Capacitors and the parallel plate capacitor

Capacitance

Narrated video. Transcript:

A capacitor is two conductors separated by an insulator. Connect them to a battery: one gets a charge +Q, the other −Q, and there is a potential difference V between them.

Double the charge, and the field everywhere between them doubles, so V doubles too. Q and V are proportional, and their ratio is the capacitance: C = Q/V.

C depends only on the shapes, sizes and separation of the conductors, and on the insulator between them. Its SI unit is the farad: one coulomb per volt.

The commonest kind is the parallel plate capacitor: two flat metal plates of area A, a small distance d apart. The charge spreads evenly, with density σ = Q/A.

Outside the plates, the fields of the two charged sheets cancel. Between them they add: E = σ/ε₀, uniform, from the positive plate to the negative one. We ignore the fringing at the edges.

So V = Ed = Qd/ε₀A. Divide Q by V, and C = ε₀A/d. Bigger plates, closer together, give a bigger capacitance.

Worked example: plates of area 100 cm², that is 0.01 m², 2 mm apart. C = 8.85 × 10⁻¹² × 0.01 ÷ 0.002 = 4.4 × 10⁻¹¹ F, about 44 picofarads.

On a 100 V battery it takes a charge Q = CV, about 4.4 × 10⁻⁹ C. The field between the plates is 100 ÷ 0.002, which is 5 × 10⁴ V/m.

A whole farad is huge. With plates 1 cm apart you would need an area of 0.01 ÷ 8.85 × 10⁻¹², about 1.1 × 10⁹ m²: a square more than 30 km on each side! Real capacitors are measured in μF and pF.

Adding a dielectric

Narrated video. Transcript:

Now fill the gap between the plates with a dielectric, keeping the charge on the plates fixed. The slab is polarised, with bound charges −σₚ and +σₚ on its faces, right next to the plates.

The field inside becomes E = (σ − σₚ)/ε₀. For a linear dielectric this is σ/Kε₀: smaller by the dielectric constant K.

So the potential difference falls to V = Qd/Kε₀A, and the capacitance rises: C = Kε₀A/d. It is K times the capacitance with vacuum, C₀.

We also write ε = Kε₀, the permittivity of the medium, so K = ε/ε₀. For vacuum K is 1, for air very nearly 1, and for every other dielectric it is more than 1.

Example: our 44 pF capacitor is charged to 100 V and the battery is disconnected. A slab with K = 4 is slid in to fill the gap. The charge is stuck at 4.4 nC, but C rises to about 177 pF.

So V = Q/C falls to a quarter, just 25 V. The field between the plates drops from 5 × 10⁴ to 1.25 × 10⁴ V/m.

Repeat with the battery still connected. Now V stays at 100 V, so the battery pushes in more charge: Q = CV becomes four times as much, about 17.7 nC.

A slab that fills only part of the gap helps too. If its thickness is t, then V = E₀(d − t) + E₀t/K, and C = ε₀A/(d − t + t/K).

For a slab three quarters of d thick, C = 4K/(K + 3) × C₀. With K = 4 that is 16/7, about 2.3 times C₀.

Try it: parallel plate capacitor lab

Interactive simulation. Things to try:

• Increase the plate area A: C grows in proportion. Double d and C halves.
• With the battery connected, pull the plates apart: E = V/d falls and charge flows back to the battery.
• Disconnect the battery, then push in a full slab (t/d = 1) with K = 4: C goes up 4 times and V drops to a quarter.
• Set t/d = 0.75 and K = 4: C/C₀ = 16/7 ≈ 2.29.

Closer, with a dielectric

An air-filled parallel plate capacitor has C = 8 pF. The plate separation is halved and the gap is filled with a dielectric of K = 6. What is the new capacitance?

1. 24 pF
2. 48 pF
3. 96 pF
4. 12 pF

Battery disconnected

A parallel plate capacitor is charged and then disconnected from the battery. A slab of K = 4 is now pushed in to fill the gap. What happens?

1. C becomes 4 times, Q stays the same, V becomes ¼
2. C becomes 4 times, Q becomes 4 times, V stays the same
3. C becomes ¼, Q stays the same, V becomes 4 times
4. C becomes 4 times, Q and V stay the same

Summary

• Capacitance: C = Q/V (1 F = 1 C/V)
• Parallel plate, vacuum: C₀ = ε₀A/d
• Filled with dielectric: C = Kε₀A/d = KC₀, K = ε/ε₀
• Slab of thickness t: C = ε₀A/(d − t + t/K)

• A capacitor is two conductors with charges +Q and −Q; C depends only on geometry and the medium.
• Between parallel plates the field is uniform, E = σ/ε₀, and V = Ed.
• A dielectric reduces the field by K and so increases C by K.
• Battery connected: V fixed, Q changes. Battery disconnected: Q fixed, V changes.