# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

##### Definition
ems.es.ep.d

Since this work is independent of the path chosen, we can define a function called the electrostatic potential

- Gr (2.21)

\[ V({\bf r}) \equiv -\int_{\cal O}^{\bf r} {\bf E} \cdot d{\bf l} \tag{es_pot}\label{es_pot} \] where \({\cal O}\) is some chosen reference point. The potential difference between two points is well-defined without the need to specify the reference point,

\[ V({\bf b}) - V({\bf a}) = - \int_{\bf a}^{\bf b} {\bf E} \cdot d{\bf l} \label{Gr(2.22)} \]

Thus, the electrostatic potential is interpreted as the potential energy which a unit charge would have obtained if brought to the specified point from the reference point, in other words the work you need to do on the unit charge to bring it there. Often, we put the reference point at infinity. The electrostatic potential coming from a single point charge \(q\) at the origin then becomes \[ V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{q}{r} \label{Gr(2.26)} \] The electrostatic potential moreover inherits the superposition principle from the electric field, so for a distribution of point charges \(q_i\) at positions \({\bf r}_i\), we have

\[ V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i} \frac{q_i}{|{\bf r} - {\bf r}_i|} \label{Gr(2.27)} \]

For a continuous charge density in a volume \({\cal V}\), we have

\[ V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\cal V} d\tau' \frac{\rho({\bf r}')}{|{\bf r} - {\bf r}'|}, \label{eq:V_from_rho} \]

whereas for a surface or line charge distribution, respectively,

\[ V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\cal S} da' \frac{\sigma({\bf r}')}{|{\bf r} - {\bf r}'|}, \hspace{2cm} V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\cal P} dl' \frac{\lambda({\bf r}')}{|{\bf r} - {\bf r}'|}. \label{Gr(2.30)} \]

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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