Dielectrics and polarization

Conductors versus dielectrics. A conductor placed in an external field has free electrons that flow until the field inside becomes exactly zero. A dielectric has no free charges — every electron stays bound to its own atom or molecule — so it can never cancel the field completely. Instead, each molecule shifts its charge distribution slightly under the field.

How polarization arises. In a non-polar molecule ($\mathrm{O_2}$, $\mathrm{CH_4}$), which has no dipole moment on its own, the external field pulls the positive and negative charge centres apart just enough to induce a small dipole. In a polar molecule ($\mathrm{H_2O}$), a permanent dipole already exists but points randomly; the field simply lines these dipoles up along itself. Either way, once every molecule in the slab has a small dipole moment aligned with the field, the bulk material is said to be polarized.

Bound charge and the reduced field. Inside the slab, the head of one molecule's dipole cancels the tail of its neighbour's, so the interior stays neutral. But at the two faces of the slab there is nothing to cancel against, so a net bound surface charge appears — negative on the face nearer the positive plate, positive on the face nearer the negative plate. This sheet of bound charge sets up its own field, $E_{\text{induced}}$, pointing opposite to the applied field $E_0$, so the net field inside the dielectric drops:

where: $E$ is the field actually present inside the dielectric, $E_0$ is the field that would exist between the same plates with vacuum (or air) in the gap, and $K$ (also written $\varepsilon_r$) is the dielectric constant, a dimensionless number with $K = 1$ for vacuum and $K > 1$ for every real dielectric — never negative, never less than one.

From field reduction to capacitance. Because the field between the plates drops by the factor $K$ when a dielectric fills the gap at constant charge $Q$, the potential difference $V = Ed$ drops by the same factor. Since $C = Q/V$, the capacitance rises by exactly that factor:

where: $C'$ is the capacitance with the dielectric filling the gap, $A$ is the plate area, $d$ is the plate separation, and $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ is the permittivity of free space. This is the single quantitative result the syllabus wants: dropping a dielectric of constant $K$ into an air/vacuum capacitor $C = \varepsilon_0 A/d$ multiplies its capacitance by $K$. Typical values: air $K \approx 1$, paper $K \approx 3.5$, mica $K \approx 6$, glass $K \approx 5$–$10$, water $K \approx 80$.

A parallel-plate capacitor has plates of area $A = 200\ \text{cm}^2 = 2.0\times10^{-2}\ \text{m}^2$ separated by $d = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}$, with air between them. (a) Find its capacitance. (b) A dielectric slab of constant $K = 5$ is slid in to completely fill the gap; find the new capacitance. (c) The capacitor, with the dielectric still in place, is connected to a $12\ \text{V}$ battery; find the charge it now carries.

Step 1. Capacitance with air. $C = \varepsilon_0\dfrac{A}{d} = (8.85\times10^{-12})\dfrac{2.0\times10^{-2}}{2.0\times10^{-3}} = 8.85\times10^{-12}\times 10 = 8.85\times10^{-11}\ \text{F} = 88.5\ \text{pF}$.

Step 2. Capacitance with the dielectric. $C' = K\varepsilon_0\dfrac{A}{d} = 5 \times 8.85\times10^{-11} = 4.425\times10^{-10}\ \text{F} = 442.5\ \text{pF}$ — exactly fivefold, as expected from $C' = KC$.

Step 3. Charge at 12 V. $Q = C'V = 4.425\times10^{-10} \times 12 = 5.31\times10^{-9}\ \text{C} = 5.31\ \text{nC}$.

Filling the gap with this dielectric lets the capacitor store 5.31 nC at 12 V — five times what the same air capacitor would store at the same voltage, which is exactly why real capacitors are built with dielectrics.

Competitive stretch

Now suppose instead the same air capacitor is first charged to $Q_0 = 6\ \mu\text{C}$ at $V_0 = 100\ \text{V}$ and the battery is then disconnected before a $K = 4$ slab is slid in. Since the battery is gone, $Q_0 = 6\ \mu\text{C}$ stays fixed on the isolated plates. The capacitance still rises to $C' = KC = 4 \times (Q_0/V_0) = 4 \times 0.06\ \mu\text{F} = 0.24\ \mu\text{F}$, but now it is the voltage that falls: $V' = Q_0/C' = 6\times10^{-6}/(0.24\times10^{-6}) = 25\ \text{V}$, matching $V' = V_0/K = 100/4 = 25\ \text{V}$. If the original field was $E_0 = 2.0\times10^4\ \text{V/m}$, the new field is $E' = E_0/K = 5.0\times10^3\ \text{V/m}$ — field and voltage both fall by the factor $K$ while charge stays put, the mirror image of Steps 1–3 above where voltage was fixed and charge rose.

• Mixing up "battery connected" with "battery disconnected." Connected: $V$ stays fixed, so inserting the dielectric raises $Q$ to $Q' = KQ$. Disconnected: $Q$ stays fixed, so $V$ falls to $V' = V/K$. Using the wrong assumption gives a wrong numeric answer even with the correct formula.
• Treating a dielectric like a conductor. A conductor drives the internal field to exactly zero; a dielectric only reduces it to $E_0/K$ — for any finite $K$ the field inside never vanishes.
• Forgetting bound charge is not free charge. The induced surface charge on a dielectric cannot be drawn off with a wire or measured by an ammeter; it exists only while the polarizing field is present and disappears the moment the field is removed.
• Blurring polar and non-polar polarization. In non-polar molecules the field induces a dipole that did not exist before; in polar molecules the field only aligns dipoles that already existed. The ISC syllabus wants this distinction stated qualitatively, not derived.

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