# Electrostatic Potential Energy

Consider a skalar potential field $\Phi(\mathbf{x}_i)$ that vanishes at infinity. If we bring a charge $q_i$, initially infinitly far away to the point $\mathbf{x}_i$ then the potential energy $W_i$ (work done by the charge) is \begin{align} W_i = q_i \cdot \Phi(\mathbf{x}_i) \end{align} For potentials arise form several charges $q_j$ with $j=1,2, \dots, n-1$ at $\mathbf{x}_j$, the potential is \begin{align} \Phi(\mathbf{x}_i)=\sum_{j=1}^{n-1} \frac{q_j}{|\mathbf{x}_i-\mathbf{x}_j|} \end{align} and hence the potential energy of the charge $q_i$ is \begin{align} W_i=q_i \cdot \sum_{j=1}^{n-1} \frac{q_j}{|\mathbf{x}_i-\mathbf{x}_j|} \end{align} The total energy of all interacting particles is given by the sum \begin{align} W= \sum_{i=1}^{n}\sum_{j<i} \frac{q_i q_j}{|\mathbf{x}_i-\mathbf{x}_j|} \end{align} For the first particle $i=1$ no work is done. For the second particle $i=2$ moves in the potential of the first particle $j=1$. The third particle in the potential of the previous two $j=1,2$ and so on. If we sum instead over all $j$'s, meaning that we assume all charges present for each charge added, we get twice the total potential. Thus summing over all $j$'s we write the total potential \begin{align} W= \frac{1}{2} \sum_{i}\sum_{j \not= i} \frac{q_i q_j}{|\mathbf{x}_i-\mathbf{x}_j|} \tag{1} \end{align} We can generalize this expression for continuous charge distributions $\rho(\mathbf{x})$ and $\rho(\mathbf{x}')$ \begin{align} W= \frac{1}{2} \int \int \frac{\rho(\mathbf{x}) \rho(\mathbf{x})'}{|\mathbf{x}-\mathbf{x}'|} d\mathbf{x} \, d\mathbf{x}' \end{align} which by $\Phi(\mathbf{x})=\int \frac{\rho(\mathbf{x})'}{|\mathbf{x}-\mathbf{x}'|} d^3\mathbf{x}'$ can be written

\begin{align} W= \frac{1}{2} \int \rho(\mathbf{x})\Phi(\mathbf{x}) d\mathbf{x} \end{align}

Using Poisson's Equation $\nabla^2\Phi= -4 \pi \rho(\mathbf{x})$ we eliminate $\rho(\mathbf{x})$ from the equation \begin{align} W= \frac{1}{8 \pi} \int \Phi(\mathbf{x}) \nabla^2\Phi(\mathbf{x}) d\mathbf{x} \end{align} By partial integration \begin{align} \int \Phi(\mathbf{x}) \nabla^2\Phi(\mathbf{x}) d^3\mathbf{x}= \underbrace{\left[ \Phi(\mathbf{x}) \nabla \Phi(\mathbf{x}) \right]_{-\infty}^{\infty}}_{=0} - \int \nabla \Phi(\mathbf{x}) \nabla \Phi(\mathbf{x}) d\mathbf{x} \end{align} we find the relation

\begin{align} W= \frac{1}{8 \pi} \int |\nabla \Phi(\mathbf{x})|^2 d^3\mathbf{x}=\frac{1}{8 \pi} \int |\mathbf{E}(\mathbf{x})|^2 d\mathbf{x} \tag{2} \end{align}

and identify the energy density $w$ as \begin{align} w= \frac{1}{8 \pi} |\mathbf{E}(\mathbf{x})|^2 \end{align} which is a positive definite value $w \leq 0$, hence the Integral for $W$ can never be positive. This is against the intuition that the potential of two opositely charged particles can of course be negative. This seemingly contradicts our finding $w \leq 0$. The problem is resolved by recognizing that the self-energy contained in (2) is not included in (1). To make this more clear, consider two point charges $q_1$ and $q_2$ at position $\mathbf{x}_1$ and $\mathbf{x}_2$. The electric field at the point $P$ specified by the vector $\mathbf{x}$ is \begin{align} \mathbf{E}(\mathbf{x})=\frac{q_1(\mathbf{x}-\mathbf{x}_1)}{|\mathbf{x}-\mathbf{x}_1|^3}+\frac{q_2(\mathbf{x}-\mathbf{x}_2)}{|\mathbf{x}-\mathbf{x}_2|^3} \end{align} the total potential energy is according to (2) \begin{align} W(\mathbf{x})=\int \left( \underbrace{\frac{q_1^2}{8 \pi|\mathbf{x}-\mathbf{x}_1|^4}+\frac{q_2^2}{8 \pi|\mathbf{x}-\mathbf{x}_2|^4}}_{\text{self-energy}}+\frac{q_1q_2(\mathbf{x}-\mathbf{x}_1)(\mathbf{x}-\mathbf{x}_2)}{4 \pi |\mathbf{x}-\mathbf{x}_1|^3 |\mathbf{x}-\mathbf{x}_2|^3} \right) d\mathbf{x} \end{align} The first two terms are the self-energy. The third term is the potential energy of the interaction, as will be shown. To do so we integrate \begin{align} W_{int}(\mathbf{x})= \frac{q_1q_2}{4 \pi} \int \frac{(\mathbf{x}-\mathbf{x}_1)(\mathbf{x}-\mathbf{x}_2)}{ |\mathbf{x}-\mathbf{x}_1|^3 |\mathbf{x}-\mathbf{x}_2|^3} d\mathbf{x} \end{align} by changing integration varibles $\vec{\rho}=(\mathbf{x}-\mathbf{x}_1)/|\mathbf{x}-\mathbf{x}_2|$ which results in \begin{align} W_{int}(\mathbf{x})= \frac{q_1q_2}{|\mathbf{x}_1-\mathbf{x}_2|} \times \frac{1}{4 \pi} \int \frac{\vec{\rho}(\vec{\rho}-\mathbf{n})}{ \rho^3|\vec{\rho}-\mathbf{n}|^3} d\mathbf{x} \end{align} where $\mathbf{n}=(\mathbf{x}_1-\mathbf{x}_2)/|\mathbf{x}_1-\mathbf{x}_2|$ is the unit vector in the direction $(\mathbf{x}_1-\mathbf{x}_2)$ and $\frac{(\vec{\rho}-\mathbf{n})}{ |\vec{\rho}-\mathbf{n}|^3}=-\nabla_{\rho} (1/|\vec{\rho}-\mathbf{n})|)$. Calculating the volume integral \begin{align} - \frac{1}{4 \pi} \int \frac{\vec{\rho}}{ \rho^3}\nabla_{\rho} \frac{1}{|\vec{\rho}-\mathbf{n}|} d\mathbf{x}\\ &=- \frac{1}{4 \pi} \int \nabla_{\rho} \left( \frac{\vec{\rho}}{ \rho^3}\frac{1}{|\vec{\rho}-\mathbf{n}|} \right) d^3\mathbf{x}+\frac{1}{4 \pi} \int \nabla_{\rho} \left( \frac{\vec{\rho}}{ \rho^3} \right) \frac{1}{|\vec{\rho}-\mathbf{n})|}d\mathbf{x}\\ &=\underbrace{-\oint_S \left(\frac{\vec{\rho}}{ \rho^3} \right) \frac{1}{|\vec{\rho}-\mathbf{n}|} d\mathbf{S}}_{\rightarrow 0 \,for \, S \rightarrow \infty} + \int \frac{1}{|\vec{\rho}-\mathbf{n}|} 4 \pi \delta(\vec{\rho}) d^3\rho= 4\pi \end{align} so that indeed the potential energy from the interaction of $q_1$ and $q_2$ is simply \begin{align} W_{int}(\mathbf{x})= \frac{q_1q_2}{|\mathbf{x}_1-\mathbf{x}_2|} \end{align}

## Force on a conductor

From Gauß's box we now that a surface with charge $\sigma$ has a leap in the electric field $\mathbf{E}$ proportional to $\sigma$ in the direction normal to the surface.

\begin{align} \mathbf{E}(\mathbf{x}) \cdot \mathbf{n} = 4\pi \, \sigma \end{align} For $\mathbf{E}(\mathbf{x})$ parallel to $\mathbf{n}$ we get for $|\mathbf{E}(\mathbf{x})|^2$

\begin{align} |\mathbf{E}(\mathbf{x}) \cdot \mathbf{n}|^2 =E(\mathbf{x})^2= (4\pi \, \sigma)^2 \end{align} with this expression the energy density becomes \begin{align} w= \frac{1}{8 \pi} |\mathbf{E}(\mathbf{x})|^2 =2 \pi \sigma^2 \end{align}

The change in potential energy, by moving the surface by $\Delta x$ outward is given by \begin{align} \Delta W= 2 \pi \sigma^2 \Delta x \Delta a \end{align} where $\Delta a$ is the area moved.

Hence the force $\mathbf{F}=W/\Delta x \cdot \mathbf{n}$ on the conductor is \begin{align} \mathbf{F}=2 \pi \sigma^2 \Delta a \cdot \mathbf{n} \end{align}