2 Electrostatic fields
2.1 Charge density
In this section we will explore the first of Maxwell’s equations,
\(\displaystyle\nabla.\vec{E}(\vec{r})=\frac{\rho(\vec{r})}{\epsilon_0}\)
and see how it relates to your existing knowledge of electric fields.
The first thing you will notice is that we’ve expressed charge in the system in terms of the charge density \(\rho\) rather than describing the field in terms of individual ‘point’ charges (\(q\)) with which you are familiar in Coulomb’s law,
\(\vec{F}=\dfrac{q_1 q_2}{4\pi\epsilon_0 r^2}\hat {r}\)
The difficulty here is what we mean by a point charge. What happens when \(r=0\)? We end up with a singularity with an infinite electric field! You are also aware that point charges don’t really exist. So, as we try to isolate a single particle with charge, say an electron, we find we can’t describe it as a simple particle but by its quantum mechanical wavefunction. So, as far as classical electromagnetism is concerned, we don’t talk about charged fundamental particles but rather the ‘spatial average’ of the macroscopic distributions of these charges. Hence \(\rho(\vec{r})\) may be thought of as the smoothed charge distribution that we see at the ‘macroscopic’ scale.
2.1 The Superposition Principle.
Coulomb’s law was derived from an experiment in which we would now consider the charges \(q_1\) an \(q_2\) as due to multiple charged particles accumulating on small but macrosopic particles. However, the experiments did reveal the important result that any distribution of charges obeys the ‘superposition principle’. Simply put, this says that for any distribution of charges the force on a small test charge \(q_t\) is given by,
\(\displaystyle\vec{F}(\vec{r})=\dfrac{1}{4\pi\epsilon_0}\sum_{i} \dfrac{q_t q_i(\vec{r}-\vec{r_i})}{|\vec{r}-\vec{r_i}|^3}\)
If we add another charge, the extra force on the test particle is not affected by the other charges in the system. Alternatively, we can write,
\(\vec{F}(\vec{r})=q_t\vec{E}(\vec{r})\)
where
\(\displaystyle\vec{E}(\vec{r})=\dfrac{1}{4\pi\epsilon_0}\sum_{i} \dfrac{q_i(\vec{r}-\vec{r_i})}{|\vec{r}-\vec{r_i}|^3}\)
Aside. You may note in the above expressions that we have a \(1/|\vec{r}-\vec{r_i}|^3\) in our formulae. Make sure that you understand that, combined with the numerator, this still gives the inverse square force law for each of the charges in the system.
If we now consider the charges to be distributed continuosly over some macroscopic region we can write the electric field as an integral,
\(\displaystyle\vec{E}(\vec{r})=\dfrac{1}{4\pi\epsilon_0}\int \dfrac{(\vec{r}-\vec{r_{dq}})}{|\vec{r}-\vec{r_{dq}}|^3}dq\)
Expressed in terms of the charge density \(\rho\), we can write \(dq = \rho d\tau'\) where \(d\tau '\) is an element of volume. With this we can write,
\(\displaystyle\vec{E}(\vec{r})=\dfrac{1}{4\pi\epsilon_0}\int\rho(\vec{r'})\dfrac{(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3}d\tau' \;\; ...\;(2.1)\)
You will find we consider many integrals like this in this course, so it is worthwhile taking sometime to make sure that you know how to interpret what it means. We will take some time in the lectures to look at some examples. You will also see this and similar formulae represented differently in different textbooks with a wide range of different symbols used to denote different vectors and scalars. I recommend you go to section 2.1.4 in Griffiths and see how the same result is derived but with what appears to be a simplified formula with an unsual typset ‘r’. If you can reconcile the two equations and understand their physical intepretation you are well placed to follow the subsequent lectures and problems. So, if you know the charge distribution then this integral will always give the electric field. However, the trouble with these integrals for real problems is they are often difficult or even not analytically solvable.
We should also note that we can write equivalent equations to (2.1) when we consider surface and line charge density (i.e. where we have restricted the charges to lie on a known surface (2D) or line (1D)). You can, but we will not do it here, derive these from equation 2.1 (see also Griffiths 2.1.4)
\(\displaystyle\vec{E}(\vec{r})=\dfrac{1}{4\pi\epsilon_0}\int\sigma(\vec{r'})\dfrac{(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3}da'\)
\(\displaystyle\vec{E}(\vec{r})=\dfrac{1}{4\pi\epsilon_0}\int\lambda(\vec{r'})\dfrac{(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3}dl'\)
2.2 Gauss’s theorem and electric flux.
If we go back to thinking of our charges we can imagine that they are \(\textit{sources}\) or \(\textit{sinks}\) of the electric field (the direction of the field goes from positive to negative). There are several ways in which we can represent the electric field (in 2D on paper) with which you should be familiar. One way is to indicate the strength of the field with an arrow pointing in the direction of the field with a length corresponding to the strength of the field, i.e.
An alternative way is to plot continuous field lines eminating from positive charges and ending on negative charges. We’ll see some examples in the lectures.
Note, how the density (separation) of the lines becomes lower as we move to regions in which the field is weaker. Before attempting a mathematical solution of the field from charges it is often useful to sketch these field lines to get an idea of the solution you expect. The concept of field lines also leads us neatly to the idea of the \(\textit{electric flux}\). If I consider some surface Sa$ I can think of the flux through the surface as the \(\textit{number of field lines}\) passing through the surface, or put mathematically,
\(\displaystyle\Phi_E=\int_S\vec{E}.d\vec{a}\)
If I take this further it would suggest that the \(\textit{total}\) flux passing through a closed surface is a measure of the total charge within the surface. Any charge outside the surface does not contribute to the next flux (the field lines from it go in and then out of the surface giving a net zero contribution).
Applying this gives the result,
\(\displaystyle\int_S\vec{E}.d\vec{a}=\frac{Q_{enc}}{\epsilon_0} \;\;...\;(2.2)\).
This is known as Gauss’s law. You might note that it is no more than the application of Coulomb’s (inverse square) law and the principle of superposition. Gauss’s law in this form can be applied directly to find the electric field for cases of high symmetry - you will see this in some examples and problems. One nice result is that if we apply this to a macroscopic sphere of charge of uniform (or at least radially symmetrical) charge density then the field outside the sphere is exactly the same as you’d expect from a point charge with a total charge equal to the total charge enclosed in the sphere. Working out the field in the sphere will be saved for a problem but you will be able to see that it goes to zero at the centre which solves our problem of the singularity with a point charge. For spherical charge densities (outside the spheres) we can always apply Coulom’s law as before.
2.3 Gauss’s Law and the Divergence theorem
In your mathematics course you will have the seen a statement (and perhaps a proof) of the Divergence Theorem,
\(\displaystyle\int_V (\nabla.\vec{v})d\tau = \oint_S\vec{v}.d\vec{a}\)
As this is true for all vectors we can hence write
\(\displaystyle\int_V (\nabla.\vec{E})d\tau = \oint_S\vec{E}.d\vec{a}\)
We also know that
\(\displaystyle Q_{enc}=\int_V \rho d\tau\)
from which we can write using Gauss’s law (2.2),
\(\displaystyle\int_V (\nabla.\vec{E})d\tau=\int_V \frac{\rho}{\epsilon_0} d\tau\)
As this holds for any volume we get the important result and the first of our Maxwell’s equations,
This is also known Gauss’s law in \(\textit{differential form}\). Note that this relates the divergence of the electric field at a \(\textit{point}\) in space to the presence or a charge (density) at that point in space. i.e If there is no charge at that point (i.e. \(\rho(\vec(\vec{r})=0)\), then the divergence is zero. You may like to check that this is true for the electric field in space away from a point charge by taking the divergence (try it in Cartesian and Spherical Polar coordinates).
2.4 The electric (static) field is a conservative field
We have established how the electric field is related to the distribution of charge. Is there anything else that characterises the electric field? Let’s consider the following line integral (you should be familiar with these from your mathematics course),
\(\displaystyle\int_a^b \vec{E}.d\vec{l}\)
where \(\vec{E}\) originates from a static point charge. If this a point charge then we see that the components of \(\vec{E}.d\vec{l}\) disappear in the \(\mathbf{\hat{\theta}}\) and \(\mathbf{\hat{\phi}}\) directions as the field points radially outward. Hence the integral reduces to a single, scalar term,
\(\displaystyle\int_a^b\frac{q\,dr}{4\pi\epsilon_0 r^2}\).
(remember \(d\vec{l}=dr\,\mathbf{\hat{r}} +r d\theta\,\mathbf{\hat{\theta}}+r\,sin\,d\phi\,\mathbf{\hat{\phi}}\).
Evaluating the integral we find,
\(\displaystyle\int_a^b\frac{q\,dr}{4\pi\epsilon_0 r^2}=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_a}-\frac{q}{r_b}\right)\)
What happens if we follow a path in a closed loop? In this case we may write,
\(\displaystyle\oint\vec{E}.d\vec{l}=\int_a^b\vec{E}.d\vec{l}+\int_b^a \vec{E}.d\vec{l}=0\)
Although, we’ve obtained this result for a single point charge, we can see, from the principle of superposition, that it will apply to the field from \(\textit{any}\) distribution of charges. Hence, we can also see that \(\vec{E}\) is a \(\textit{conservative}\) field. You should have come across the description and definitions of conservative fields in your mathematics course. In that course you should also have come across Stoke’s theorem, that states for any vector \(\vec{X}\)
\(\displaystyle\oint\vec{X}.d\vec{l}=\int_S(\nabla\times\vec{X}).d\vec{S}\),
where \(\displaystyle\int_S\) means the surface integral over any surface bounded by the path of the line integral. Applying this theorem to the electric field we find,
Indeed, if the curl of \(\textit{any}\) field is zero, then the field is conservative. You will see that this result is close to , but not the same as, the second Maxwell equation we wrote down at the start. So it is important to remember that this result applies specifically to \(\textit{electrostatic}\) fields. We will come back later to the case of \(\textit{dynamic}\) (time varying) fields.
2.5 Summary of the electrostatic field.
From the results in the previous sections we can summarise the governing equations for electrostatic fields as,
and
These properties of the electrostatic field were found by careful experiments in th 19th century, but if we accept these two equations as the basis for electrostatics then Gauss’s law and the general solution
can be used to find the electrostatic field from any charge distribution.
The general solution we can always be used to find the electric field if we know the distribution of charges by solving this integral. Apart from simple cases (that you will see) it is generally not possible to find analytical results for \(\vec{E}\). However, with modern computers, it is relatively straightforward and feasible to calculate the electric fields numerically in relatively short time by essentially this direct method. What is not so easy is to ask the question “What is the charge distribution I need to produce the electric field I require?” You might like to ponder this question.