##### Field in terms of the potential
ems.es.ep.fp

Relation p gives us the potential in terms of the field. Let us now try to find a relationship giving the field in terms of the potential. For this, we look again at p_diff. By the fundamental theorm for gradients, this also equals

\[ \phi({\bf b}) - \phi({\bf a}) = \int_{\bf a}^{\bf b} ({\boldsymbol \nabla} \phi) \cdot d{\bf l} \]

so

\[ \int_{\bf a}^{\bf b} ({\boldsymbol \nabla} \phi) \cdot d{\bf l} = - \int_{\bf a}^{\bf b} {\bf E} \cdot d{\bf l} \]

Since this is true for any choice of \({\bf a}\) and \({\bf b}\), we have

- Gr (2.23)

\[ {\bf E} = -{\boldsymbol \nabla} \phi \tag{Emgp}\label{Emgp} \]

We can explicitly check Emgp by 'extracting' the \({\boldsymbol \nabla}\) operator from the integral in the definition of the field in e.g. E_pcd :

\begin{align} {\bf E} ({\bf r}) &= \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n q_i \frac{{\bf r} - {\bf r}_i}{|{\bf r} - {\bf r}_i|^3} = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n q_i (-1) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}_i|} \nonumber \\ &= - {\boldsymbol \nabla} \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{|{\bf r} - {\bf r}_i|} = - {\boldsymbol \nabla} \phi \end{align}

by using div1or and p_pcd.

When compared to the electrostatic field, the electrostatic potential has a substantial practical advantage: in view of Emgp, it's much simpler to first calculate the potential (which is a scalar field, so a single function), and to then calculate the electrostatic field (which is a vector field, i.e. three functions).

You might wonder how a vector field (3 components, so 3 functions) can be generated by a scalar field (one component, so one function)? Well it's important to count your degrees of freedom properly. Equation curlE0, when written out in components, actually represents 3 equations constraining the 3 components of the electric field. We might thus think that we have no freedom left, but a function which is a gradient of a scalar automatically fulfills the curlless constraint, which is the only freedom left.

About the reference point for defining the potential: once again, this isn't physical, it's purely practical (expect perhaps if your charge distribution extends to infinity). Feel free to choose another reference point if you want or need to.

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42