4 Electrostatic fields - boundary conditions and energy
4.1 Boundary conditions
In our consideration of electric (and magnetic) fields an important consideration is what happens to the fields as we pass across boundaries of charge distribution (or current densities). I am going to state the results here. It is not difficult to show the results using Gauss’ law and the arguments are presented in Griffiths if you wish to see the details.
In the most compact form we may write the boundary conditions for the electric field as,
where \(\hat{\vec{n}}\) is the normal to the surface. In words, this says there is no change in \(\vec{E}\) in the direction parallel to the surface, as I pass through the surface. On the other hand, the electric field perpendicular to surface goes through a discontinuous change (it changes sign) as you go through the surface.
As you approach the surface the value of the potentials above and below the surface will converge to the be same at the surface, i.e
\(V_{above}=V_{below}\)
but if we think of the gradient of the potential at the surface (\(\nabla V\)) then this is equivalent to \(\vec{E}\) at the surface so for potential we have,
where we have defined the ‘normal’ derivative as,
\(\displaystyle \frac{\partial V}{\partial n}=\nabla V.\hat{\vec{n}}\)
i.e the component of \(\nabla V\) in the direction of \(\hat{\vec{n}}\).
4.2 Work and energy
From your mechanics course in the first year you will have seen that the work done against a force (and hence a store of poetential energy) is given by,
\(\displaystyle W= \int_a^b \vec{F}.d\vec{l}\)
If we apply this the result to the work needed to move a charge in an electric field we get,
\(\displaystyle W = Q\int_a ^b \vec{E}.d\vec{l}=Q\left[V(b)-V(a)\right] = Q \Delta V\)
Hence we can also use the potential difference to describe the work done to move a charge \(Q\) in the field.
So what is the energy stored in a distribution of point charges? Well we can imagine building up an assembly of charges by bringing each one suequentially in from infinity. There is no energy needed to bring the first charge in. For the second charge we need,
\(\displaystyle W_2 = \frac{q_1q_2}{4\pi\epsilon_0 r_{12}}\),
the third one needs to do work against the other two charges (we’re OK to use superposition here as we can treat the fields from the first two charges as the sum of the individual charges), so,
\(\displaystyle W_3 = \frac{q_3q_2}{4\pi\epsilon_0 r_{32}}+\frac{q_3q_1}{4\pi\epsilon_0 r_{31}}\),
\(\displaystyle W_4 = \frac{q_4}{4\pi\epsilon_0}\left( \frac{q_3}{r_{43}}+\frac{q_2}{r_{42}}+\frac{q_1}{r_{41}}\right)\).
For \(n\) charges this can be written as,
\(\displaystyle W_{total}=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^n \sum_{j\gt i}^n \frac{q_iq_j}{r_{ij}}=\frac{1}{8\pi\epsilon_0}\sum_{i=1}^n \sum_{j\ne i}^n \frac{q_iq_j}{r_{ij}}\).
Note, in the second expression we’ve allowed counting a pair twice (i.e. i-j and j-i) and adjusted by dividing the result by 2. Now we can write,
\(\displaystyle W=\frac{1}{2}\sum_{i=1}^n q_i\left(\sum_{j,j \ne i}^n \frac{1}{4\pi\epsilon_0}\frac{q_j}{r_{ij}}\right)\)
or, finally,
Here we note that \(V(\vec{r_i})\) is the potential from all the other charges at the point \(\vec{r_i}\), the position of \(q_i\).
If we extend this argument to a continuous distribution of charge the 4.3 becomes
There is an interesting way to get the energy stored in terms of the electric field (see Griffiths section 2.4.3) to give,
or we can also write \(\vec{E}.\vec{E} = |E|^2\) in the kernel of the integral.
So we have two ways of thinking about the way the energy is stored, as the work done in assembling the charges or as associated with the field generated (the field energy) of the charges.
IMPORTANT. We can’t use the superposition principle to sum energies. For example take two separate charge distributions for which we have calculated their energy as \(W_1=E_1^2\) and \(W_2=E_2^2\). We might be tempted to say that bringing them together gives a total energy \(W_1+W_2\). Well there is clearly extra work done bringing them together, or in terms of the field, the new field is \(\vec{E}_{new}=\vec{E_1}+\vec{E_2}\) so in 4.5 the kernel becomes \(E_{new}^2\) which is not the same as \((\vec{E_1}+\vec{E_2}).(\vec{E_1}+\vec{E_2})=E_1^2+E_2^2 +2\vec{E_1}.\vec{E_2}\) where the cross term is not zero.