Capacitance & Parallel Plates
Two conductors, equal and opposite charges, and Q = CV.
Lesson at a glanceShow the key points
- Capacitor: two close conductors with +Q and −Q; Q = CV (31.1).
- C depends on geometry and medium; unit farad (C/V); practical units µF, nF, pF.
- An ideal battery keeps its terminals at a fixed potential difference, its emf.
- Parallel plates: E = Q/ε₀A, V = Ed, C = ε₀A/d (31.2); ε₀ in F/m.
Put two conductors close together, push charge onto one and pull the same amount off the other, and you have a device that stores charge and energy: a capacitor.
Capacitor and capacitance
- The charge on the capacitor means Q, the charge on the positive plate. (The total charge, +Q − Q, is zero.)
- The potential of the capacitor means the potential difference V = V₊ − V₋ between the plates.
For a given capacitor, double the charge and the potential difference doubles too: Q is proportional to V.
The constant C, the capacitance, depends only on the shape, size and arrangement of the conductors and on what fills the space between them. Its unit is the coulomb per volt, called the farad (F). A farad is enormous; everyday capacitors are measured in microfarads (µF), nanofarads (nF) or picofarads (pF).
Charging with a battery
The easy way to put ±Q on the plates is to connect them to a battery. For now we treat the battery as ideal:
- It has two terminals with a fixed potential difference between them, equal to its emf ℰ. The higher-potential terminal is positive.
- A conductor joined to a terminal takes that terminal’s potential, so a capacitor across the battery ends up with V = ℰ.
- Its net charge stays zero: if the + terminal supplies +Q, the − terminal supplies −Q.
- Moving charge Q through itself from − to +, the battery does work Qℰ.
In circuit diagrams a battery is drawn as a long and a short line (the long line is positive), and a capacitor as two equal parallel lines.
A 6.0 V battery puts 24 µC on a capacitor. What is its capacitance?
Finding a capacitance
The recipe is always the same: place +Q and −Q on the plates, find the field between them (usually with Gauss’s law), integrate it to get V, then C = Q/V.
The parallel-plate capacitor
Two large flat plates, each of facing area A, a small distance d apart, with vacuum between. The charges sit on the facing surfaces with density σ = Q/A. If the plates are large compared with d, the field between them is uniform and perpendicular to the plates (except near the edges).
A pillbox with one face inside the positive plate (no field there) and one face in the gap encloses charge σΔA, so Gauss’s law gives EΔA = σΔA/ε₀, i.e. E = Q/ε₀A. Going across the gap,
Find the capacitance of two 30 cm × 30 cm plates 0.50 mm apart in vacuum.
Plates of area 100 cm², 0.50 mm apart, are given 2.0 nC. Find the potential difference. The plates are then pulled to 1.5 mm apart with the charge unchanged. What is the new potential difference?
- 1A capacitor: two close conductors with +Q and −Q; Q = CV (31.1).
- 2C depends only on geometry and the medium; unit farad = C/V.
- 3An ideal battery fixes the potential difference at its emf.
- 4Parallel plates: E = Q/ε₀A, V = Qd/ε₀A, C = ε₀A/d (31.2).
Check yourself
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