Electric Potential

Step 1: State the two fundamental relationships

From Gauss's law (differential form) and the definition of potential:

$$\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0} \qquad \text{and} \qquad \mathbf{E} = -\nabla V$$

Step 2: Substitute $\mathbf{E} = -\nabla V$ into Gauss's law

$$\nabla\cdot(-\nabla V) = \frac{\rho}{\epsilon_0}$$

Step 3: Recognize the Laplacian

The divergence of a gradient is the Laplacian operator: $\nabla\cdot(\nabla V) = \nabla^2 V$. In Cartesian coordinates:

$$\nabla^2 V = \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} + \frac{\partial^2 V}{\partial z^2}$$

Therefore:

$$\boxed{\nabla^2 V = -\frac{\rho}{\epsilon_0}} \qquad \text{(Poisson's equation)}$$

Step 4: The charge-free case — Laplace's equation

In regions where there is no charge ($\rho = 0$), Poisson's equation reduces to:

$$\boxed{\nabla^2 V = 0} \qquad \text{(Laplace's equation)}$$

Step 5: Verify with the point charge potential

For $V = q/(4\pi\epsilon_0 r)$, compute $\nabla^2 V$ in spherical coordinates. For $r \neq 0$:

$$\nabla^2 V = \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dV}{dr}\right) = \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\frac{d}{dr}\left(r^2\cdot\left(-\frac{1}{r^2}\right)\right) = \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\frac{d}{dr}(-1) = 0$$

This confirms Laplace's equation holds away from the charge. At $r = 0$, the correct result is $\nabla^2(1/r) = -4\pi\delta^3(\mathbf{r})$, giving $\nabla^2 V = -q\delta^3(\mathbf{r})/\epsilon_0 = -\rho/\epsilon_0$.

Step 6: Physical meaning of $\nabla^2 V$

The Laplacian measures how $V$ at a point differs from the average of $V$ on a small sphere around that point. Specifically, for a small sphere of radius $\varepsilon$:

$$V(\mathbf{r}) - \langle V\rangle_{\rm sphere} = -\frac{\varepsilon^2}{6}\nabla^2 V + O(\varepsilon^4) = \frac{\varepsilon^2}{6}\frac{\rho}{\epsilon_0}$$

Where $\rho > 0$ (positive charge), $V$ exceeds its local average — it is a local maximum relative to its neighborhood. This is why solutions to Laplace's equation ($\nabla^2 V = 0$) have no local maxima or minima — a key property called the mean value theorem.

Step 1: Start with the exact potential of a charge distribution

$$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\,d\tau'$$

We need to expand $1/|\mathbf{r} - \mathbf{r}'|$ for $r \gg r'$ (field point far from the source).

Step 2: Apply the law of cosines

Let $\alpha$ be the angle between $\mathbf{r}$ and $\mathbf{r}'$. Then:

$$|\mathbf{r} - \mathbf{r}'|^2 = r^2 + r'^2 - 2rr'\cos\alpha = r^2\left(1 - 2\frac{r'}{r}\cos\alpha + \frac{r'^2}{r^2}\right)$$

Define $\epsilon = (r'/r)$ and write:

$$\frac{1}{|\mathbf{r} - \mathbf{r}'|} = \frac{1}{r}\frac{1}{\sqrt{1 - 2\epsilon\cos\alpha + \epsilon^2}}$$

Step 3: Expand using the generating function for Legendre polynomials

The generating function for Legendre polynomials $P_\ell$ is:

$$\frac{1}{\sqrt{1 - 2\epsilon\cos\alpha + \epsilon^2}} = \sum_{\ell=0}^{\infty}\epsilon^\ell P_\ell(\cos\alpha)$$

Therefore:

$$\frac{1}{|\mathbf{r} - \mathbf{r}'|} = \frac{1}{r}\sum_{\ell=0}^{\infty}\left(\frac{r'}{r}\right)^\ell P_\ell(\cos\alpha)$$

Step 4: Substitute into the potential and identify terms

$$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\sum_{\ell=0}^{\infty}\frac{1}{r^{\ell+1}}\int (r')^\ell P_\ell(\cos\alpha)\,\rho(\mathbf{r}')\,d\tau'$$

Writing out the first few terms explicitly:

Step 5: Monopole term ($\ell = 0$)

Since $P_0(\cos\alpha) = 1$:

$$V_{\rm mono} = \frac{1}{4\pi\epsilon_0}\frac{1}{r}\int\rho\,d\tau' = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}$$

where $Q = \int\rho\,d\tau'$ is the total charge. This is just the point-charge potential — the leading term at large $r$.

Step 6: Dipole term ($\ell = 1$)

Since $P_1(\cos\alpha) = \cos\alpha = \hat{r}\cdot\hat{r}'$:

$$V_{\rm dip} = \frac{1}{4\pi\epsilon_0}\frac{1}{r^2}\int r'\cos\alpha\,\rho\,d\tau' = \frac{1}{4\pi\epsilon_0}\frac{\hat{r}}{r^2}\cdot\int\mathbf{r}'\rho\,d\tau' = \frac{1}{4\pi\epsilon_0}\frac{\mathbf{p}\cdot\hat{r}}{r^2}$$

where $\mathbf{p} = \int\mathbf{r}'\rho(\mathbf{r}')\,d\tau'$ is the dipole moment. This falls off as $1/r^2$ and dominates when $Q = 0$.

Step 7: Quadrupole term ($\ell = 2$)

Since $P_2(\cos\alpha) = \frac{1}{2}(3\cos^2\alpha - 1)$:

$$V_{\rm quad} = \frac{1}{4\pi\epsilon_0}\frac{1}{r^3}\int\frac{1}{2}(r')^2(3\cos^2\alpha - 1)\,\rho\,d\tau'$$

This can be written in terms of the traceless quadrupole tensor $Q_{ij} = \int(3r'_ir'_j - r'^2\delta_{ij})\rho\,d\tau'$:

$$V_{\rm quad} = \frac{1}{4\pi\epsilon_0}\frac{1}{2r^3}\sum_{i,j}Q_{ij}\hat{r}_i\hat{r}_j$$

This falls off as $1/r^3$ and dominates when both $Q = 0$ and $\mathbf{p} = 0$.

Step 8: The complete expansion

$$\boxed{V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\left[\frac{Q}{r} + \frac{\mathbf{p}\cdot\hat{r}}{r^2} + \frac{1}{2}\sum_{i,j}\frac{Q_{ij}\hat{r}_i\hat{r}_j}{r^3} + \cdots\right]}$$

At large distances, only the lowest non-vanishing term matters. The multipole expansion is the systematic way to characterize charge distributions by their "shape" at long range.