a charge exert a force on itself?

To reiterate, when calculating net force on a collection of charged particles, we don’t include the field coming from any of the particles themselves. Here’s why:

While parts of a nucleus may push on other parts, there is always an equal and opposite force pushing back the other way so that the net force from charges within the nucleus itself sums to precisely zero (see ‘conservation of momentum’ below).
To be clear, this only applies to the net force; individual forces can still be tremendous so you might expect the nucleus to explode.
The nucleus doesn’t explode, because there are also nuclear forces and they hold it together.

You might also now wonder about why, when calculating force on even a single charged particle, we don’t include its own contribution to the electric field. If you look at Coulomb’s law and imagine calculating the force of a charged particle on itself, the $\vert r \vert^2$ term would be zero. Since this appears on the denominator, the strength of the force of a particle on itself would appear to be infinite! How, then can we just ignore it?

The real answer to this requires quantum mechanics³, I’m afraid. But just like with the nucleus, at least you can see that the net force on a particle from itself is zero. So as long as it has a way to stop itself from exploding (which you can take for granted for now), it’s all good.

Here is what you should remember:

it is valid to ignore the contribution of a given charge to the electric field when you’re calculating the force on that charge;
it is even valid to ignore the contribution of a whole set of charges to the electric field when you’re calculating the net force on that set of charges;
it is the total net charge of a nucleus (or other object) that matters when calculating the net force on that object from the electric field.

Conservation of momentum

I said above that if a charge 1 pushes on charge 2, then charge 2 must push back on charge 1 with an equal and opposite force.

This is because of Newton’s third law:

every force has an equal and opposite force
or, in other words, momentum is conserved

which remains true in electromagnetism.⁴

In the lecture, we will check this explicitly for the case of two charges, by calculating Coulomb’s law for the force on each by the other.

Superposition

Let’s now suppose there are two charges $q_1$ and $q_2$ and we want to calculate their force on a further charge $q_3$.

Since the forces add as vectors, we can say the force on charge 3 is given by adding two instances of Colulomb’s law.
We will write this out in the lecture and see that it is also equal to $q_3$ times the sum of the electric fields produced by charge $q_1$ and $q_2$.
In other words the final effect on $q_3$ is of a single electric field made up by adding (as vectors) the electric fields from the other two charges.

Here is an example to play with, where there is one positive and one negative charge. You can either show the electric fields associated with the individual charges, or sum them to produce the total electric field. Watch the arrows at different points in space and toggle the separate fields on and off. See if you can make sense of how the final field which ‘curves’, arcing from one charge to the other, comes from adding two fields which just point radially away from/towards the charges.

More generally, to get the true electric field, you add up (as vectors) the electric field from all the charges present (other than the one for which you are calculating the force, see below).

The strength of electromagnetism

The force of gravity and electromagnetism are cosmetically similar, both following an inverse square law. Example 21.1 in YF looks at this similarity a bit. In the lecture I will take a slightly different approach and calculate the gravitational force between Earth and moon as having magnitude:

\[\vert\vec{F}\vert \simeq 2 \times 10^{20}\, {\rm N}.\]

That sounds like a big force, as you might expect.

For electromagnetism, we have the very similar force law from Coulomb. But the Earth and the moon, along with all other known astronomical bodies, are very nearly neutral, so $q_1\simeq 0$, $q_2\simeq 0$ and the electrostatic force is also negligibly small.

But what if the Moon gave up a tiny fraction (say, 0.1%) of its electrons to the Earth? I’ll estimate the force in this fictional scenario in the lecture, and we’ll see it is about:

\[\vert\vec{F}\vert\simeq 10^{48}\,{\rm N}\]

This is mind-bogglingly larger than the gravitational force, and would obviously lead the moon to collapse down onto the earth.

Electromagnetism is a far stronger force than gravity, but is less obvious because most objects (both astronomical and in everyday life) are very nearly neutral. The principle of superposition then means there is no overall effect.

Discussion point: It now seems pretty lucky that astronomical bodies don’t get significant electrical charges. But why don’t they?

Footnotes

(Remember, none of the material in footnotes is examinable, so this is just for your interest.)

The experimental approach to showing the charge on protons and electrons is equal is to check that (unionized) atoms don’t respond to an electric field. Here is an example research article. The limit to how precisely the charges are known to be equal is set by how precisely one can measure the behaviour of the atoms. In the modern perspective, the theoretical reason for the precise matching is probably a result of how particles emerge from symmetry breaking in the early universe. ↩
Franklin wasn’t the first to think of this conservation property of electrical charge, but he did experiments that tested it and then wrote about it in a way that made it clear to others. Here’s an extract from a letter he wrote in 1747:

We suppose every particle… carries off with it a portion of the electrical fire [i.e., charge], but that the same still subsits in those particles, till they communicate it to something else, and that it is never really destroyed.

Here’s an extract from what Franklin wrote about his particularly ill-advised experiment:

This Kite is to be raised when a Thunder Gust appears to be coming on… the wire will draw the Electric Fire from them, and the Kite, with all the Twine, will be electrified, and the loose Filaments of the Twine will stand out every Way, and be attracted by an approaching Finger.

Please don’t try this yourself. Instead, if interested, safely read about it at the National Archives.

From a modern perspective, charge conservation is automatic in the case of electrons because they are completely stable particles (so far as anyone can tell) and so they just forever carry around their unchanging charge. In the case of protons, it is less obvious because the protons can decay into other particles – but when that happens, they emit a positron (the antimatter version of an electron) which carries away the charge. So charge is still conserved overall.

The role of the positron in all this is a case where the theory came before experimental evidence. Paul Dirac predicted the existence of ‘anti-particles’ in 1928; positrons were then discovered experimentally by studying cosmic rays (very energetic particles arriving from space) in 1932 by Carl Anderson. Positrons are now used routinely, albeit indirectly, for example in medical imaging (positron emission tomography). ↩
In quantum electrodynamics, one really does consider the effect of a charged particle on itself, and finds a highly principled way to show it can be ignored known as renormalization. ↩
We are only going to prove it here for the ‘electrostatic’ case, which means there are point charges moving slowly. In more advanced electromagnetism problems in later years, things might not be so simple. Momentum can be carried away by electromagnetic waves, for example. So while the conservation of momentum remains true, exactly how it is achieved becomes a bit more subtle. ↩