# 5.12: Electric Potential Field Due to Point Charges

- Page ID
- 6313

The electric field intensity due to a point charge \(q\) at the origin is (see Section 5.1 or 5.5)

\[{\bf E} = \hat{\bf r}\frac{q}{4\pi\epsilon r^2} \label{eEPPCE}\]

In Sections 5.8 and 5.9, it was determined that the potential difference measured from position \({\bf r}_1\) to position \({\bf r}_2\) is

\[V _ { 21 } = - \int _ { \mathbf { r } _ { 1 } } ^ { \mathbf { r } _ { 2 } } \mathbf { E } \cdot d \mathbf { l } \label{m0064_eV12}\]

This method for calculating potential difference is often a bit awkward. To see why, consider an example from circuit theory, shown in Figure \(\PageIndex{1}\). In this example, consisting of a single resistor and a ground node, we’ve identified four quantities:

- The resistance \(R\)
- The current \(I\) through the resistor
- The node voltage \(V_1\), which is the potential difference measured from ground to the left side of the resistor
- The node voltage \(V_2\), which is the potential difference measured from ground to the right side of the resistor

Let’s say we wish to calculate the potential difference \(V_{21}\) across the resistor. There are two ways this can be done:

- \(V_{21}=-IR\)
- \(V_{21}=V_2-V_1\)

The advantage of the second method is that it is not necessary to know \(I\), \(R\), or indeed anything about what is happening between the nodes; it is only necessary to know the node voltages. The point is that it is often convenient to have a common *datum* – in this example, ground – with respect to which the potential differences at all other locations of interest can be defined. When we have this, calculating potential differences reduced to simply subtracting predetermined node potentials.

* Figure \(\PageIndex{1}\): A resistor in a larger circuit, used as an example to demonstrate the concept of node voltages. Image used with permission (CC BY SA 4.0; K. *Kikkeri).

So, can we establish a datum in general electrostatic problems that works the same way? The answer is yes. The datum is arbitrarily chosen to be a sphere that encompasses the universe; i.e., a sphere with radius \(\to\infty\). Employing this choice of datum, we can use Equation \ref{m0064_eV12} to define \(V({\bf r})\), the potential at point \({\bf r}\), as follows:

\[\boxed{ V({\bf r}) \triangleq - \int_{\infty}^{\bf r} {\bf E} \cdot d{\bf l} } \label{m0064_eVP}\]

The electrical potential at a point, given by Equation \ref{m0064_eVP}, is defined as the potential difference measured beginning at a sphere of infinite radius and ending at the point \({\bf r}\). The potential obtained in this manner is with respect to the potential infinitely far away.

In the particular case where \({\bf E}\) is due to the point charge at the origin:

\[V({\bf r}) = - \int_{\infty}^{\bf r} \left[ \hat{\bf r}\frac{q}{4\pi\epsilon r^2} \right] \cdot d{\bf l}\]

The principle of *independence of path* (Section 5.9) asserts that the path of integration doesn’t matter as long as the path begins at the datum at infinity and ends at \({\bf r}\). So, we should choose the easiest such path. The radial symmetry of the problem indicates that the easiest path will be a line of constant \(\theta\) and \(\phi\), so we choose \(d{\bf l}=\hat{\bf r}dr\). Continuing: so

\[\boxed{ V({\bf r}) = + \frac{q}{4\pi\epsilon r} } \label{m0064_eV}\]

(Suggestion: Confirm that Equation \ref{m0064_eV} is dimensionally correct.) In the context of the circuit theory example above, this is the “node voltage” at \({\bf r}\) when the datum is defined to be the surface of a sphere at infinity. Subsequently, we may calculate the potential difference from any point \({\bf r}_1\) to any other point \({\bf r}_2\) as \[V_{21} = V({\bf r}_2)-V({\bf r}_1)\] and that will typically be a *lot* easier than using Equation \ref{m0064_eV12}.

It is not often that one deals with systems consisting of a single charged particle. So, for the above technique to be truly useful, we need a straightforward way to determine the potential field \(V({\bf r})\) for arbitrary distributions of charge. The first step in developing a more general expression is to determine the result for a particle located at a point \({\bf r}'\) somewhere other than the origin. Since Equation \ref{m0064_eV} depends only on charge and the distance between the field point \({\bf r}\) and \({\bf r}'\), we have

\[V({\bf r};{\bf r}') \triangleq + \frac{q'}{4\pi\epsilon \left|{\bf r}-{\bf r}'\right|} \label{m0064_eVd}\]

where, for notational consistency, we use the symbol \(q'\) to indicate the charge. Now applying superposition, the potential field due to \(N\) charges is

\[V({\bf r}) = \sum_{n=1}^N { V({\bf r};{\bf r}_n) }\]

Substituting Equation \ref{m0064_eVd} we obtain:

\[\boxed{ V({\bf r}) = \frac{1}{4\pi\epsilon} \sum_{n=1}^N { \frac{q_n}{\left|{\bf r}-{\bf r}_n\right|} } } \label{m0064_eVN}\]

Equation \ref{m0064_eVN} gives the electric potential at a specified location due to a finite number of charged particles.

The potential field due to continuous distributions of charge is addressed in Section 5.13.

## Contributors

Ellingson, Steven W. (2018) Electromagnetics, Vol. 1. Blacksburg, VA: VT Publishing. https://doi.org/10.21061/electromagnetics-vol-1 Licensed with CC BY-SA 4.0 https://creativecommons.org/licenses/by-sa/4.0. Report adoption of this book here. If you are a professor reviewing, adopting, or adapting this textbook please help us understand a little more about your use by filling out this form.