The Electrostatic Field: Postulates

Gauss ’ s Law and the Electric Potential

4

I have had my results for a long time but I do not yet know how I am to arrive at them.

—Johann Carl Friedrich Gauss (1777–1855) (A. Arber, The Mind and the Eye, 1954)

To calculate the divergence of the electric field intensity, we apply the divergence theorem to this field:

ð

v

∇

^E

ð Þdv¼ þ

s

E

^ds¼þ

s

R^ Q 4πε0

R²

^{R^ds¼}_ε^Q

0

ð4:2Þ

where the element of surface was taken by definition to point out of the volume. In fact, we obtained two results:

ð

v

∇

^E

ð Þdv¼Q ε0

ð4:3Þ

and

þ

s

E

^ds¼_ε^Q

0

ð4:4Þ

which are different ways of describing the same quantity.

Reverting back to the divergence theorem [Eq. (4.2)], we see that the closed surface integral or the volume integral is related to the total flux ofEthrough the surfacesenclosing the volumev. This means that the chargeQinEq. (4.3)is the total charge in volumev. Assuming the charge is distributed throughout the volumev, with a charge densityρv, we have in free space(seeFigure4.1b)

ð

v

∇

^E

ð Þdv¼ 1 ε0

ð

v

ρvdv ð4:5Þ

or, equating the integrands,

∇

^E¼ρ_ε^v

0

ð4:6Þ

Equation (4.6)applies to any charge distribution since it gives a differential or point relation. Similarly, if the divergence of an electric field is zero at a given point, the charge density at that point is zero. Note also that the result inEq. (4.5) indicates that the two charge distributions inFigures4.1aand4.1bproduce identical fields at identical distance if the two surfaces enclose identical total charge.

The result in Eq. (4.6) [or Eq. (4.5)] is important in two ways: First, it defines the sources of the electric field intensity; second, it provides a means of calculating the electric field intensity. However, more important is the fact that we have established the divergence of the electric field intensity as one of the conditions required to define the vector field.

Q R

E

a b

ρv

.

^{. R E}

ρ Q= _vdv

∫

v

Figure 4.1 Two charge distributions that produce identical fields outside the sphere of radiusR. (a) Electric field of a positive point charge. (b) Charge density in the volume. The total charge in the two cases is the same

140 4 Gauss’s Law and the Electric Potential

The second condition is the curl of the electric field intensity, which is defined through application of Stokes’theorem to the field of the point charge:

ð

s

∇E

ð Þ

^ds¼þ

C

E

^dl¼þ

C

Q 4πε0

R²R^

^{dl ð4:7Þ}

For the case given here, any contourCthat lies on the sphere of radiusRproduces an open, circular surface as shown in Figure4.2; that is, if we were to cut a section of the sphere, the rim of the cut is the closed contour. The electric field intensity is radial, therefore perpendicular everywhere to this surface. This means that the scalar productR^

^dl¼0 and, therefore, the right-hand side ofEq. (4.7)is zero. Thus,

∇E¼0 ð4:8Þ

However, we have taken a very special example: that of a point charge and a surface on the sphere at which the electric field has constant magnitude. The question that remains to be answered is: Is this relation general? In other words, can we, in fact, state that the curl of the static electric field is always zero? The answer to this is twofold. First, it is always zero for the electrostatic field. We can obviously say nothing about any time dependency that might exist since this possibility was never considered. Second, we will give more general proof to the correctness of this relation, including a very simple physical meaning. For now, we will accept this to be correct. Thus, we have at this point both the divergence [Eq. (4.6)] and curl [Eq.(4.8)] of the electric field.

According to the Helmholtz theorem(Section2.5.1), the electrostatic field isirrotational(curl-free) andnonsolenoidal (nonzero divergence); that is, the electrostatic field is generated by a scalar source alone: a charge or a charge density. These two equations are, in fact, all that is necessary to define the electrostatic vector field. For this reason, they are often taken as the basic postulates¹for the electrostatic field. From these, we can derive all previous relations. The postulates were derived here for two purposes. One is the simple fact that they describe the fundamental properties of the static electric field as a vector and in very compact form. The second is to show how the properties of vectors can be used to derive useful field relations. This process, while not as intuitive as, say, Coulomb’s law, is extremely useful in field relations and we will make considerable use of both relations inEqs. (4.6)and(4.8).

To derive the differential relations inEqs. (4.6)and(4.8), we made use of the divergence and Stokes’theorems, both of which are integral relations. If we now return toEqs. (4.4), (4.7)and(4.8), we can also write directly

þ

s

E

^ds¼_ε^Q

0

, þ

C

E

^{dl¼0 ð4:9Þ}

E E

C C

Figure 4.2 A closed contour on the surface of a sphere

1Postulates are axioms that have not been disproved. As mentioned before, the postulates, or fundamental relations, are experimentally evaluated quantities, and since experiments have not shown anything to the contrary, we can use them as postulates. In the very unlikely event that one or more of our postulates turn out to be wrong, we would have then to“adjust”all derived relations. There is very little danger of that happening, and, until it does, we can safely use the current relations.

The relations inEq. (4.9)are the integral forms of the postulates from which the differential forms inEqs. (4.6)and(4.8) were obtained. They are, therefore, the integral form of the postulates.

To summarize, the two required postulates for the electrostatic field are

∇

^E¼ρ_ε^v

0

, ∇E¼0 in differential form orþ

s

E

^ds¼_ε^Q

0

, þ

C

E

^dl¼0 in integral form

ð4:10Þ

Any electrostatic field must satisfy these conditions.

Example 4.1The Electrostatic Field The following vector fields are given in free space:

(1) A1¼x^x, (2)A2 ¼x^5, (3)A3¼A1þA2 ¼x^ðxþ5Þ (a) Are these electrostatic fields?

(b) For those fields that are electrostatic, calculate the equivalent charge density that generates these fields in free space.

Solution: For a vector field to be an electrostatic field, it must satisfy the two postulates inEq. (4.10); that is, its curl must be zero and its divergence must be nonzero within the charge distribution, if any:

(a) The divergence of each of the three vector fields is

∇

^A1¼∂A_∂x^xþ∂A_∂y^yþ∂A_∂z^{z¼d x}_dx^{ð Þ¼1, ∇}

^{A2¼0, ∇}

^A3¼1

The first condition, namely, that the divergence be nonzero (or zero if there are no charge densities), is satisfied. The curl ofAis

∇A1¼x^ ∂Az

∂y ∂Ay

∂z 0

@

1

Aþy^ ∂Ax

∂z ∂Az

∂x 0

@

1

Aþz^ ∂Ay

∂x ∂Ax

∂y 0

@

1

A¼x^ð00Þ þy^ ∂ð Þx

∂z 0 0

@

1

Aþz^ 0∂ð Þx

∂y 0

@

1 A¼0

∇A2 ¼0, ∇A3¼0

Thus, since both postulates are satisfied, the vector fieldsA₁,A₂, andA₃represent electrostatic fields. Note in particular that vector fieldA₂has zero divergence. This means that within the domain considered, there are no charge densities and the field is generated by charges outside the solution domain. How this is done will be shown later in this chapter. In this respect, it is worth recalling that the Helmholtz theorem defines the vector field to within an additive constant: the divergence or the curl is not changed by adding a constant toA₁,A₂, orA₃.

(b) The charge density everywhere is calculated from the first postulate [Eq.(4.6)]:

∇

^A1¼ρ_ε¹

0

¼1 ! ρ1¼ε0¼18:85410¹² C m³

Similarly,

ρ2 ¼0, ρ3¼ε0¼18:85410¹² C m³

These volume charge densities are uniform throughout space.

Exercise 4.1 Is the vector fieldA¼ϕ^r²z^5zan electrostatic field? Explain.

142 4 Gauss’s Law and the Electric Potential

Answer No, because its curl is nonzero:∇

^{A¼ 5,∇A¼z^3r.}