# 4.3 Coulomb’s Law

One of the primary goals of this chapter is to develop dexterity in applying the expressions for the electric field intensity E and associated electric flux density D induced by a specified distribution of charge. Our discussion will be limited to electrostatic fields induced by stationary charge densities.

We begin by reviewing the expression for the electric field introduced in Section 1-3.2 on the basis of the results of Coulomb’s experiments on the electrical force between charged bodies. Coulomb’s law, which was first introduced for electrical charges in air and later generalized to material media, implies that:

1. An isolated charge q induces an electric field E at every point in space, and at any specific point P, E is given by

   E=R ˆ q 4 π ε R 2(V/m), (4.13)

   where R ˆ is a unit vector pointing from q to P (Fig. 4-3), R is the distance between them, and ε is the electrical permittivity of the medium containing the observation point P.

   # Figure 4-3 Electric field lines due to a charge q.

   Figure 4 3 Full Alternative Text

2. In the presence of an electric field E at a given point in space that may be due to a single charge or a distribution of charges, the force acting on a test charge q′ when placed at P is

   F=q′E(N). (4.14)

With F measured in newtons (N) and q′ in coulombs (C), the unit of E is (N/C), which will be shown later in Section 4-5 to be the same as volt per meter (V/m).

For a material with electrical permittivity ε, the electric field quantities D and E are related by

D=ε E(4.15)

with

ε=ε r ε 0,(4.16)

where

ε 0=8.85×10−12≈(1/36 π)×10−9(F/m)

is the electrical permittivity of free space and ε r=ε/ε 0 is called the relative permittivity (or dielectric constant) of the material. For most materials and under a wide range of conditions, ε is independent of both the magnitude and direction of E [as implied by Eq. (4.15)].

Materials usually do not exhibit nonlinear permittivity behavior except when the amplitude of E is very high (at levels approaching dielectric breakdown conditions discussed later in Section 4-7), and anisotropy is present only in certain materials with peculiar crystalline structures. Hence, except for unique materials under very special circumstances, the quantities D and E are effectively redundant; for a material with known ε, knowledge of either D or E is sufficient to specify the other in that material.

# 4.3.1 Electric Field Due to Multiple Point Charges

The expression given by Eq. (4.13) for the field E due to a single point charge can be extended to multiple charges. We begin by considering two point charges, q 1 and q 2, with position vectors R 1 and R 2 (measured from the origin in Fig. 4-4). The electric field E is to be evaluated at a point P with position vector R. At P, the electric field E 1 due to q 1 alone is given by Eq. (4.13) with R, which is the distance between q 1 and P, replaced with |R−R 1| and the unit vector R ˆ replaced with (R−R 1)/|R−R 1|. Thus,

E 1=q 1(R−R 1)4 π ε|R−R 1|3(V/m). (4.17a)

Similarly, the electric field at P due to q 2 alone is

E 2=q 2(R−R 2)4 π ε|R−R 2|3(V/m). (4.17b)

# Figure 4-4 The electric field E at P due to two charges is equal to the vector sum of E 1 and E 2.

Figure 4 4 Full Alternative Text

Hence, the total electric field E at P due to q 1 and q 2 together is determined as

E=E 1+E 2=1 4 π ε q 1(R−R 1)|R−R 1|3+q 2(R−R 2)|R−R 2|3.(4.18)

Generalizing the preceding result to the case of N point charges, the electric field E at point P with position vector R due to charges q 1,q 2,…,q N located at points with position vectors R 1,R 2,…,R N equals the vector sum of the electric fields induced by all the individual charges. Thus,

# Example 4-3: Electric Field Due to Two Point Charges

Two point charges with

q 1=2×10−5 C

and

q 2=−4×10−5 C

are located in free space at points with Cartesian coordinates (1,3,−1) and (−3,1,−2), respectively. Find (a) the electric field E at (3,1,−2) and (b) the force on a 8×10−5 C charge located at that point. All distances are in meters.

Solution: (a) From Eq. (4.18), the electric field E with ε=ε 0 (free space) is

E=1 4 π ε 0 q 1(R−R 1)|R−R 1|3+q 2(R−R 2)|R−R 2|3(V/m).

The vectors R 1, R 2, and R are

R 1=x ˆ+y ˆ 3−z ˆ,R 2=−x ˆ 3+y ˆ−z ˆ 2,R=x ˆ 3+y ˆ−z ˆ 2.

Hence,

E=1 4 π ε 0 2(x ˆ 2−y ˆ 2−z ˆ)27−4(x ˆ 6)216×10−5=x ˆ−y ˆ 4−z ˆ 2 108 π ε 0×10−5(V/m).

(b) The force on q 3 is

F=q 3 E=8×10−5×x ˆ−y ˆ 4−z ˆ 2 108 π ε 0×10−5=x ˆ 2−y ˆ 8−z ˆ 4 27 π ε 0×10−10(N).

# 4.3.2 Electric Field Due to a Charge Distribution

We now extend the results obtained for the field due to discrete point charges to continuous charge distributions. Consider a volume υ′ that contains a distribution of electric charge with volume charge density ρ v, which may vary spatially within υ′ (Fig. 4-5). The differential electric field at a point P due to a differential amount of charge d q=ρ v d υ′ contained in a differential volume d υ′ is

d E=R ˆ′d q 4 π ε R′2=R ˆ′ρ v d υ′4 π ε R′2,(4.20)

where R′ is the vector from the differential volume d υ′ to point P. Applying the principle of linear superposition, the total electric field E is obtained by integrating the fields due to all differential charges in υ′. Thus,

It is important to note that, in general, both R′ and R ˆ′ vary as a function of position over the integration volume υ′.

# Figure 4-5 Electric field due to a volume charge distribution.

Figure 4 5 Full Alternative Text

If the charge is distributed across a surface S′ with surface charge density ρ s, then d q=ρ s d s′, and if it is distributed along a line l′ with a line charge density ρ ℓ, then d q=ρ ℓ d l′. Accordingly, the electric fields due to surface and line charge distributions are

# Example 4-4: Electric Field of a Ring of Charge

A ring of charge of radius b is characterized by a uniform line charge density of positive polarity ρ ℓ. The ring resides in free space and is positioned in the x–y plane, as shown in Fig. 4-6. Determine the electric field intensity E at a point P=(0,0,h) along the axis of the ring at a distance h from its center.

Solution: We start by considering the electric field generated by a differential ring segment with cylindrical coordinates (b, ϕ,0) in Fig. 4-6(a). The segment has length d l=b d ϕ and contains charge d q=ρ ℓ d l=ρ ℓ b d ϕ. The distance vector R 1′ from segment 1 to point P=(0,0,h) is

R 1′=−r ˆ b+z ˆ h,

from which it follows that

R 1′=|R 1′|=b 2+h 2,R ˆ 1′=R 1′|R 1′|=−r ˆ b+z ˆ h b 2+h 2.

The electric field at P=(0,0,h) due to the charge in segment 1 therefore is

d E 1=1 4 π ε 0 R ˆ 1′ρ ℓ d l R 1′2=ρ ℓ b 4 π ε 0(−r ˆ b+z ˆ h)(b 2+h 2)3/2 d ϕ.

# Figure 4-6 Ring of charge with line density ρ ℓ. (a) The field d E 1 due to infinitesimal segment 1 and (b) the fields d E 1 and d E 2 due to segments at diametrically opposite locations (Example 4-4).

Figure 4 6-a Full Alternative Text
Figure 4 6-b Full Alternative Text

The field d E 1 has component d E 1 r along −r ˆ and component d E 1 z along z ˆ. From symmetry considerations, the field d E 2 generated by differential segment 2 in Fig. 4-6(b), which is located diametrically opposite to segment 1, is identical to d E 1 except that the r ˆ component of d E 2 is opposite that of d E 1. Hence, the r ˆ components in the sum cancel and the z ˆ contributions add. The sum of the two contributions is

d E=d E 1+d E 2=z ˆ ρ ℓ b h 2 π ε 0 d ϕ(b 2+h 2)3/2.(4.22)

Since for every ring segment in the semicircle defined over the azimuthal range 0≤ϕ≤π (the right-hand half of the circular ring) there is a corresponding segment located diametrically opposite at (ϕ+π), we can obtain the total field generated by the ring by integrating Eq. (4.22) over a semicircle as

E=z ˆ ρ ℓ b h 2 π ε 0(b 2+h 2)3/2∫0 π d ϕ=z ˆ ρ ℓ b h 2 ε 0(b 2+h 2)3/2=z ˆ h 4 π ε 0(b 2+h 2)3/2 Q,(4.23)

where Q=2 π b ρ ℓ is the total charge on the ring.

# Example 4-5: Electric Field of a Circular Disk of Charge

Find the electric field at point P with Cartesian coordinates (0,0,h) due to a circular disk of radius a and uniform charge density ρ s residing in the x–y plane (Fig. 4-7). Also, evaluate E due to an infinite sheet of charge density ρ s by letting a→∞.

# Figure 4-7 Circular disk of charge with surface charge density ρ s. The electric field at P=(0,0,h) points along the z direction (Example 4-5).

Figure 4 7 Full Alternative Text

Solution: Building on the expression obtained in Example 4-4 for the on-axis electric field due to a circular ring of charge, we can determine the field due to the circular disk by treating the disk as a set of concentric rings. A ring of radius r and width d r has an area d s=2 π r d r and contains charge d q=ρ s d s=2 π ρ s r d r. Upon using this expression in Eq. (4.23) and also replacing b with r, we obtain the following expression for the field due to the ring:

d E=z ˆ h 4 π ε 0(r 2+h 2)3/2(2 π ρ s r d r).

The total field at P is obtained by integrating the expression over the limits r=0 to r=a:

E=z ˆ ρ s h 2 ε 0∫0 a r d r(r 2+h 2)3/2=±z ˆ ρ s 2 ε 0 1−|h|a 2+h 2,(4.24)

with the plus sign for h>0 (P above the disk) and the minus sign when h<0 (P below the disk).

For an infinite sheet of charge with a=∞,

We note that for an infinite sheet of charge E is the same at all points above the x–y plane, and a similar statement applies for points below the x–y plane.

Concept Question 4-4: When characterizing the electrical permittivity of a material, what do the terms linear and isotropic mean?

Concept Question 4-5: If the electric field is zero at a given point in space, does this imply the absence of electric charges?

Concept Question 4-6: State the principle of linear superposition as it applies to the electric field due to a distribution of electric charge.