Electric Potential Energy

Consider a charge $q$ placed in a uniform electric field ${\bf E}$ (e.g., the field between two oppositely charged, parallel conducting plates). Suppose that we very slowly displace the charge by a vector displacement ${\bf r}$ in a straight-line. How much work must we perform in order to achieve this? Well, the force ${\bf F}$ we must exert on the charge is equal and opposite to the electrostatic force $q\,{\bf E}$ experienced by the charge (i.e., we must overcome the electrostatic force on the charge before we are free to move it around). The amount of work $W$ we would perform in displacing the charge is simply the product of the force ${\bf F}=- q\,{\bf E}$ we exert, and the displacement of the charge in the direction of this force. Suppose that the displacement vector subtends an angle $\theta$ with the electric field ${\bf E}$. It follows that

$$W = {\bf F} \!\cdot\! {\bf r} = -q\,{\bf E}\!\cdot\! {\bf r} = -q\, E \,r\, \cos\theta.$$ (76)

Thus, if we move a positive charge in the direction of the electric field then we do negative work (i.e., we gain energy). Likewise, if we move a positive charge in the opposite direction to the electric field then we do positive work (i.e., we lose energy).

Consider a set of point charges, distributed in space, which are rigidly clamped in position so that they cannot move. We already know how to calculate the electric field ${\bf E}$ generated by such a charge distribution (see Sect. 3). In general, this electric field is going to be non-uniform. Suppose that we place a charge $q$ in the field, at point $A$, say, and then slowly move it along some curved path to a different point $B$. How much work must we perform in order to achieve this? Let us split up the charge's path from point $A$ to point $B$ into a series of $N$ straight-line segments, where the $i$th segment is of length ${\mit\Delta}r_i$ and subtends an angle $\theta_i$ with the local electric field $E_i$. If we make $N$ sufficiently large then we can adequately represent any curved path between $A$ and $B$, and we can also ensure that $E_i$ is approximately uniform along the $i$th path segment. By a simple generalization of Eq. (76), the work $W$ we must perform in moving the charge from point $A$ to point $B$ is

$$W =- q\sum_{i=1}^N E_i\,{\mit\Delta}r_i\,\cos\theta_i.$$ (77)

Finally, taking the limit in which $N$ goes to infinity, the right-hand side of the above expression becomes a line integral:

$$W = - q\,\int_A^B{\bf E}\cdot d{\bf r}.$$ (78)

Let us now consider the special case where point $B$ is identical with point $A$. In other words, the case in which we move the charge around a closed loop in the electric field. How much work must we perform in order to achieve this? It is, in fact, possible to prove, using rather high-powered mathematics, that the net work performed when a charge is moved around a closed loop in an electric field generated by fixed charges is zero. However, we do not need to be mathematical geniuses to appreciate that this is a sensible result. Suppose, for the sake of argument, that the net work performed when we take a charge around some closed loop in an electric field is non-zero. In other words, we lose energy every time we take the charge around the loop in one direction, but gain energy every time we take the charge around the loop in the opposite direction. This follows from Eq. (77), because when we switch the direction of circulation around the loop the electric field $E_i$ on the $i$th path segment is unaffected, but, since the charge is moving along the segment in the opposite direction, $\theta_i \rightarrow 180^\circ +\theta_i$, and, hence, $\cos\theta_i\rightarrow -\cos\theta_i$. Let us choose to move the charge around the loop in the direction in which we gain energy. So, we move the charge once around the loop, and we gain a certain amount of energy in the process. Where does this energy come from? Let us consider the possibilities. Maybe the electric field of the movable charge does negative work on the fixed charges, so that the latter charges lose energy in order to compensate for the energy which we gain? But, the fixed charges cannot move, and so it is impossible to do work on them. Maybe the electric field loses energy in order to compensate for the energy which we gain? (Recall, from the previous section, that there is an energy associated with an electric field which fills space). But, all of the charges (i.e., the fixed charges and the movable charge) are in the same position before and after we take the movable charge around the loop, and so the electric field is the same before and after (since, by Coulomb's law, the electric field only depends on the positions and magnitudes of the charges), and, hence, the energy of the field must be the same before and after. Thus, we have a situation in which we take a charge around a closed loop in an electric field, and gain energy in the process, but nothing loses energy. In other words, the energy appears out of ``thin air,'' which clearly violates the first law of thermodynamics. The only way in which we can avoid this absurd conclusion is if we adopt the following rule:

The work done in taking a charge around a closed loop in an electric field generated by fixed charges is zero.

One corollary of the above rule is that the work done in moving a charge between two points $A$ and $B$ in such an electric field is independent of the path taken between these points. This is easily proved. Consider two different paths, 1 and 2, between points $A$ and $B$. Let the work done in taking the charge from $A$ to $B$ along path 1 be $W_1$, and the work done in taking the charge from $A$ to $B$ along path 2 be $W_2$. Let us take the charge from $A$ to $B$ along path 1, and then from $B$ to $A$ along path 2. The net work done in taking the charge around this closed loop is $W_1-W_2$. Since we know this work must be zero, it immediately follows that $W_1=W_2$. Thus, we have a new rule:

The work done in taking a charge between two points in an electric field generated by fixed charges is independent of the path taken between the points.

A force which has the special property that the work done in overcoming it in order to move a body between two points in space is independent of the path taken between these points is called a conservative force. The electrostatic force between stationary charges is clearly a conservative force. Another example of a conservative force is the force of gravity (the work done in lifting a mass only depends on the difference in height between the beginning and end points, and not on the path taken between these points). Friction is an obvious example of a non-conservative force.

Suppose that we move a charge $q$ very slowly from point $A$ to point $B$ in an electric field generated by fixed charges. The work $W$ which we must perform in order to achieve this can be calculated using Eq. (78). Since we lose the energy $W$ as the charge moves from $A$ to $B$, something must gain this energy. Let us, for the moment, suppose that this something is the charge. Thus, the charge gains the energy $W$ when we move it from point $A$ to point $B$. What is the nature of this energy gain? It certainly is not a gain in kinetic energy, since we are moving the particle slowly: i.e., such that it always possesses negligible kinetic energy. In fact, if we think carefully, we can see that the gain in energy of the charge depends only on its position. For a fixed starting point $A$, the work $W$ done in taking the charge from point $A$ to point $B$ depends only on the position of point $B$, and not, for instance, on the route taken between $A$ and $B$. We usually call energy a body possess by virtue of its position potential energy: e.g., a mass has a certain gravitational potential energy which depends on its height above the ground. Thus, we can say that when a charge $q$ is taken from point $A$ to point $B$ in an electric field generated by fixed charges its electric potential energy $P$ increases by an amount $W$:

$$P_B - P_A = W.$$ (79)

Here, $P_A$ denotes the electric potential energy of the charge at point $A$, etc. This definition uniquely defines the difference in the potential energy between points $A$ and $B$ (since $W$ is independent of the path taken between these points), but the absolute value of the potential energy at point $A$ remains arbitrary.

We have seen that when a charged particle is taken from point $A$ to point $B$ in an electric field its electric potential energy increases by the amount specified in Eq. (79). But, how does the particle store this energy? In fact, the particle does not store the energy at all. Instead, the energy is stored in the electric field surrounding the particle. It is possible to calculate this increase in the energy of the field directly (once we know the formula which links the energy density of an electric field to the magnitude of the field), but it is a very tedious calculation. It is far easier to calculate the work $W$ done in taking the charge from point $A$ to point $B$, via Eq. (78), and then use the conservation of energy to conclude that the energy of the electric field must have increased by an amount $W$. The fact that we conventionally ascribe this energy increase to the particle, rather than the field, via the concept of electric potential energy, does not matter for all practical purposes. For instance, we call the money which we have in the bank ``ours,'' despite the fact that the bank has possession of it, because we know that the bank will return the money to us any time we ask them. Likewise, when we move a charged particle in an electric field from point $A$ to point $B$ then the energy of the field increases by an amount $W$ (the work which we perform in moving the particle from $A$ to $B$), but we can safely associate this energy increase with the particle because we know that if the particle is moved back to point $A$ then the field will give all of the energy back to the particle without loss. Incidentally, we can be sure that the field returns the energy to the particle without loss because if there were any loss then this would imply that non-zero work is done in taking a charged particle around a closed loop in an electric field generated by fixed charges. We call a force-field which stores energy without loss a conservative field. Thus, an electric field, or rather an electrostatic field (i.e., an electric field generated by stationary charges), is conservative. It should be clear, from the above discussion, that the concept of potential energy is only meaningful if the field which generates the force in question is conservative.

A gravitational field is another example of a conservative field. It turns out that when we lift a body through a certain height the increase in gravitational potential energy of the body is actually stored in the surrounding gravitational field (i.e., in the distortions of space-time around the body). It is possible to determine the increase in energy of the gravitational field directly, but it is a very difficult calculation involving General Relativity. On the other hand, it is very easy to calculate the work done in lifting the body. Thus, it is convenient to calculate the increase in the energy of the field from the work done, and then to ascribe this energy increase to the body, via the concept of gravitational potential energy.

In conclusion, we can evaluate the increase in electric potential energy of a charge when it is taken between two different points in an electrostatic field from the work done in moving the charge between these two points. The energy is actually stored in the electric field surrounding the charge, but we can safely ascribe this energy to the charge, because we know that the field stores the energy without loss, and will return the energy to the charge whenever it is required to do so by the laws of Physics.