Dielectrics

Fill the gap of a charged, isolated capacitor with a sheet of glass and a voltmeter across the plates drops: no charge left the plates, yet the potential difference fell. The glass is a dielectric, an insulator placed in an electric field, and it weakens the field between the plates without cancelling it. That is how real capacitors pack microfarads into a component the size of a grain of rice.

How an insulator polarizes

In a conductor (topic 10.1), electrons roam the whole object and pile up on its surface until the field inside is exactly zero. In an insulator, every electron stays bound to its own atom or molecule. An external field can only shift charge a tiny distance within each molecule, in one of two ways:

• Induced dipoles. In a nonpolar molecule (like the long chains in Teflon or polystyrene), the field pushes the electron cloud one way and the nuclei the other, stretching the molecule into a small dipole aligned with the field.
• Aligned permanent dipoles. A polar molecule, such as water, is already a dipole. The field pulls its positive end one way and its negative end the other, which twists the molecule partly into line. Thermal jostling keeps the alignment from being complete.

Either way the material becomes polarized: its molecules point their negative ends toward the positive plate and their positive ends toward the negative plate. Deep inside, each molecule's positive end sits next to its neighbor's negative end, and the charges cancel on average. At the two faces they do not: the face next to the positive plate is left with a layer of negative charge, and the face next to the negative plate a layer of positive charge. These bound charges can't leave their molecules, unlike the free charges on the plates, which can flow through wires. The dielectric as a whole stays neutral.

The dielectric's field opposes the applied field

The layer of bound charge acts like a second, weaker pair of plates inside the first, with the signs reversed. Its field, $\vec{E}_{\text{ind}}$, points from the bound positive face to the bound negative face, which is opposite to the applied field, the field $\vec{E}_0$ of the free charge on the plates. The net field in the dielectric is smaller than $\vec{E}_0$ but in the same direction, because the bound charge is always less than the free charge (the molecules only partly cancel the field that is polarizing them).

The dielectric constant $\kappa$ (kappa) measures how much a material weakens the field. For a dielectric filling the gap of an isolated capacitor,

The free surface charge density $\sigma$ on the plates and the bound density $\sigma_b$ on the dielectric's faces add like two pairs of sheets, so $E = \frac{\sigma - \sigma_b}{\varepsilon_0}$. Setting this equal to $\frac{\sigma}{\kappa\varepsilon_0}$ gives $\sigma_b = \sigma\left(1 - \frac{1}{\kappa}\right)$: with $\kappa = 4$, the bound charge is three quarters of the free charge and the field is a quarter as strong.

Some values of $\kappa$ at room temperature:

Every material has $\kappa > 1$, because polarization can only oppose the applied field. Air is so close to vacuum that the exam treats air-filled capacitors as having $\kappa = 1$. Water's large value comes from its strongly polar molecules, which turn readily into line.

The dielectric constant also defines the material's permittivity, $\varepsilon = \kappa\varepsilon_0$. Inside a dielectric, every formula you derived with $\varepsilon_0$ carries over with $\varepsilon$ in its place.

What changes when a dielectric goes in

The capacitance follows directly from the weaker field. With the plates' charge $Q$ fixed, $\Delta V = Ed = \frac{E_0 d}{\kappa} = \frac{\Delta V_0}{\kappa}$, so $C = \frac{Q}{\Delta V} = \kappa C_0$. The capacitance is a property of the device (topic 10.3), so it rises by $\kappa$ in every case. What happens to the other quantities depends on whether the capacitor is isolated or stays connected to a battery:

Isolated: the charge has nowhere to go, so the dielectric's field weakens the net field and the potential difference falls by $\kappa$. The energy $\frac{Q^2}{2C}$ falls by $\kappa$ as well. The missing energy went into the dielectric: the plates' field pulls a dielectric slab into the gap (the bound charges near the edge are attracted toward the plates of opposite sign), so you must hold it back as it goes in, and the field does positive work on it.

Connected: the battery fixes $\Delta V$, so the field $\frac{\Delta V}{d}$ between the plates can't change. As the dielectric starts to weaken the field, the battery pushes more charge onto the plates until the field is back to $\frac{\Delta V}{d}$; the free charge ends at $\kappa Q_0$. The energy $\frac{1}{2}C(\Delta V)^2$ rises by $\kappa$, supplied by the battery.

The general statement is that inserting a dielectric may change each quantity. Decide first what is held fixed, as in topic 10.3's Example 2.

Worked examples

Worked exampleExample 3: measuring a dielectric constant (experimental design)

Students want to find the dielectric constant of a plastic that comes in sheets $0.50\ \text{mm}$ thick. They have two square metal plates of area $0.0400\ \text{m}^2$, a stack of the sheets, and a capacitance meter.

(a) Describe a procedure that would let them determine $\kappa$ from a graph. Say what they vary, what they measure and what they keep the same.

(b) Their data are below. What should they graph to get a straight line, and how is $\kappa$ found from it?

Sheets	$d$ (mm)	$C$ (nF)
1	0.50	2.30
2	1.00	1.12
3	1.50	0.76
4	2.00	0.56
5	2.50	0.46

(c) Use the data to calculate $\kappa$. Use $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$.

(d) In one trial, thin air gaps were left between the sheets. Would that make the measured value of $\kappa$ too high or too low? Explain.

Show solution

(a) Stack 1, 2, 3, 4 and 5 sheets between the plates, pressing the plates flat so the plastic fills the gap with no air. For each stack, record the plate separation (number of sheets times $0.50\ \text{mm}$, or measured with calipers) and the capacitance from the meter. Keep the plates and the plastic the same throughout, and keep the plates square to each other so the area of overlap stays $0.0400\ \text{m}^2$. Take several readings per stack and average them.

(b) The model is $C = \frac{\kappa\varepsilon_0 A}{d}$, so $C$ is proportional to $\frac{1}{d}$. Graph $C$ against $\frac{1}{d}$: the points should fall on a line through the origin with slope $\kappa\varepsilon_0 A$, so $\kappa = \frac{\text{slope}}{\varepsilon_0 A}$. A graph of $C$ against $d$ would be a curve, which is much harder to read a constant from.

(c) The $\frac{1}{d}$ values are $2000$, $1000$, $667$, $500$ and $400\ \text{m}^{-1}$. Draw the best-fit line through the plotted points and read two well-separated points on the line (not data points), such as $(400\ \text{m}^{-1},\ 0.45\ \text{nF})$ and $(2000\ \text{m}^{-1},\ 2.29\ \text{nF})$:

$$\text{slope} = \frac{(2.29 - 0.45)\times10^{-9}\ \text{F}}{(2000 - 400)\ \text{m}^{-1}} = 1.15\times10^{-12}\ \text{F}\cdot\text{m}.$$

Then

$$\kappa = \frac{1.15\times10^{-12}\ \text{F}\cdot\text{m}}{(8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2))(0.0400\ \text{m}^2)} = 3.2.$$

(Units: $\varepsilon_0 A$ is in F$\cdot$m, so $\kappa$ is dimensionless, as it must be.) Taking the slope from the line, rather than from two of the data points, averages out the scatter in the measurements.

(d) Too low. In an air gap the field is not reduced by the plastic, so, as in Example 2, a gap partly filled with plastic has less capacitance than one completely filled. The measured $C$ for that stack comes out smaller than $\frac{\kappa\varepsilon_0 A}{d}$, which pulls the slope, and the calculated $\kappa$, down.

Where students lose points

On the exam

Dielectric questions ask you to predict changes in $C$, $Q$, $\Delta V$, $E$ and $U$ when a dielectric is inserted or removed, to explain polarization and the direction of the induced field, and to design or analyze an experiment that measures $\kappa$ (linearize the data, take the slope from a best-fit line, carry units). A calculator is allowed throughout.