Dielectric slabs in capacitors

Why is there plastic inside a real capacitor?

Open an old paper capacitor, or look at a modern film capacitor. Between the metal foils there is a thin sheet of paper, plastic or oxide. It is not there only to keep the foils apart. A dielectric between the plates makes the capacitor hold much more charge. We shall see why.

A dielectric that fills the gap

Charge an air capacitor to a charge $Q$, so that the voltage is $V_0$ and the field is $E_0 = V_0/d$. Now let a dielectric of constant $K$ fill the whole gap, and keep the charge $Q$ on the plates. The field is cut to $E = E_0/K$. The voltage is $V = Ed$, so it is cut too:

The same charge now makes a smaller voltage, so the capacitance $C = Q/V$ is larger:

The product $K\varepsilon_0$ is written $\varepsilon$ and called the permittivity of the dielectric. So a dielectric multiplies the capacitance by $K$, and this is always so, whether the battery is connected or not.

The battery connected, or the battery removed

When the dielectric goes in, other things change, and they depend on what is joined to the capacitor. There are two cases to learn. Take a capacitor $C_0$ charged to $V_0$, so that its charge is $Q_0 = C_0V_0$.

The battery stays connected. The battery holds the voltage fixed.

• The voltage stays $V_0$, and so the field $E = V_0/d$ stays the same.
• The capacitance becomes $KC_0$.
• The charge becomes $Q = KC_0V_0 = KQ_0$. The extra charge $(K - 1)Q_0$ comes from the battery.

The battery is removed first. The charge cannot leave, so it is fixed.

• The charge stays $Q_0$.
• The capacitance becomes $KC_0$.
• The voltage falls to $V_0/K$, and the field falls to $E_0/K$.

The simulation shows both cases. Read the lists again after you have used it. The two are mirror images: in one the voltage is fixed and the charge changes, and in the other the charge is fixed and the voltage changes.

A worked example. An air capacitor of $150$ pF is charged to $20$ V. A dielectric of constant $4.0$ fills the gap. Find what happens in each case.

1. The start. $Q_0 = C_0V_0 = 150 \times 10^{-12}\ \text{F} \times 20\ \text{V} = 3.0\ \text{nC}$.
2. The capacitance. $C = KC_0 = 4.0 \times 150\ \text{pF} = 600\ \text{pF}$, in both cases.
3. Battery connected. The voltage stays $20$ V, and $Q = CV = 600\ \text{pF} \times 20\ \text{V} = 12\ \text{nC}$. The battery sent in $12\ \text{nC} - 3.0\ \text{nC} = 9.0\ \text{nC}$.
4. Battery removed. The charge stays $3.0$ nC, and $V = \dfrac{Q}{C} = \dfrac{3.0\ \text{nC}}{600\ \text{pF}} = 5.0\ \text{V}$.

Set $K = 1$. Nothing changes, as it should. Now raise $K$. Which line rises and which falls? Can the dashed line ever be higher than $1$?

A slab that fills only part of the gap

Often a slab is thinner than the gap. Let the plates have an area $A$ and a gap $d$. A dielectric slab of thickness $t$ and constant $K$ sits between them, with air in the rest of the gap. The picture shows it, and it also shows the potential.

Put a charge $Q$ on the plates. The charge on each unit of area is $\sigma = Q/A$. In the air the field is the old one, $E_0 = \dfrac{\sigma}{\varepsilon_0}$. In the slab it is $\dfrac{E_0}{K}$. The potential difference is the sum over the gap:

With $E_0 = \dfrac{Q}{\varepsilon_0 A}$, the capacitance $C = Q/V$ is

Read this as a rule. A slab of thickness $t$ and constant $K$ does the work of an air layer of thickness $t/K$. The rest of the gap, $d - t$, stays air. Three things follow.

• The slab does not need to touch a plate. Its position makes no difference to the sum, so the capacitance is the same wherever it sits. We only need the slab to be parallel to the plates.
• If the slab fills the gap, $t = d$ and $C = K\varepsilon_0 A/d$, as before.
• A metal slab of thickness $t$ has no field inside. It acts like $K = \infty$, with $t/K = 0$. So $C = \dfrac{\varepsilon_0 A}{d - t}$. A metal slab simply takes away a length $t$ from the gap.

A worked example. The plates of a capacitor have an area of $200\ \text{cm}^2$ and a gap of $5.0$ mm. A slab of constant $6.0$ and thickness $3.0$ mm is put between them. Find $C$ with and without the slab, and with a metal slab of the same thickness.

1. Without the slab. $C_0 = \dfrac{\varepsilon_0 A}{d} = \dfrac{8.85 \times 10^{-12} \times 2.0 \times 10^{-2}}{5.0 \times 10^{-3}} = 35.4\ \text{pF}$.
2. The effective gap. $d - t + \dfrac{t}{K} = 5.0\ \text{mm} - 3.0\ \text{mm} + 0.50\ \text{mm} = 2.5\ \text{mm}$.
3. With the slab. $C = \dfrac{8.85 \times 10^{-12} \times 2.0 \times 10^{-2}}{2.5 \times 10^{-3}} = 70.8\ \text{pF}$. This is $2.0$ times $C_0$.
4. With a metal slab. The gap is $5.0\ \text{mm} - 3.0\ \text{mm} = 2.0\ \text{mm}$, so $C = \dfrac{8.85 \times 10^{-12} \times 2.0 \times 10^{-2}}{2.0 \times 10^{-3}} = 88.5\ \text{pF}$.

Slide the thickness from $0$ to $4.5$ mm. The curve rises slowly at first and steeply at the end. Why? Think of the effective gap. Then set $K$ to $10$. Does the curve come close to the dashed one?

A dielectric also protects against sparks

A dielectric has a limit too. Every material has a dielectric strength, the largest field it can take before it breaks down. For air it is about $3 \times 10^{6}$ V/m. For paper it is several times more, about $16 \times 10^{6}$ V/m. So a dielectric does two jobs. It raises $C$ by the factor $K$. It also lets the plates come closer, or the voltage go higher, before a spark jumps.

Practise with new numbers

Try it in Python

The program finds the capacitance for slabs of different thickness, a dielectric slab of constant $6.0$ and a metal slab, between plates of area $200\ \text{cm}^2$ and a gap of $5.0$ mm.

Common mistakes

• Mixing the two cases. With the battery connected, $V$ stays fixed. With the battery removed, $Q$ stays fixed. Decide first which one holds.
• Saying the voltage rises when a dielectric goes in. With the battery removed it falls to $V_0/K$.
• Using $KC_0$ for a slab that fills only part of the gap. Use $C = \varepsilon_0 A/(d - t + t/K)$.
• Adding $t$ and $K$. The slab acts as an air layer of thickness $t/K$. It is a division.
• Thinking the position of the slab matters. Between the plates, near one plate or in the middle, the capacitance is the same.
• Forgetting the metal case. A metal slab takes away its thickness $t$ from the gap. Here $K = \infty$ and $t/K = 0$.

Remember these

• A dielectric that fills the gap: $C = KC_0 = \dfrac{K\varepsilon_0 A}{d}$.
• Battery connected: $V$ and $E$ stay the same. $C$ and $Q$ grow by the factor $K$.
• Battery removed: $Q$ stays the same. $C$ grows by $K$, and $V$ and $E$ fall by the factor $K$.
• A slab of thickness $t$: $C = \dfrac{\varepsilon_0 A}{d - t + t/K}$. It does not matter where the slab sits.
• A metal slab: $C = \dfrac{\varepsilon_0 A}{d - t}$.
• A dielectric has a strength too, the largest field it takes before breakdown.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A capacitor stays joined to a battery. A dielectric slab fills the gap. Which of $Q$, $C$, $V$ and $E$ change, and which stay the same?

2. A charged capacitor has been disconnected. A dielectric of constant $5.0$ fills the gap. The capacitor had $40$ V across it. What is the voltage now?

3. A metal slab of thickness $d/2$ is slid into the gap of a parallel-plate capacitor, parallel to the plates. By what factor does the capacitance change?

4. A dielectric slab of thickness $t$ is moved from the middle of the gap to the surface of one plate. Does the capacitance change?