"The Big Cheese #4",
sculpture in welded aluminum by Bruce Gray.

A conductor is a material with a seemingly unlimited supply of charge carriers which are free to move. Implications:

  1. $\myv E = 0$ inside a conductor. For, if $\myv E \ne 0$ somewhere, then charges will move.

    In an external field, charges will move until the net field inside vanishes.

  2. $\rho = 0$ everywhere inside a conductor. One of the fundamental properties of the electric field is $\myv \grad \cdot \myv E = \rho/\epsilon_0$.
  3. Any non-zero charge may only be found at the surface.
  4. A conductor is an equipotential. Consider for any two points $V(\myv b) -V(\myv a) = \int_{\myv a}^{\myv b} \myv E \cdot d \myv l$.
  5. Just outside the surfac, $\myv E$ is perpendicular to the surface: Since $\myv E-\myv \grad V$, $\myv E$ is always perpendicular to any equipotentials

Induced charge

A charge outside a conductor gives rise to a non-zero electric field, and this in turn induces a charge on a conductor (on the surface of a conductor).

A charge inside a conductor also induces a charge on a conductor.


What's the electric field at a point $P$ at displacement $\myv r$ outside a spherical conductor with a charge inside the odd-shaped cavity shown?

The answer is... $\myv E = \frac{1}{4 \pi \epsilon_0}\frac{q}{r^2}\uv{r}$. Why?

Force on a conducting surface

When we looked at boundary conditions near an accumulation of surface charge, we found that the perpendicular component of the electric field is not continuous across some accumulation of surface charge, but rather changes by $$ E_{\perp}^{\text{above}} - E_{\perp}^{\text{below}}=\sigma / \epsilon_0.$$

...So applying Gauss' law, the field due to a surface charge density above an infinite plane has magnitude $$E_z=\frac{\sigma}{2\epsilon_0}.$$

To find the force on the small "pillbox" pictured we should multiply the charge in the box $\sigma A \times$ the electric field...but should we use the field above or below the box?

[For two point charges... distinguish between $\myv E$ and field of each particle...]

The electric field that we want to use in calculating the force is actually not either one, but rather that portion of the electric field which is due to charges "external" to the box. This will be continuous across the interface.

Part of the field is due to the surface charges themselves, and we should subtract that part out. Indeed, close to the surface, the field due to the surface charge $\myv E_\sigma$ is pointing normally away from the surface both above and below. So, we could write the field above and below as: $$\myv E^{\text{above}} = \myv E^{\text{ext}}+\frac{\sigma}{2 \epsilon_0} \uv{n}$$ and
$$\myv E ^{\text{below}} = \myv E^{\text{ext}}-\frac{\sigma}{2 \epsilon_0} \uv{n}$$

Adding these two equations, $$\begineq \myv E^{\text{ext}} &=& \frac{1}{2}(\myv E^{\text{above}} +\myv E^{\text{below}})\\ &=& \myv E^{\text{ave}}.\endeq$$ On the surface of a conductor:

  • the field in the conductor (below) has to be zero, and,
  • the field just outside (above) has to be $\frac{\sigma}{\epsilon_0} \uv{n}$, and so
  • $\myv E^{\text{ave}} = \frac{\sigma}{2 \epsilon_0} \uv{n}$, and so
  • the force per unit area (pressure) is $F/A = qE^{\text{ave}}/A$ which is...
$$ P =(\sigma A E^{\text{ave}})/A=\frac{\sigma^2}{2\epsilon_0}.$$

Or, since the field just outside the conductor is $E=\sigma/\epsilon_0$, $$P= \frac{\epsilon_0}{2} E^2.$$


Imagine placing charges $+q$ and $-q$ on two adjacent conductors. Since they're conductors, the potential on one is everywhere the same, and the potential difference, $V$ between them can be found from $$V = V_+-V_- = \int_-^+ \myv E \cdot d \myv l.$$

Since $$\myv E =\frac{1}{4 \pi \epsilon_0}\int \frac{\rho}{\rr^2}\uv \rr\, d \tau,$$ it seems that doubling the charge doubles the potential difference, that is, that this ratio is a constant:

$$\frac{q}{V} \equiv C.$$ Where $C$ is a constant which depends on the geometry of the arrangement of the conductors, but not on the charge per se.

$C$ has units of coulombs / volt = farads (F).

Since the work to move an object with small charge $dq$ from potential 1 to potential 2, is $dq (V_1-V_2)$: $$dW = dq V =dq \frac{q}{C},$$

The work needed to "charge up" a capacitor from 0 to total charge $q$ is: $$W=\int_0^q \frac{q'}{C}dq' = \frac{1}{2}\frac{q^2}{C} = \frac{1}{2}CV^2.$$

Example Derive the capacitance of a flat-plate capacitor, assuming (only slight approximation) that the field between plates is uniform, and zero everywhere else...

We still have the same discontinuity in the electric field above and below any surface charge density: $$ E_{\perp}^{\text{above}} - E_{\perp}^{\text{below}}=\sigma / \epsilon_0.$$

But if the surface below the boundary is a conductor, then $\myv E_\text{below}=0$, and so the field above must be $\sigma/\epsilon_0$ (twice the field on either side of an infinite sheet of charge).

The answer is $C=A\epsilon_0/d$.

Image credits

Brian Yap