3. Electric potential
In the previous section we derived the equations that define the electrostatic field in terms of the distribution of the electric charges in the system. This arose naturally by starting from Coulomb’s law for the force of attraction between two charged particles. However, in general terms they are of limited use as, to obtain a desired electric field we need to not only calculate where the charge distribution needs to be but we also need to know how we would achieve this in practise. We might also ask how we might expect to get stable charge densities. If I have a positive charge density what would stop the indvidual charges in the system blowing itself apart? In a sense our examples were quite contrived as we never explained how we generated the charge distribution in the first place!
To make progress, we are going to introduce a new property of the Electrostatic field - the Electric potential. You should already be familiar with this concept but we will now apply some of the mathematics you have now learnt to decribe the electric potential in a more succint and concise form.
3.1 The electrostatic potential
In section 2 we showed that \(\vec{E}\) is a conservative field. Hence, we can apply the properties of such fields to define the electrostatic potential,
\(\displaystyle V(\vec{a}) = -\int_O ^\vec{a}\vec{E}.d\vec{l}\).
Note, this should not be confused with potential energy (that we will discuss later). Note also the minus sign. We can also see that value of \(V(a)\) will depend on our choice of origin \(O\) - i.e. it is not really possible to define an \(\textit{absolute}\) value of potential.
However, the \(\textit{potential difference}\) between two points \(a\) and \(b\) is of importance so let’s see what we mean by potential difference.
We can write the potental diffirence between two points as follows,
\(\displaystyle V(\vec{b})-V(\vec{a}) = - \int_O^\vec{b} \vec{E}.d\vec{l}+\int_O^\vec{a}\vec{E}.d\vec{l}\)
or
\(\displaystyle V(\vec{b})-V(\vec{a}) = - \int_O^\vec{b} \vec{E}.d\vec{l}-\int_\vec{a}^O\vec{E}.d\vec{l}\)
so that
Now, we note, again remembering the property of conservative fields, that this line integral is independent of the path taken.
3.2 The electric field expressed as the gradient of a scalar potential
In your mathematics course you will have seen seen the result,
\(\displaystyle \int_a^b \nabla \phi.d\vec{l}=-\int_a^b\vec{X}.d\vec{l}\)
where \(\phi\) is a scalar quantity (a scalar field) that gives rise to the vector quantity (vector field) \(\nabla\phi\). If we apply this result to the electrostatic field we obtain the result.
3.2.1 Some comments about the scalar potential.
Finding scalar fields (potentials) is easier than finding vector fields as we only have one component (the scalar) to find rather than the three components of a vector.
Note the concept of the potential,
only applies to conservative fields,
there is no absolute definition of potential - we can only find differences in potential,
we can define our position of zero potential in an arbitary way (as we can only need differences in potential) and
potentials obey the principle of superposition (i.e. \(V=V_1+V_2+V_3 +\)).
The S.I. unit of potential is \(J\cdot C^{-1}\) or \(N\cdot m\cdot C^{-1}\) that is given the name Volt.
You should be very familiar with the Volt, it is a unit people use everyday (think batteries, power supplies …). It maybe gives a hint as to the importance of the potential in electrical and electronic systems.
3.2 Poisson’s equation.
With our knowledge of vector calculus we can very quickly find the relationship between the charge density and the potential
\(\displaystyle \nabla.\vec{E}=-\nabla. \nabla V = -\nabla^2V=\frac{\rho}{\epsilon_0}\)
From this we find \(\textit{Poisson's equation}\),
3.3 Lapace’s equation
In regions of space where there are no charges (such that \(\rho(\vec{r})=0\) Poisson’s equation reduces to,
that is known as \(\textit{Laplace's equation}\).
It is perhaps not obvious why equations 3.3 and 3.4 are any more useful than calculating \(\vec{E}\) directly from the charge distribution. i.e. we have just substituted a first order vector equation for a second order scalar equation. We will discuss potentials in more detail shortly but, you will see (or you have seen already), that in the world around us metals are equipotentals. We can change the potential of a metal component with, for example, a battery, so we can have two pieces of metal at different potentials if we connect them to different sides of our battery . These metal surfaces, at different potentials, provide the boundary conditions for equation 3.3. Therefore the potential determines the charge density and the electric field due to our metal surfaces. This is much more useful and practical and we’ll see why this works later.
3.4 The general solution to Poisson’s equation
As we discussed in the previous section, it is the potential that is most easily controlled in the world around us. So, if we know the potential we can calculate the field without needing to know the charge density. Nevertheless, if we know the charge density distribution we can calculate the electric potential directly. The result for the general solution of Poisson’s equation is,
Although I haven’t formally proved this result, you should be able to see how it can be obtained from the potential of single point charge (that I leave you to show),
\(\displaystyle V(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{q}{r}\)
and then using the principle of super position and spatial averaging to generate the integral over a continuous charge distribution. (The method is similar to the arguments we used for the electric field). Note also in the case of line (\(\lambda)\)) and surface (\(\sigma\)) charge densities we may use the equivalent results,
\(\displaystyle V(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_{path}\frac{\lambda(\vec{r'})}{|\vec{r}-\vec{r'}|})dl'\)
and
\(\displaystyle V(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_{surface}\frac{\sigma(\vec{r'})}{|\vec{r}-\vec{r'}|}da'\)
3.5 Electric field and potential summary
At this point we can summarise our knowledge of Electrostatics in the following triangle of relationships.
With these and given, either \(\rho\), \(\vec{E}\) or \(V\) it is possible to derive the other two quantities. However, apart from simple cases, of which a few will be given in the lectures, it is not easy to solve the various integrals to find \(\vec{E}\) or \(V\). For the electric field you will find you will either use Gauss’ law for highly symmetric charge distributions or the integral methods where it is possible to cast the solution in terms of an analytically solvable integral (again largely in cases of symmetrical charge distributions). If this is not possible you can revert to numerical methods. For the case of the electric potential there are a few other useful methods that we will explore in more detail later. Remember also you can apply the superposition principle.