26.5.1 The Parallel-Plate Capacitor

A parallel-plate capacitor consists of two flat electrodes (the plates) facing each other with a plate separation d that is small compared to the sizes of the plates. You learned in Chapter 25 that the potential difference across a parallel-plate capacitor is related to the electric field inside by $Δ V_{C} = E d$ . And you know from Chapter 23 that the electric field inside a parallel-plate capacitor is

E = \frac{Q}{ϵ_{0} A}

(26.16)

where A is the surface area of the plates. Combining these gives

Q = \frac{ϵ_{0} A}{d} Δ V_{C}

(26.17)

You can see that the charge is proportional to the potential difference, as expected. So from the definition of capacitance, Equation 26.14, we find that the capacitance of a parallel-plate capacitor is

C = \frac{Q}{Δ V_{C}} = \frac{ϵ_{0} A}{d} (parallel - plate capacitor)

(26.18)

The capacitance is a purely geometric property of the electrodes, depending only on their surface area and spacing. Capacitors of other shapes will have different formulas for their capacitance, but all will depend entirely on geometry. A cylindrical capacitor is the topic of the Chapter 26 Challenge Example, and a homework problem will let you analyze a spherical capacitor.

A photograph of a condenser microphone with its outer shell removed. — A *condenser microphone*, the type of microphone typically used in a recording studio, is simply a parallel-plate capacitor in which the front electrode—seen in the photo with the microphone’s outer shell removed—is a very thin, lightweight metal foil. (“Condenser” is an old-fashioned term for a capacitor.) Sound waves cause the foil to vibrate as a diaphragm, which slightly changes the spacing between the electrodes and thus the capacitance. The charge on a capacitor is $Q = C Δ V_{C}$ , so charge flows to and from the capacitor as its capacitance oscillates. This oscillating flow of charge—an oscillating current—is sensed and amplified to become the microphone’s output voltage.

Full Alternative Text

EXAMPLE 26.6 Charging a capacitor

The spacing between the plates of a $1.0 μ F$ capacitor is 0.050 mm.

What is the surface area of the plates?
How much charge is on the plates if this capacitor is charged to 1.5 V?

MODEL Assume the capacitor is a parallel-plate capacitor.

SOLVE

From the definition of capacitance,

$A = \frac{d C}{ϵ_{0}} = 5.65 m^{2}$
The charge is $Q = C Δ V_{C} = 1.5 \times 10^{- 6} C = 1.5 μ C$ .

REVIEW The surface area needed to construct a $1.0 μ F$ capacitor (a fairly typical value) is enormous. We’ll see in Section 26.7 how the area can be reduced by inserting an insulator between the capacitor plates.