# 4.9 Capacitance

When separated by an insulating (dielectric) medium, any two conducting bodies, regardless of their shapes and sizes, form a capacitor. If a dc voltage source is connected across them (Fig. 4-25) the surfaces of the conductors connected to the positive and negative source terminals accumulate charges +Q and −Q, respectively.

# Figure 4-25 A dc voltage source connected to a capacitor composed of two conducting bodies.

Figure 4 25 Full Alternative Text

The capacitance of a two-conductor configuration is defined as

C=Q V(C/V or F),(4.115)

where V is the potential (voltage) difference between the conductors. Capacitance is measured in farads (F), which is equivalent to coulombs per volt (C/V).

The presence of free charges on the conductors’ surfaces gives rise to an electric field E (Fig. 4-25) with field lines originating on the positive charges and terminating on the negative ones. Since the tangential component of E always vanishes at a conductor’s surface, E is always perpendicular to the conducting surfaces. The normal component of E at any point on the surface of either conductor is given by

where ρ s is the surface charge density at that point, n ˆ is the outward normal unit vector at the same location, and ε is the permittivity of the dielectric medium separating the conductors. The charge Q is equal to the integral of ρ s over surface S (Fig. 4-25):

Q=∫S ρ s d s=∫S ε n ˆ⋅E d s=∫S ε E⋅d s,(4.117)

where use was made of Eq. (4.116). The voltage V is related to E by Eq. (4.39):

V=V 12=−∫P 2 P 1 E⋅d l,(4.118)

where points P 1 and P 2 are any two arbitrary points on conductors 1 and 2, respectively. Substituting Eqs. (4.117) and (4.118) into Eq. (4.115) gives

where l is the integration path from conductor 2 to conductor 1. To avoid making sign errors when applying Eq. (4.119), it is important to remember that surface S is the +Q surface and P 1 is on S. [Alternatively, if you compute C and it comes out negative, just change its sign.] Because E appears in both the numerator and denominator of Eq. (4.119), the value of C obtained for any specific capacitor configuration is always independent of E’s magnitude. In fact, C depends only on the capacitor geometry (sizes, shapes and relative positions of the two conductors) and the permittivity of the insulating material.

If the material between the conductors is not a perfect dielectric (i.e., if it has a small conductivity σ), then current can flow through the material between the conductors, and the material exhibits a resistance R. The general expression for R for a resistor of arbitrary shape is given by Eq. (4.81):

R=−∫l E⋅d l∫S σ E⋅d s(Ω).(4.120)

For a medium with uniform σ and ε, the product of Eqs. (4.119) and (4.120) gives

This simple relation allows us to find R if C is known, and vice versa.

# Example 4-14: Capacitance of Parallel-Plate Capacitor

Obtain an expression for the capacitance C of a parallel-plate capacitor comprised of two parallel plates each of surface area A and separated by a distance d. The capacitor is filled with a dielectric material with permittivity ε.

Solution: In Fig. 4-26, we place the lower plate of the capacitor in the x–y plane and the upper plate in the plane z=d. Because of the applied voltage difference V, charges +Q and −Q accumulate on the top and bottom capacitor plates. If the plate dimensions are much larger than the separation d, then these charges distribute themselves quasi-uniformly across the plates, giving rise to a quasi-uniform field between them pointing in the −z ˆ direction. In addition, a fringing field will exist near the capacitor edges, but its effects may be ignored because the bulk of the electric field exists between the plates. The charge density on the upper plate is ρ s=Q/A. Hence, in the dielectric medium

E=−z ˆ E,

and from Eq. (4.116), the magnitude of E at the conductor–dielectric boundary is E=ρ s/ε=Q/ε A. From Eq. (4.118), the voltage difference is

V=−∫0 d E⋅d l=−∫0 d(−z ˆ E)⋅z ˆ d z=E d,(4.122)

and the capacitance is

where use was made of the relation E=Q/ε A.

# Figure 4-26 A dc voltage source connected to a parallel-plate capacitor (Example 4-14).

Figure 4 26 Full Alternative Text

# Example 4-15: Capacitance per Unit Length of Coaxial Line

Obtain an expression for the capacitance of the coaxial line shown in Fig. 4-27.

# Figure 4-27 Coaxial capacitor filled with insulating material of permittivity ε (Example 4-15).

Figure 4 27 Full Alternative Text

Solution: For a given voltage V across the capacitor, charges +Q and −Q accumulate on the surfaces of the outer and inner conductors, respectively. We assume that these charges are uniformly distributed along the length and circumference of the conductors with surface charge density ρ s′=Q/2 π b l on the outer conductor and ρ s′′=−Q/2 π a l on the inner one. Ignoring fringing fields near the ends of the coaxial line, we can construct a cylindrical Gaussian surface in the dielectric in between the conductors with the radius r such that a<r<b. Symmetry implies that the E-field is identical at all points on this surface, which is directed radially inward. From Gauss’s law, it follows that the field magnitude equals the absolute value of the total charge enclosed, divided by the surface area. That is,

E=−r ˆ Q 2 π ε r l.(4.124)

The potential difference V between the outer and inner conductors is

V=−∫a b E⋅d l=−∫a b−r ˆ Q 2 π ε r l⋅(r ˆ d r)=Q 2 π ε l ln b a.(4.125)

The capacitance C is then given by

and the capacitance per unit length of the coaxial line is

C′=C l=2 π ε ln(b/a)(F/m).(4.127)

Concept Question 4-24: How is the capacitance of a two-conductor structure related to the resistance of the insulating material between the conductors?

Concept Question 4-25: What are fringing fields and when may they be ignored?