Coulomb's Law & Electric Field

Step 1: From discrete to continuous — the volume charge density

For $N$ point charges, the field is $\mathbf{E} = \frac{1}{4\pi\epsilon_0}\sum_{i=1}^N \frac{q_i}{\mathscr{r}_i^2}\hat{\mathscr{r}}_i$. In the continuum limit, replace $q_i \to \rho(\mathbf{r}')\,d\tau'$ where $\rho$ is the volume charge density [C/m$^3$], and the sum becomes an integral:

$$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int_{\mathcal{V}} \frac{\rho(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^2}\,\hat{\mathscr{r}}\,d\tau'$$

Step 2: Surface charge density

If charge is confined to a 2D surface, the volume element $d\tau'$ collapses to a surface element $da'$. Define $\sigma(\mathbf{r}')$ as the surface charge density [C/m$^2$], so $dq = \sigma\,da'$:

$$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int_{\mathcal{S}} \frac{\sigma(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^2}\,\hat{\mathscr{r}}\,da'$$

Step 3: Line charge density

For charge distributed along a 1D curve, define $\lambda(\mathbf{r}')$ as the line charge density [C/m], so $dq = \lambda\,dl'$:

$$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int_{\mathcal{L}} \frac{\lambda(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^2}\,\hat{\mathscr{r}}\,dl'$$

Step 4: Unified notation using the separation vector

In all three cases, $\boldsymbol{\mathscr{r}} = \mathbf{r} - \mathbf{r}'$ is the vector from the source point $\mathbf{r}'$ to the field point $\mathbf{r}$. We can write each integral compactly as:

$$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int \frac{\boldsymbol{\mathscr{r}}}{\mathscr{r}^3}\,dq' = \frac{1}{4\pi\epsilon_0}\int \frac{(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3}\,dq'$$

where $dq'$ is $\rho\,d\tau'$, $\sigma\,da'$, or $\lambda\,dl'$ as appropriate.

Step 5: Relationship between the three densities

The three descriptions are related by dimensional reduction. A surface charge $\sigma$ can be viewed as a volume charge $\rho = \sigma\,\delta(n)$ where $n$ is the coordinate normal to the surface. Similarly, a line charge $\lambda$ is $\rho = \lambda\,\delta(x)\delta(y)$ for a wire along $z$. The most general form is always the volume integral:

$$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int_{\text{all space}} \frac{\rho(\mathbf{r}')\,(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3}\,d\tau'$$

Step 1: Set up the geometry

Place the field point at $\mathbf{r} = s\,\hat{s}$ (in the $xy$-plane, distance $s$ from the $z$-axis). A source element is at $\mathbf{r}' = z'\,\hat{z}$. The separation vector is:

$$\boldsymbol{\mathscr{r}} = \mathbf{r} - \mathbf{r}' = s\,\hat{s} - z'\,\hat{z}, \qquad \mathscr{r} = \sqrt{s^2 + z'^2}$$

Step 2: Write the field contribution from element $dz'$

The charge element $dq = \lambda\,dz'$ produces a field contribution:

$$d\mathbf{E} = \frac{1}{4\pi\epsilon_0}\frac{\lambda\,dz'}{(s^2 + z'^2)}\,\hat{\mathscr{r}} = \frac{\lambda\,dz'}{4\pi\epsilon_0}\frac{s\,\hat{s} - z'\,\hat{z}}{(s^2 + z'^2)^{3/2}}$$

Step 3: Exploit symmetry to eliminate the $z$-component

For every element at $+z'$, there is a corresponding element at $-z'$ that contributes an equal and opposite $\hat{z}$-component. Therefore, the $z$-component integrates to zero:

$$\int_{-\infty}^{\infty}\frac{-z'\,dz'}{(s^2+z'^2)^{3/2}} = 0 \quad \text{(odd integrand)}$$

Step 4: Evaluate the radial component

Only the $\hat{s}$-component survives. We need:

$$E_s = \frac{\lambda s}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\frac{dz'}{(s^2+z'^2)^{3/2}}$$

Step 5: Evaluate the integral via trigonometric substitution

Let $z' = s\tan\theta$, so $dz' = s\sec^2\theta\,d\theta$ and $(s^2+z'^2)^{3/2} = s^3\sec^3\theta$:

$$\int_{-\infty}^{\infty}\frac{dz'}{(s^2+z'^2)^{3/2}} = \int_{-\pi/2}^{\pi/2}\frac{s\sec^2\theta\,d\theta}{s^3\sec^3\theta} = \frac{1}{s^2}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta = \frac{1}{s^2}\Big[\sin\theta\Big]_{-\pi/2}^{\pi/2} = \frac{2}{s^2}$$

Step 6: Assemble the final result

Substituting the integral back:

$$E_s = \frac{\lambda s}{4\pi\epsilon_0}\cdot\frac{2}{s^2} = \frac{\lambda}{2\pi\epsilon_0 s}$$

Therefore the electric field of an infinite line charge is:

$$\boxed{\mathbf{E} = \frac{\lambda}{2\pi\epsilon_0 s}\,\hat{s}}$$

The field falls off as $1/s$, not $1/s^2$, because the line source is one-dimensional — integrating the $1/r^2$ contribution over the infinite line yields a net $1/s$ dependence.

Step 1: Decompose the disk into concentric rings

A ring at radius $s'$ with width $ds'$ has area $da' = 2\pi s'\,ds'$ and carries charge $dq = \sigma \cdot 2\pi s'\,ds'$. From the ring formula (Section 1.4), the axial field of this ring at height $z$ is:

$$dE_z = \frac{1}{4\pi\epsilon_0}\frac{z\,dq}{(s'^2 + z^2)^{3/2}} = \frac{\sigma z}{2\epsilon_0}\frac{s'\,ds'}{(s'^2 + z^2)^{3/2}}$$

Step 2: Integrate over all rings from $s' = 0$ to $s' = R$

$$E_z = \frac{\sigma z}{2\epsilon_0}\int_0^R \frac{s'\,ds'}{(s'^2 + z^2)^{3/2}}$$

Step 3: Evaluate the integral

Let $u = s'^2 + z^2$, so $du = 2s'\,ds'$:

$$\int_0^R \frac{s'\,ds'}{(s'^2 + z^2)^{3/2}} = \frac{1}{2}\int_{z^2}^{R^2+z^2} u^{-3/2}\,du = \frac{1}{2}\left[-2u^{-1/2}\right]_{z^2}^{R^2+z^2} = \frac{1}{|z|} - \frac{1}{\sqrt{R^2+z^2}}$$

Step 4: Write the result for the disk

For $z > 0$:

$$\boxed{E_z = \frac{\sigma}{2\epsilon_0}\left(1 - \frac{z}{\sqrt{R^2 + z^2}}\right)}$$

Step 5: Limit — infinite plane ($R \to \infty$)

As $R \to \infty$, the term $z/\sqrt{R^2+z^2} \to 0$, giving:

$$E_z = \frac{\sigma}{2\epsilon_0}$$

This is the field of an infinite plane of charge — uniform, independent of distance, directed perpendicular to the plane. This result will also follow from Gauss's law.

Step 6: Limit — far from the disk ($z \gg R$)

Expand $1/\sqrt{R^2+z^2} = (1/z)(1+R^2/z^2)^{-1/2} \approx (1/z)(1 - R^2/(2z^2))$:

$$E_z \approx \frac{\sigma}{2\epsilon_0}\left(1 - 1 + \frac{R^2}{2z^2}\right) = \frac{\sigma R^2}{4\epsilon_0 z^2} = \frac{1}{4\pi\epsilon_0}\frac{\sigma\pi R^2}{z^2} = \frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}$$

where $Q = \sigma\pi R^2$ is the total charge — the disk looks like a point charge at large distances, as expected.

Step 1: Define the geometry

Place $+q$ at $\mathbf{r}' = +\frac{d}{2}\hat{z}$ and $-q$ at $\mathbf{r}' = -\frac{d}{2}\hat{z}$. The dipole moment is $\mathbf{p} = qd\,\hat{z}$. The potential at field point $\mathbf{r}$ is:

$$V(\mathbf{r}) = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{|\mathbf{r} - \frac{d}{2}\hat{z}|} - \frac{1}{|\mathbf{r} + \frac{d}{2}\hat{z}|}\right)$$

Step 2: Expand $1/|\mathbf{r} - \mathbf{r}'|$ for $r \gg d$

Using the law of cosines, $|\mathbf{r} \mp \frac{d}{2}\hat{z}|^2 = r^2 \mp rd\cos\theta + d^2/4$. For $r \gg d$:

$$\frac{1}{|\mathbf{r} \mp \frac{d}{2}\hat{z}|} = \frac{1}{r}\left(1 \mp \frac{d\cos\theta}{2r} + \frac{d^2}{4r^2}\right)^{-1/2} \approx \frac{1}{r}\left(1 \pm \frac{d\cos\theta}{2r} + \cdots\right)$$

Step 3: Subtract to get the dipole potential

The monopole ($1/r$) terms cancel (total charge is zero), leaving the dipole term:

$$V = \frac{q}{4\pi\epsilon_0}\left[\frac{1}{r}\left(1+\frac{d\cos\theta}{2r}\right) - \frac{1}{r}\left(1-\frac{d\cos\theta}{2r}\right)\right] = \frac{qd\cos\theta}{4\pi\epsilon_0 r^2}$$

Step 4: Express in terms of the dipole moment

With $\mathbf{p} = qd\,\hat{z}$ and $\mathbf{p}\cdot\hat{r} = p\cos\theta$:

$$\boxed{V_{\rm dip}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{\mathbf{p}\cdot\hat{r}}{r^2} = \frac{p\cos\theta}{4\pi\epsilon_0 r^2}}$$

Step 5: Compute $\mathbf{E} = -\nabla V$ in spherical coordinates

Taking the gradient in spherical coordinates ($\nabla V = \frac{\partial V}{\partial r}\hat{r} + \frac{1}{r}\frac{\partial V}{\partial\theta}\hat{\theta}$):

$$E_r = -\frac{\partial V}{\partial r} = \frac{2p\cos\theta}{4\pi\epsilon_0 r^3}$$

$$E_\theta = -\frac{1}{r}\frac{\partial V}{\partial\theta} = \frac{p\sin\theta}{4\pi\epsilon_0 r^3}$$

Step 6: Write the coordinate-free dipole field

Combining components and using $\hat{r}\cos\theta - \hat{\theta}\sin\theta = \hat{z}$ (not quite — we use the vector identity), the coordinate-free form is:

$$\boxed{\mathbf{E}_{\rm dip}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{1}{r^3}\left[3(\mathbf{p}\cdot\hat{r})\hat{r} - \mathbf{p}\right]}$$

Key features: the field falls off as $1/r^3$ (one power faster than a point charge). Along the axis ($\theta = 0$), $E = 2p/(4\pi\epsilon_0 r^3)$; in the equatorial plane ($\theta = \pi/2$), $E = p/(4\pi\epsilon_0 r^3)$, antiparallel to $\mathbf{p}$.