The Relationship between \EE V and \rho

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Section 20.1 The Relationship between \(\EE\text{,}\) \(V\text{,}\) and \(\rho\)

Starting with the electrostatic potential for a point charge, we used the superposition principle in Section 9.5 to obtain an integral formula for the potential due to any continuous charge distribution, namely

\begin{equation} V(\rr) = \int\limits_{\textrm{all charge}} {1\over 4\pi\epsilon_0} {\rho(\rrp)\, \delta\tau'\over|\rr-\rrp|} .\tag{20.1.1} \end{equation}

We also obtained a similar formula for the electric field in in Section 11.6, namely

\begin{equation} \EE(\rr) = \int\limits_{\textrm{all charge}} {1\over 4\pi\epsilon_0} {\rho(\rrp)\,(\rr-\rrp)\,\delta\tau'\over|\rr-\rrp|^3} .\tag{20.1.2} \end{equation}

These relationships correspond, respectively, to the arrows numbered 1 and 2 in Figure 20.1.1.

Figure 20.1.1. The relationships between \(V\text{,}\) \(\EE\text{,}\) and \(\rho\text{.}\) Each numbered arrow is discussed in the text; moving down the diagram corresponds to differentiation.

Furthermore, as discussed in Section 11.2, the electric field is just (minus) the gradient of the potential, that is

\begin{equation} \EE(\rr)=-\grad V(\rr)\tag{20.1.3} \end{equation}

which, as discussed in Section 16.8, can be inverted to obtain the potential as an integral of the electric field, namely

\begin{equation} V(\rr) = -\int\limits_{\rr_0}^{\rr}\EE(\rr')\cdot d\rr'\tag{20.1.4} \end{equation}

where \(\rr_0\) is the point where you have chosen the potential to be zero and the integral is along any (definite) path from \(\rr_0\) to \(\rr\text{.}\) These last two relationships are arrows 5 and 3, respectively. We have also seen in Section 15.1 that the charge density can be recovered as the divergence of the electric field, namely

\begin{equation} \frac{\rho}{\epsilon_0}=\grad\cdot\EE\tag{20.1.5} \end{equation}

which is arrow number 4.

It remains to show how to recover the charge density from the potential (arrow number 6), at which point we are able to compute any one of the quantities \(\EE\text{,}\) \(V\text{,}\) and \(\rho\) from any of the others.

But from Equation (20.1.3) and Equation (20.1.5) it is easy to compute

\begin{equation} \frac{\rho}{\epsilon_0} = -\grad\cdot\grad V\tag{20.1.6} \end{equation}

which is the desired relation; this is Poisson's equation.

The special case of Poisson's equation with no source, namely

\begin{equation} \nabla^2 V = 0\tag{20.1.7} \end{equation}

is called Laplace's equation. This equation arises in many contexts, not just in electrodynamics.