# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

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

Relation es_pot 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 \ref{Gr(2.22)}. By the fundamental theorm for gradients, this also equals \[ V({\bf b}) - V({\bf a}) = \int_{\bf a}^{\bf b} ({\boldsymbol \nabla} V) \cdot d{\bf l} \] so \[ \int_{\bf a}^{\bf b} ({\boldsymbol \nabla} V) \cdot d{\bf l} = - \int_{\bf a}^{\bf b} {\bf E} \cdot d{\bf l} \] Since this is true for any choice of \({\bf a}\) and \({\bf b}\), we have

\[ {\bf E} = -{\boldsymbol \nabla} V \label{Gr(2.23)} \]

We can explicitly check \ref{Gr(2.23)} by 'extracting' the \({\boldsymbol \nabla}\) operator from the integral in the definition of the field in e.g. \ref{Gr(2.4)}: \[ {\bf E} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n q_i \frac{{\bf r} - {\bf r}_i}{|{\bf r} - {\bf r}_i|^3} = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n q_i (-1) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}_i|} = - {\boldsymbol \nabla} \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{|{\bf r} - {\bf r}_i|} = - {\boldsymbol \nabla} V \] by using \ref{Gr(1.101)} and \ref{Gr(2.27)}.

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

Validate