Skip to main content
\(\require{cancel}\)
Physics LibreTexts

2.2: Electrostatic Potential

  • Page ID
    17355
  • Test Charges

    An alternative way to look at electric fields from what we did in Section 1.2 is from the perspective of a test charge. The idea is to use a charged point particle as a means of measuring electric force vectors at various points in space. When the force vectors are all mapped-out, we then divide them by the charge of the point particle, and the new vectors are then the electric field vectors. A common (but somewhat strange) way to write this mathematically is:

    \[\overrightarrow E\left(\overrightarrow r\right) = \lim\limits_{q_{test}\rightarrow 0} \dfrac{\overrightarrow F_{on\;q_{test}}}{q_{test}},\;\;\;\;\;\;\text{where } \overrightarrow r\text{ is the position of }q_{test} \]

    A region around a collection of charge can similarly be tested with a charged point particle. At every point in space, the potential energy of the point charge can be measured, and then the amount of testing charge can be divided out, so that all that remains is a function of the source charges. We write it this way:

    \[V\left(\overrightarrow r\right) = \lim\limits_{q_{test}\rightarrow 0} \dfrac{U\left(q_{test}\right)}{q_{test}},\;\;\;\;\;\;\text{where } \overrightarrow r\text{ is the position vector of }q_{test} \]

    This process maps out a scalar field, since at every point in space is associated a number (not a vector, like in the case of electric field). Just as electric field vectors are not the same as force vectors, the values in this scalar field are not potential energies – indeed, this can be see even in the units of these numbers, which are joules divided by coulombs. The ratio of joules per coulomb is given its own name: volts. The scalar field we have invented this way is called electrostatic potential. Like an electric field vectors, this is a quantity that is defined at every point in space in the vicinity of some electric charge. Unlike electric field vectors, these quantities are scalars – they have no direction.

    Alert

    Possibly the most confusing thing to students new to electrostatics is use of the word "potential" in "electrostatic potential." This name derives from the fact that it is related to electric potential energy, but these quantities are very different, and the reader is admonished to keep this in mind.

    Superposition

    When there is more than one source of electric field in the vicinity of a point in space, the contributions of those sources to the field at that point can be added together. This can be seen simply from the text charge approach – clearly the forces on the test charge can be added together, and when the test charge is divided out, the sum of the electric field vectors remains. We see the same thing for electrostatic potential:

    \[U\left(q_{test}\right) = \dfrac{q_1q_{test}}{4\pi\epsilon_or_1}+\dfrac{q_2q_{test}}{4\pi\epsilon_or_2}+\dfrac{q_3q_{test}}{4\pi\epsilon_or_3}\dots \;\;\; \Rightarrow \;\;\; V\left(\overrightarrow r\right)=\dfrac{U\left(q_{test}\right)}{q_{test}}=\dfrac{q_1}{4\pi\epsilon_or_1}+\dfrac{q_2}{4\pi\epsilon_or_2}+\dfrac{q_3}{4\pi\epsilon_or_3}\dots\]

    Here \(r_i\) is the distance from the \(i^{th}\) source charge to the position in space indicated by the position vector \(\overrightarrow r\). Notice that by adopting the \(U\left(\infty\right)=0\) convention, we have also done so for the electrostatic potential. And like the potential energy, the position that we choose to call the electric potential zero is arbitrary.

    All of the things we developed for electric fields also apply to potentials, with the only difference being that potentials superpose as scalars, not vectors (which actually makes them easier to deal with in many cases). The main point is that when we have a collection of source charges – including a continuous distribution – we can define a potential at every point in space, and if we place a point charge there, we can determine its potential energy by multiplying the charge by the electric potential:

    \[U=qV\left(\overrightarrow r\right),\;\;\;\;\;\;\text{where }\overrightarrow r= \text{position vector of the charge }q\]

    The similarity with Equation 1.2.2 is obvious – we have simply replaced force and field with energy and potential.

    Bridging the Scalar and Vector Fields

    It's clear that both the scalar potential field and the vector electric field are determined by the source charge distribution, so these fields must be related to each other somehow. Indeed they are! To see how, we once again look back to our study of mechanics, where we related potential energy and force. We've already mentioned one such relation, through the computation of work:

    \[\Delta U = U_B-U_A = -W_{A\rightarrow B} = -\int\limits_A^B\overrightarrow F\cdot \overrightarrow{dl}\]

    If we just plug in \(U=qV and \overrightarrow F=q \overrightarrow E, we get a direct relation between the change of potential and the electric field:

    \[ U_A-U_B = \int\limits_A^B\overrightarrow E\cdot \overrightarrow{dl}\]

    Note that the signs have been flipped on both sides of the equation. The quantity on the left is usually referred to as the potential drop from A to B. It is actually common to use the units of \(Vm^{-1}\) for electric field, rather than our previous \(NC^{-1}\).

    Whenever we have an integral relationship like this, then (as we saw for Gauss's law), a differential (local) relation is also available. We actually saw this back in our study of mechanics, and it comes through here as well:

    \[\overrightarrow F = -\overrightarrow \nabla U \;\;\; \Rightarrow \;\;\; \overrightarrow E = -\overrightarrow \nabla V \]

    While this is interesting, the reader can be forgiven for asking what use it has. This last relation is particularly powerful for the following reason. Suppose we wish to compute the electric field of a charge distribution. Assuming we don't have a clever way of using Gauss's law to do this, we have to perform a calculation like we did back in Section 1.3. Part of what makes that computation challenging is keeping track of three different components of the electric field vector (i.e. three separate integrals). If we instead compute the potential field (one integral, with no vectors involved), we can then take derivatives (the gradient) to get the electric field. We will see how one calculates the potential field from a distribution of charge in the next section.

    Alert

    The relation between field and potential is often misunderstood, in yet another incarnation of confusing a quantity with a change in that quantity (like mistaking acceleration with velocity. Just as zero instantaneous velocity does not mean the acceleration is zero, a zero potential at a point in space does not mean that the field there is zero. Indeed, we can define the potential to be zero anywhere, no matter what the field is! It is the rate of change of the potential that determines the field, not the value of the potential.

    Example \(\PageIndex{1}\)

    In a certain region of space around the origin, the electrostatic potential field satisfies:

    \[V\left(x,y,z\right)=\alpha\; x + \beta \;y^2 + \gamma\; z^3 \nonumber\]

    Find the charge density at the origin in terms of \(\alpha\), \(\beta\), and \(\gamma\).

    Solution

    We can find the electric field from the potential field:

    \[\overrightarrow E = -\overrightarrow \nabla V = \dfrac{\partial V}{\partial x}\;\widehat i+ \dfrac{\partial V}{\partial y}\;\widehat j+ \dfrac{\partial V}{\partial z}\;\widehat k = \alpha\;\widehat i + 2\beta \;y\;\widehat j + 3\gamma \;z^2\;\widehat k \nonumber\]

    Now the divergence of the field gives us the charge density (Gauss's law):

    \[\dfrac{\rho\left(x,y,z\right)}{\epsilon_o} = \nabla \cdot \overrightarrow E = \dfrac{\partial E_x}{\partial x}+\dfrac{\partial E_y}{\partial y}+\dfrac{\partial E_z}{\partial z}=0+2\beta+6\gamma \;z \;\;\; \Rightarrow \;\;\; \boxed{\rho\left(0,0,0\right)=2\beta} \nonumber\]

    Notice that at the origin the potential is zero, but the electric field is not, nor is the charge density.

    Equipotential Surfaces

    A consequence of the gradient relation is that their relationship is geometric in nature. The first manifestation of this is that the gradient of a scalar field points in the direction where the scalar values are increasing the fastest. With the presence of the negative sign, we therefore conclude that the electric field points in the direction of the fastest descent of electric potential. Put more concisely, electric field points from high electric potential to low electric potential.

    We can demonstrate this geometrical relationship through a diagram. Let's imagine starting at a certain point in space, and measuring the potential there (after designating the zero point). Then we sample nearby points, and find a direction we can move our detector so that the potential doesn't change. If we keep following this procedure, and map the entire space where the potential doesn't change, we will find that it is a surface. As this imaginary surface exists at a single, equal potential, it called an equipotential surface. Here is a two-dimensional depiction of a collection of such surfaces:

    Figure 2.2.1 – Equipotential Surfaces

    equipotential_surfaces.png

    With a positive source charge, the field lines are pointing outward, which is indeed pointing from higher potential to lower potential, but there is something more specific that we can conclude about the geometric relationship of the field and potential. When we follow a path that remains on an equipotential, the potential never changes, so if we traverse such a path from position \(A\) to position \(B\), we find:

    \[V_A-V_B=0=\int\limits_A^B \overrightarrow E\cdot \overrightarrow{dl} \]

    Certainly the electric field is not zero everywhere we go, and the distance we travel isn't zero, so how can this integral come out to be zero? Maybe parts of it cancel other parts? No, because it happens on every single path we take, between any two points, so long as that path stays on an equipotential. The answer is that the only way this integral can be zero is if at every point on the equipotential, the electric field is perpendicular to \(\overrightarrow {dl}\). In other words, the electric field is perpendicular to equipotentials everywhere. This completes the geometric picture the field points from equipotential surfaces of higher potential toward those with lower potentials, and it does so perpendicular to the surfaces.

    That electric fields are perpendicular to equipotential surfaces sounds very familiar. We said the same thing about conducting surfaces for electrostatics. Indeed, we immediately conclude that for electrostatics, conductors are equipotentials.