Electrostatic Energy

Consider a collection of $N$ static point electric charges $q_i$ located at displacements ${\bf r}_i$. What is the electrostatic energy stored in such a collection? In other words, how much work would we have to perform in order to assemble the charges, starting from an initial state in which they are all at rest and very widely separated?

We know that a static electric field is conservative, and can consequently be written in terms of a scalar potential:

$${\bf E} = - \nabla\phi.$$ (2.65)

[See Equation (2.17).] We also know that the electrical force acting on a charge $q$ located at displacement ${\bf r}$ is written

$${\bf f} = q\, {\bf E}({\bf r}).$$ (2.66)

[See Equation (2.10).] The work that we would have to do against electrical forces in order to slowly move the charge from point $P$ to point $Q$ is simply

$$W = \int_P^Q (-{\bf f}) \cdot d{\bf r} =- q \int_P^Q {\bf E} \cdo... ...bf r} =q\int_P^Q \nabla\phi \cdot d{\bf r} = q \left[ \phi(Q) - \phi(P)\right],$$ (2.67)

where $d{\bf r}$ is an element of the path taken between the two points. (See Section 1.3.2.) The negative sign in the previous expression comes about because we would have to exert a force $-{\bf f}$ on the charge, in order to counteract the force exerted by the electric field. Recall, finally, that the scalar potential field generated by a point charge $q$ located at position ${\bf r}'$ is

$$\phi({\bf r})= \frac{1}{4\pi\,\epsilon_0} \frac{q}{\vert{\bf r} - {\bf r}'\vert}.$$ (2.68)

[See Equation (2.21).]

Let us build up our collection of charges one by one. It takes no work to bring the first charge from infinity, because there is no electric field to fight against. Let us clamp this charge in position at displacement ${\bf r}_1$. In order to bring the second charge into position at displacement ${\bf r}_2$, we have to do work against the electric field generated by the first charge. According to Equations (2.67) and (2.68), this work is given by

$$W_2 = \frac{1}{4\pi\,\epsilon_0} \frac{q_2 \,q_1}{\vert{\bf r}_2 - {\bf r}_1\vert}.$$ (2.69)

Let us now bring the third charge into position. Because electric fields and scalar potentials are superposable, the work done while moving the third charge from infinity to displacement ${\bf r}_3$ is simply the sum of the works done against the electric fields generated by charges 1 and 2, taken in isolation:

$$W_3 = \frac{1}{4\pi\,\epsilon_0} \left( \frac{q_3\, q_1}{\vert{\b... ...3 - {\bf r}_1\vert} + \frac{q_3\, q_2}{\vert{\bf r}_3 - {\bf r}_2\vert}\right).$$ (2.70)

Thus, the total work done in assembling the collection of three charges is given by

$$W = \frac{1}{4\pi\,\epsilon_0}\left( \frac{q_2 \,q_1}{\vert{\bf r... ...3 - {\bf r}_1\vert} + \frac{q_3 \,q_2}{\vert{\bf r}_3 - {\bf r}_2\vert}\right).$$ (2.71)

This result can easily be generalized to a collection of $N$ charges:

$$W = \frac{1}{4\pi\,\epsilon_0} \sum_{i=1,N} \sum_{j=1,N}^{j<i}\frac{q_i \,q_j}{\vert{\bf r}_i -{\bf r}_j\vert}.$$ (2.72)

The restriction that $j$ must be less than $i$ makes the previous summation rather messy. If we were to sum without restriction (other than $j\neq i$) then each pair of charges would be counted twice. It is convenient to do just this, and then to divide the result by two. Thus, we obtain

$$W =\frac{1}{2} \frac{1}{4\pi\,\epsilon_0} \sum_{i=1,N} \sum_{j=1,N}^{j\neq i}\frac{q_i\, q_j}{\vert{\bf r}_i-{\bf r}_j\vert}.$$ (2.73)

This is the electric potential energy (i.e., the difference between the total energy and the kinetic energy) of a collection of point electric charges. We can think of this quantity as the work required to bring stationary charges from infinity and assemble them in the required formation. Alternatively, it is the kinetic energy that would be released if the collection were dissolved, and the charges returned to infinity. But where is this potential energy stored? Let us investigate further.

Equation (2.73) can be written

$$W = \frac{1}{2} \sum_{i=1,N} q_i \,\phi_i,$$ (2.74)

where

$$\phi_i = \frac{1}{4\pi\,\epsilon_0}\sum_{j=1,N}^{j\neq i} \frac{q_j}{\vert{\bf r}_i - {\bf r}_j\vert}$$ (2.75)

is the scalar potential experienced by the $i$ th charge due to the other charges in the distribution. [See Equation (2.20).]

Let us now consider the potential energy of a continuous charge distribution. It is tempting to write

$$W = \frac{1}{2} \int \rho\,\phi\,dV,$$ (2.76)

by analogy with Equations (2.74) and (2.75), where

$$\phi({\bf r}) = \frac{1}{4\pi\,\epsilon_0}\int \frac{\rho({\bf r}')} {\vert{\bf r} - {\bf r}'\vert}\,dV'$$ (2.77)

is the familiar scalar potential generated by a continuous charge distribution of charge density $\rho({\bf r})$ [see Equation (2.18)], and where the volume integrals are over all space. Let us try this scheme out. We know from Equation (2.54) that

$$\rho = \epsilon_0 \,\nabla\cdot {\bf E},$$ (2.78)

so Equation (2.76) can be written

$$W =\frac{\epsilon_0}{2}\int\phi\, \nabla\cdot{\bf E} \,dV.$$ (2.79)

Now,

$$\nabla \cdot({\bf E} \,\phi) \equiv \phi\,\nabla\cdot{\bf E} +{\bf E} \cdot \nabla\phi.$$ (2.80)

(See Section A.24.) However, $\nabla\phi = - {\bf E}$, so we obtain

$$W = \frac{\epsilon_0}{2} \left[\int \nabla\cdot ({\bf E}\,\phi)\,dV + \int E^2\,dV\right]$$ (2.81)

Application of the divergence theorem (see Section A.20) gives

$$W = \frac{\epsilon_0}{2} \left(\oint_S \phi\,{\bf E} \cdot d{\bf S}+ \int_V E^2\,dV\right),$$ (2.82)

where $V$ is some volume that encloses all of the charges, and $S$ is its bounding surface. Let us assume that $V$ is a sphere, centered on the origin, and let us take the limit in which the radius $r$ of this sphere goes to infinity. We know that, in general, the electric field at large distances from a bounded charge distribution looks like the field of a point charge, and, therefore, falls off like $1/r^2$. Likewise, the potential falls off like $1/r$. (See Section 2.1.7.) However, the surface area of the sphere increases like $r^2$. Hence, it is clear that, in the limit as $r\rightarrow \infty$, the surface integral in Equation (2.82) falls off like $1/r$, and is consequently zero. Thus, Equation (2.82) reduces to

$$W = \frac{\epsilon_0}{2} \int E^2\,dV,$$ (2.83)

where the volume integral is over all space. This is a very interesting result. It tells us that the potential energy of a continuous charge distribution is stored in the electric field generated by that distribution. Of course, we now have to assume that an electric field possesses an energy density

$$U = \frac{\epsilon_0}{2} \,E^2.$$ (2.84)

Incidentally, the fact that an electric field possess an energy density demonstrates that it has a real physical existence, and is not just an aid to the calculation of electrostatic forces.

We can easily check that Equation (2.83) is correct. Suppose that we have an electric charge $Q$ that is uniformly distributed within a sphere of radius $a$ centered on the origin. Let us imagine building up this charge distribution from a succession of thin spherical layers of infinitesimal thickness. At each stage, we gather a small amount of charge $dq$ from infinity, and spread it over the surface of the sphere in a thin layer extending from $r$ to $r+dr$. We continue this process until the final radius of the sphere is $a$. If $q(r)$ is the sphere's charge when it has attained radius $r$ then the work done in bringing a charge $dq$ to its surface is

$$dW = \frac{1}{4\pi\,\epsilon_0} \frac{ q(r)\,dq}{r}.$$ (2.85)

This follows from Equation (2.69), because the electric field generated outside a spherical charge distribution is the same as that of a point charge $q(r)$ located at its geometric center ($r=0$). (See Section 2.1.7.) If the constant charge density of the sphere is $\rho$ then

$$q(r) = \frac{4\pi}{3} \,r^3\,\rho,$$ (2.86)

and

$$dq = 4\pi\, r^2\, \rho\,dr.$$ (2.87)

Thus, Equation (2.85) becomes

$$dW = \frac{4\pi}{3\,\epsilon_0}\, \rho^2\, r^4\,dr.$$ (2.88)

The total work needed to build up the sphere from zero radius to radius $a$ is plainly

$$W = \frac{4\pi}{3\,\epsilon_0}\, \rho^2 \int_0^a r^4\,dr = \frac{4\pi}{15\,\epsilon_0}\, \rho^2\, a^5.$$ (2.89)

This can also be written in terms of the total charge $Q = (4\pi/3)\,a^3\, \rho$ as

$$W = \frac{3}{5} \frac{Q^2}{4\pi\,\epsilon_0\, a}.$$ (2.90)

Now that we have evaluated the potential energy of a spherical charge distribution by the direct method, let us work it out using Equation (2.83). We shall assume that the electric field is both radial and spherically symmetric, so that ${\bf E} = E_r(r)\,{\bf e}_r$. Here, $r$ is a standard spherical polar coordinate. (See Section A.23.) Application of Gauss's law,

$$\oint_S {\bf E} \cdot d{\bf S} = \frac{1}{\epsilon_0} \int_V \rho \,dV,$$ (2.91)

where $V$ is a sphere of radius $r$, centered on the origin, gives

$$E_r(r) = \frac{Q}{4\pi\,\epsilon_0} \frac{r}{a^3}$$ (2.92)

for $r<a$, and

$$E_r(r) = \frac{Q}{4\pi\,\epsilon_0\,r^2}$$ (2.93)

for $r\geq a$. Equations (2.83), (2.92), and (2.93) yield

$$W = \frac{Q^2}{8\pi\,\epsilon_0} \left( \frac{1}{a^6} \int_0^a r^4\,dr + \int_a^\infty \frac{dr}{r^2} \right),$$ (2.94)

which reduces to

$$W = \frac{Q^2}{8\pi\,\epsilon_0\, a} \left( \frac{1}{5} + 1\right)= \frac{3}{5} \frac{Q^2}{4\pi\,\epsilon_0 \,a}.$$ (2.95)

Thus, Equation (2.83) gives the correct answer.

The reason that we have checked Equation (2.83) so carefully is that, on close inspection, it is found to be inconsistent with Equation (2.74), from which it was supposedly derived. For instance, the energy given by Equation (2.83) is manifestly positive definite, whereas the energy given by Equation (2.74) can be negative (it is certainly negative for a collection of two point charges of opposite sign). The inconsistency was introduced into our analysis when we replaced Equation (2.75) by Equation (2.77). In Equation (2.75), the self-interaction of the $i$ th charge with its own electric field is specifically excluded, whereas it is included in Equation (2.77). Thus, the potential energies (2.74) and (2.83) are different because in the former we start from ready-made point charges, whereas in the latter we build up the whole charge distribution from scratch. Hence, if we were to work out the potential energy of a point charge distribution using Equation (2.83) then we would obtain the energy (2.74) plus the energy required to assemble the point charges. What is the energy required to assemble a point electric charge? In fact, it is infinite. To see this, let us suppose, for the sake of argument, that our point charges actually consist of electric charge uniformly distributed in small spheres of radius $b$. According to Equation (2.90), the energy required to assemble the $i$ th point charge is

$$W_i = \frac{3}{5} \frac{q_i^{\,2}}{4\pi\,\epsilon_0 \,b}.$$ (2.96)

We can think of this as the self-energy of the $i$ th charge. Thus, we can write

$$W = \frac{\epsilon_0}{2} \int E^2 \,dV = \frac{1}{2} \sum_{i=1,N} q_i \,\phi_i + \sum_{i=1,N} W_i$$ (2.97)

which enables us to reconcile Equations (2.74) and (2.83). Unfortunately, if our point charges really are point charges then $b\rightarrow 0$, and the self-energy of each charge becomes infinite. Thus, the potential energies predicted by Equations (2.74) and (2.83) differ by an infinite amount. What does this all mean? We have to conclude that the idea of locating electrostatic potential energy in the electric field runs into conceptual difficulties in the presence of point electric charges. One way out of this dilemma would be to say that elementary electric charges, such as protons and electrons, are not point objects, but instead have finite spatial extents. Regrettably, although protons have finite spatial extents (of about $10^{-15}\,{\rm m}$), electrons really do seem to be point objects.