Conductors & Dielectrics

Step 1: Set up the geometry

Two infinite parallel conducting plates separated by distance $d$, with surface charge $+\sigma$ on the top plate and $-\sigma$ on the bottom plate. The total charge on each plate of area $A$ is $Q = \sigma A$.

Step 2: Find the electric field between the plates using Gauss's law

Choose a Gaussian pillbox straddling the positively charged plate. The field inside the conductor is zero, and the field between the plates points from $+\sigma$ to $-\sigma$:

$$E\cdot A = \frac{\sigma A}{\epsilon_0} \quad\Longrightarrow\quad E = \frac{\sigma}{\epsilon_0}$$

Alternatively, superpose the fields of two infinite sheets: each contributes $\sigma/(2\epsilon_0)$ in the same direction between the plates, giving $E = \sigma/\epsilon_0$. Outside, they cancel.

Step 3: Compute the potential difference

The voltage between the plates is the line integral of $\mathbf{E}$ from the negative plate to the positive plate:

$$V = -\int_{-}^{+}\mathbf{E}\cdot d\boldsymbol{\ell} = Ed = \frac{\sigma d}{\epsilon_0} = \frac{Qd}{\epsilon_0 A}$$

Step 4: Apply the definition of capacitance

$$C = \frac{Q}{V} = \frac{Q}{\frac{Qd}{\epsilon_0 A}} = \frac{\epsilon_0 A}{d}$$

$$\boxed{C = \frac{\epsilon_0 A}{d}}$$

Step 5: Energy stored in the capacitor

Using $W = \frac{1}{2}CV^2$ or equivalently the field energy $W = \frac{\epsilon_0}{2}\int E^2\,d\tau$. The field is uniform with magnitude $E = V/d$ in volume $Ad$:

$$W = \frac{\epsilon_0}{2}E^2\cdot Ad = \frac{\epsilon_0}{2}\frac{V^2}{d^2}\cdot Ad = \frac{\epsilon_0 A}{2d}V^2 = \frac{1}{2}CV^2$$

Both approaches yield the same result, confirming the energy density interpretation $u = \epsilon_0 E^2/2$.

Step 1: The field at the surface is not the field acting on the surface charge

Just outside the conductor, $E = \sigma/\epsilon_0$. But this is the total field, which includes the contribution of the patch of surface charge we are considering. The force on that patch is due to the field of everything else.

Step 2: Decompose the field into "self" and "other" contributions

A small patch of surface charge $\sigma$ produces a field $\sigma/(2\epsilon_0)$ pointing away from the surface on both sides (like an infinite sheet locally). Call this $\mathbf{E}_{\rm patch}$. The rest of the system produces $\mathbf{E}_{\rm other}$.

Just outside: $E_{\rm other} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$, so $E_{\rm other} = \frac{\sigma}{2\epsilon_0}$.

Just inside: $E_{\rm other} - \frac{\sigma}{2\epsilon_0} = 0$, so $E_{\rm other} = \frac{\sigma}{2\epsilon_0}$. (Consistent.)

Step 3: Compute the average field

The field of "everything else" at the location of the surface charge is the average of the field just above and just below:

$$E_{\rm other} = \frac{1}{2}(E_{\rm above} + E_{\rm below}) = \frac{1}{2}\left(\frac{\sigma}{\epsilon_0} + 0\right) = \frac{\sigma}{2\epsilon_0}$$

Step 4: Compute the force per unit area

The force on a surface element $dA$ carrying charge $\sigma\,dA$ is due to $E_{\rm other}$:

$$dF = \sigma\,dA\cdot E_{\rm other} = \sigma\,dA\cdot\frac{\sigma}{2\epsilon_0}$$

The electrostatic pressure (force per unit area) is:

$$\boxed{P = \frac{dF}{dA} = \frac{\sigma^2}{2\epsilon_0}}$$

Step 5: Express in terms of the field

Since $E_{\rm outside} = \sigma/\epsilon_0$ at the surface, we can write $\sigma = \epsilon_0 E$, and:

$$P = \frac{\sigma^2}{2\epsilon_0} = \frac{\epsilon_0}{2}E^2$$

This is exactly the energy density $u = \epsilon_0 E^2/2$! The electrostatic pressure on a conductor surface equals the local energy density of the field just outside. The force is always outward (pulling the conductor apart), regardless of the sign of $\sigma$.

Step 1: Start with the potential of a polarized material

A small volume $d\tau'$ of dielectric with polarization $\mathbf{P}(\mathbf{r}')$ has dipole moment $d\mathbf{p} = \mathbf{P}\,d\tau'$. The potential of a dipole is $V = \frac{1}{4\pi\epsilon_0}\frac{\mathbf{p}\cdot\hat{\mathscr{r}}}{\mathscr{r}^2}$, so the total potential is:

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

Step 2: Use the identity $\hat{\mathscr{r}}/\mathscr{r}^2 = \nabla'(1/\mathscr{r})$

The gradient with respect to the source coordinates $\mathbf{r}'$ gives:

$$\nabla'\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) = \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} = \frac{\hat{\mathscr{r}}}{\mathscr{r}^2}$$

So:

$$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int_{\mathcal{V}}\mathbf{P}(\mathbf{r}')\cdot\nabla'\left(\frac{1}{\mathscr{r}}\right)d\tau'$$

Step 3: Integrate by parts

Using the product rule $\nabla'\cdot(f\mathbf{A}) = f(\nabla'\cdot\mathbf{A}) + \mathbf{A}\cdot(\nabla' f)$, with $f = 1/\mathscr{r}$ and $\mathbf{A} = \mathbf{P}$:

$$\mathbf{P}\cdot\nabla'\left(\frac{1}{\mathscr{r}}\right) = \nabla'\cdot\left(\frac{\mathbf{P}}{\mathscr{r}}\right) - \frac{1}{\mathscr{r}}\nabla'\cdot\mathbf{P}$$

Step 4: Apply the divergence theorem to the first term

$$V = \frac{1}{4\pi\epsilon_0}\left[\int_{\mathcal{V}}\nabla'\cdot\left(\frac{\mathbf{P}}{\mathscr{r}}\right)d\tau' - \int_{\mathcal{V}}\frac{\nabla'\cdot\mathbf{P}}{\mathscr{r}}\,d\tau'\right]$$

By the divergence theorem, the first integral becomes a surface integral:

$$V = \frac{1}{4\pi\epsilon_0}\left[\oint_S \frac{\mathbf{P}\cdot\hat{n}'}{\mathscr{r}}\,da' - \int_{\mathcal{V}}\frac{\nabla'\cdot\mathbf{P}}{\mathscr{r}}\,d\tau'\right]$$

Step 5: Identify the bound charge densities

Comparing with the potentials of surface and volume charge distributions $V = \frac{1}{4\pi\epsilon_0}\oint\frac{\sigma'}{{\mathscr{r}}}\,da' + \frac{1}{4\pi\epsilon_0}\int\frac{\rho'}{{\mathscr{r}}}\,d\tau'$, we read off:

$$\boxed{\sigma_b = \mathbf{P}\cdot\hat{n}} \qquad \boxed{\rho_b = -\nabla\cdot\mathbf{P}}$$

Step 6: Physical interpretation

Where $\nabla\cdot\mathbf{P} \neq 0$, the polarization is non-uniform and the displacement of positive and negative bound charges does not cancel locally — this creates a net volume bound charge. At the surface, the polarization terminates abruptly, leaving uncompensated charges with density $\mathbf{P}\cdot\hat{n}$. For uniform polarization, $\nabla\cdot\mathbf{P} = 0$ and only surface bound charges appear.

Step 7: Verify total bound charge is zero

The total bound charge must vanish (dielectric is neutral):

$$Q_b = \int_{\mathcal{V}}\rho_b\,d\tau + \oint_S\sigma_b\,da = -\int_{\mathcal{V}}\nabla\cdot\mathbf{P}\,d\tau + \oint_S\mathbf{P}\cdot\hat{n}\,da = 0$$

by the divergence theorem. Polarization redistributes charge but does not create or destroy it.

Step 1: Total charge = free + bound

In a dielectric medium, the total charge density is the sum of free charge (placed deliberately) and bound charge (induced by polarization):

$$\rho_{\rm total} = \rho_f + \rho_b = \rho_f - \nabla\cdot\mathbf{P}$$

Step 2: Write Gauss's law with total charge

Gauss's law always involves the total charge:

$$\nabla\cdot\mathbf{E} = \frac{\rho_{\rm total}}{\epsilon_0} = \frac{\rho_f - \nabla\cdot\mathbf{P}}{\epsilon_0}$$

Step 3: Rearrange to isolate free charge

Multiply through by $\epsilon_0$ and move $\nabla\cdot\mathbf{P}$ to the left:

$$\nabla\cdot(\epsilon_0\mathbf{E}) + \nabla\cdot\mathbf{P} = \rho_f$$

$$\nabla\cdot(\epsilon_0\mathbf{E} + \mathbf{P}) = \rho_f$$

Step 4: Define the displacement field

Define $\mathbf{D} \equiv \epsilon_0\mathbf{E} + \mathbf{P}$. Then Gauss's law takes the elegant form:

$$\boxed{\nabla\cdot\mathbf{D} = \rho_f} \qquad \text{or equivalently} \qquad \boxed{\oint_S\mathbf{D}\cdot d\mathbf{a} = Q_{f,\rm enc}}$$

The displacement field $\mathbf{D}$ "sees" only free charges. Bound charges are automatically accounted for through $\mathbf{P}$.

Step 5: Linear dielectrics — constitutive relation

For a linear, isotropic dielectric, $\mathbf{P} = \epsilon_0\chi_e\mathbf{E}$, so:

$$\mathbf{D} = \epsilon_0\mathbf{E} + \epsilon_0\chi_e\mathbf{E} = \epsilon_0(1+\chi_e)\mathbf{E} = \epsilon_0\epsilon_r\mathbf{E} = \epsilon\mathbf{E}$$

where $\epsilon_r = 1+\chi_e$ is the relative permittivity and $\epsilon = \epsilon_0\epsilon_r$ is the absolute permittivity.

Step 6: Boundary conditions for $\mathbf{D}$

From $\nabla\cdot\mathbf{D} = \rho_f$ and a pillbox argument, the normal component of $\mathbf{D}$ satisfies:

$$D_{\rm above}^\perp - D_{\rm below}^\perp = \sigma_f$$

In the absence of free surface charge ($\sigma_f = 0$), $D^\perp$ is continuous across the interface. Note: $\mathbf{D}$ has no curl equation of its own ($\nabla\times\mathbf{D} \neq 0$ in general when $\mathbf{P}$ has a curl), so $\mathbf{D}$ is less fundamental than $\mathbf{E}$. Its utility lies in symmetric problems where free charge is known.