Electric Potential Energy

Suppose that a particle of electric charge $q$ is taken along some path from point $P$ to point $Q$. The net work done on the particle by electrical forces is

$$W= \int_P^Q {\bf f}\cdot d{\bf r},$$ (2.22)

where ${\bf f}({\bf r})$ is the electrical force, and $d{\bf r}$ is an element of the path. (See Section 1.3.2.) Making use of Equations (2.10) and (2.17), we obtain

$$W = q \int_P^Q {\bf E}\cdot d{\bf r} = - q\int_P^Q \nabla\phi\cdot d{\bf r} = -q \left[\phi(Q)- \phi(P)\right].$$ (2.23)

(See Section A.18.) Thus, the work done on the particle is simply minus the product of its charge and the difference in electric potential between the end point and the beginning point. This work is clearly independent of the path taken between points $P$ and $Q$. Thus, we conclude that an electric field generated by stationary charges is an example of a conservative force field. (See Section 1.3.3.) The work done on the particle when it is taken around a closed loop is zero, so

$$\oint_C {\bf E}\cdot d{\bf r} = 0$$ (2.24)

for any closed loop $C$. This implies from the curl theorem that

$$\nabla\times {\bf E} = {\bf0}$$ (2.25)

for any electric field generated by stationary charges. (See Section A.22.) Equation (2.25) also follows directly from Equation (2.17), because $\nabla\times\nabla \phi \equiv {\bf0}$ for any scalar potential $\phi$. (See Section A.22.)

The SI unit of electric potential is the volt (V), which is equivalent to a joule per coulomb. Thus, according to Equation (2.23), the electrical work done on a particle when it is taken between two points is the product of minus its electric charge and the voltage difference between the points.

We are familiar with the idea that a particle moving in a gravitational field possesses potential energy, as well as kinetic energy. (See Section 1.3.5.) If the particle moves from point $P$ to a lower point $Q$ then the gravitational field does work on the particle, causing its kinetic energy to increase. The increase in kinetic energy of the particle is balanced by an equal decrease in its potential energy, so that the overall energy of the particle is a conserved quantity. Therefore, the work done on the particle as it moves from $P$ to $Q$ is minus the difference in its gravitational potential energy between points $Q$ and $P$. Of course, it only makes sense to talk about gravitational potential energy because the gravitational field is conservative. Thus, the work done in taking a particle between two points is path independent, and, therefore, well defined. This implies that the difference in potential energy of the particle between the beginning and end points is also well defined. We have already seen that an electric field generated by stationary charges is conservative. In follows that we can define an electric potential energy of a particle moving in such a field. By analogy with gravitational fields, the work done in taking a particle of electric charge $q$ from point $P$ to point $Q$ is equal to minus the difference in the electric potential energy of the particle between points $Q$ and $P$. It follows from Equation (2.23) that the electric potential energy of the particle at a general point $Q$, relative to some reference point $P$ (where the potential energy is set to zero), is given by

$$W(Q)= q \,\phi(Q),$$ (2.26)

where $\phi(Q)$ is the electric scalar potential at point $Q$. Free particles tend to move down gradients of potential energy, in order to attain a minimum potential energy state. (See Section 1.3.6.) Thus, free particles in the Earth's gravitational field tend to fall downward. Likewise, positive charges moving in an electric field tend to migrate towards regions with the most negative voltage, and vice versa for negative charges.

The scalar electric potential is undefined to an additive constant. In other words, the transformation

$$\phi({\bf r}) \rightarrow \phi({\bf r}) + c,$$ (2.27)

where $c$ is a spatial constant, leaves the electric field unchanged according to Equation (2.17). The scalar potential can be fixed unambiguously by specifying its value at a single point. The usual convention is to say that the potential is zero at infinity. This convention is implicit in Equation (2.18), where it can be seen that $\phi\rightarrow 0$ as $\vert{\bf r}\vert \rightarrow\infty$, provided that the total electric charge $\int_{V'} \rho({\bf r}')\,dV'$ is finite.