Dielectrics and polarisation
Molecules in a field
Dielectrics are insulators: glass, mica, paper, water. Unlike metals, they have no free charges. So what happens when we put one in an electric field?
Recall a conductor: its free charges move until the field inside is cancelled completely. In a dielectric the charges cannot move freely, so the field is only reduced, not cancelled.
The reason lies in the molecules. In a non-polar molecule, like oxygen or hydrogen, the centres of positive and negative charge coincide. It has no dipole moment of its own.
In an external field, the positive charge is pushed one way and the negative charge the other. The two centres separate a little, and the molecule gets an induced dipole moment along the field.
In a polar molecule, like water or HCl, the centres are apart already: each molecule is a permanent dipole. Without a field, thermal motion points them every which way, so on average they cancel.
Switch on a field. Each dipole feels a torque and tends to line up with it, while thermal agitation keeps knocking it off. On average, a net dipole moment appears along the field.
Either way, the dielectric is polarised. The polarisation P is the dipole moment per unit volume.
For many materials, P is proportional to the field: P = χₑE. The constant χₑ is the electric susceptibility of the material.
Bound charges reduce the field
Take a rectangular slab of dielectric in a uniform field E₀ pointing left to right. Every small volume of it becomes a tiny dipole, all pointing along the field.
Inside the slab, the plus end of each dipole sits right next to the minus end of its neighbour. They cancel, so the inside carries no net charge.
But at the right face there are plus ends with nothing to cancel them, and at the left face minus ends are left uncovered. So the slab carries surface charges, +σₚ and −σₚ.
These are bound charges: they cannot be taken off the molecules. For a uniformly polarised slab, the surface charge density equals the polarisation: σₚ = P.
The surface charges make a field inside the slab, from plus to minus, opposite to E₀. So the net field inside the dielectric is smaller than E₀.
Example: a slab fills the gap of a charged capacitor, whose plates carry free charge density σ. Without the slab, E₀ = σ/ε₀. With it, E = (σ − σₚ)/ε₀.
Suppose σₚ is three quarters of σ. Then E = σ/4ε₀, a quarter of E₀. If E₀ was 4 × 10⁵ V/m, the field inside the slab is only 10⁵ V/m.
The factor by which the field is reduced, here 4, is the dielectric constant K of the material. For glass it is between 5 and 10; for water, about 80. We will use it for capacitors next.
Try it: polarise a slab
- The battery is disconnected, so the free charge on the plates is fixed. The slab fills the gap.
- Slide K from 1 up to 10: the molecules line up, the bound charges on the slab’s faces grow, and the field in the slab falls to E₀/K.
- Make the slab thinner (t/d): the field in the air gap stays σ/ε₀; only the field inside the slab is reduced.
- Count the field lines: only one in K carries on through the slab. The rest end on bound charges.
Water with no field
Water molecules are polar: each has a permanent dipole moment. What is the polarisation of a glass of water when there is no external field?
- Large, because every molecule is a dipole
- Zero, because the dipoles point in random directions
- Zero, because water molecules have no dipole moment
- It depends on the shape of the glass
Answer
B. Each molecule is a dipole, but thermal motion points them randomly, so their moments cancel on average and P = 0. Only an applied field lines them up partly and gives a net dipole moment per unit volume.
How strong are the bound charges?
A charged capacitor (battery disconnected) has a field of 6 × 10⁵ V/m between its plates. A dielectric slab filling the gap reduces it to 2 × 10⁵ V/m. What are K and the ratio σₚ/σ?
- K = 3, σₚ/σ = 2/3
- K = 3, σₚ/σ = 1/3
- K = 1/3, σₚ/σ = 2/3
- K = 2, σₚ/σ = 1/2
Answer
A. The field is reduced by the factor K = 6/2 = 3. From E = (σ − σₚ)/ε₀ = σ/3ε₀ we get σ − σₚ = σ/3, so σₚ = 2σ/3. The bound charge cancels two thirds of the free charge’s field, but never all of it.
Summary
- Polarisation:
P = dipole moment per unit volume = χₑE - Bound surface charge:
σₚ = P - Field in a slab between plates:
E = (σ − σₚ)/ε₀ = E₀/K
- Non-polar molecules (O₂, H₂) get induced dipoles; polar molecules (H₂O, HCl) have permanent dipoles that partly line up in a field.
- A polarised slab has bound charges on its faces but no net charge inside.
- The bound charges reduce the field inside a dielectric; unlike a conductor, they do not cancel it.
- The factor of reduction is the dielectric constant K.