# Charge Distributions.

## 1. - Discrete Charge Distributions

A point charge is a point particle[1A point particle defines a particle which lacks spatial extension.] which has an electric charge. There are two types of electric charges: positive and negative. The elementary charge, a fundamental physical constant usually denoted as e, is the electric charge carried by a single proton, whereas a single electron has charge −e. In the International System of Units (SI) the elementary charge has a measured value of approximately 1.602 176 6208(98)×10−19 C. In the natural system of Hartree atomic units (usually abbreviated “au” or “a.u.”)[2Definition of Hartree atomic units.] the elementary charge functions as the unit of electric charge, that is e is equal to 1.

A point charge or monopole is characterized by a strength q0 and is located at a certain position r0. The amount of electric charge per unit volume is given by the charge density, ρ(r), whose SI units are C⋅m−3. The atomic unit of charge density (e/a03, where a0 = 0.529 177 210 67(12) × 10−10 m is the radius of the first Bohr orbit for hydrogen) is equal to 1.081 202 3770(67) × 1012 C ⋅ m−3. The volume charge density can be expressed by the Dirac delta function, δ(rr0), that is equal to zero everywhere but a single point r = r0:[4Dirac, P. “The Principles of Quantum Mechanics”
Oxford at the Clarendon Press: 1958.
, 5Messiah, A. “Quantum Mechanics”, Volume I
North-Holland Publishing Company, Amsterdam: 1970, pp. 179-184 and 468-471.
, 6Three-Dimensional Dirac Delta Function.]

 ρ(r) = q0 δ(r − r0) (1.1)

At the point r = r0 the Dirac delta function δ(rr0) is singular in such a manner that

 ∫   δ(r − r0) d3r = 1 (1.2)

So, integrating eq (1.1) over any volume that contains r0 yields the electric charge of the distribution

 ∫   ρ(r) d3r = q0 (1.3)

It is assumed that the vector q0r0 is pointing towards the charge q0 if this is positive, and away from the charge if this is negative.

In the general case of an ensemble of N point charges qi (i = 1, 2, …, N), the total charge Q of the distribution is the sum of the N individual point charges:

 Q = N ∑ i=1 qi (1.4)

and the total charge density is the sum of the N individual charge densities

 ρ(r) = N ∑ i=1 qi δ(r − ri) (1.5)

Integrating eq. (1.5) over any volume that contains all the ri yields, because of eq. (1.2), the total charge of the distribution

 ∫   ρ(r) d3r = N ∑ i=1 qi ∫   δ(r − ri) d3r = Q (1.6)

The sum of the vectors qi ri defines the dipole moment vector of the distribution

 p = N ∑ i=1 qi ri (1.7)

In particular, an ensemble of two point-like particles carrying charges of the same strength q and opposite sign represents an electric dipole. An electric dipole can be characterized by its electric dipole moment, a vector quantity that points from the negative charge (q1 = −q at location r1 = r) towards the positive charge (q2 = +q at location r2 = r+), so that it is defined by the difference

The magnitude of the electric dipole moment is given by the value of the charge q times the distance of the opposite charges. It is a measure of the polarity of the charge distribution, that is of the degree of local charge separation. In other words, a dipole exists when the centers of positive and negative charge distribution do not coincide. Though the position vectors r+ and r are defined with respect to some reference system of coordinates, it is clear from eq. (1.8) that the electric dipole moment of a pair of opposite charges does not depend on the choice of origin. This fact holds also in the general case of eq. (1.7), provided the overall charge of the system is zero. Indeed, if all the position vectors, ri, are moved to a new origin r0 (i.e. ririr0), then eq. (1.7) becomes:

 p(r0) = N ∑ i=1 qi (ri − r0) = N ∑ i=1 qi ri − Q r0 (1.9)

which is independent of r0 only if Q = 0.[7Conventional choice of the origin for charged systems.]

The SI units for electric dipole moment are coulomb-meter (C⋅m), however the most commonly used unit is the debye (D), which is a CGS unit. A debye is 10−18 esu⋅cm and about 3.336 × 10−30 C⋅m. An atomic unit of electric dipole moment is e⋅a0 and is about 8.478 × 10−29 C⋅m.

A general distribution of electric point-charges may be characterized, in addition to its net charge and its dipole moment, by its electric quadrupole moment and, eventually, by higher order moments. The electric quadrupole moment is a symmetric rank-two tensor (i.e. a 3×3 matrix with 9 components, but because of symmetry only 6 of these are independent).

Θ = N i=1 qi ri ri = N i=1 qi
 xixi xiyi xizi yixi yiyi yizi zixi ziyi zizi
(1.10)

As with any multipole moment, if a lower-order moment (monopole or dipole in this case) is non-zero, then the value of the quadrupole moment depends on the choice of the coordinate origin. If all the position vectors, ri in eq. (1.10), are moved to a new origin r0 (i.e. ririr0), eq. (1.10) becomes:

 Θ(r0) = N ∑ i=1 qi (ri − r0) (ri − r0)† = N ∑ i=1 qi ri ri† − p r0† − r0 p† + Q r0 r0† (1.11)

which states that Θ is independent of r0 (i.e. does not depend on the choice of origin) only if p = 0 and Q = 0. For example, the quadrupole moment of an ensemble of four same-strength point-charges, arranged in a square with alternating signs, is coordinate independent.

The SI units for electric quadrupole moment are C⋅m2, however the most commonly used unit is the buckingham (B), which is a CGS units. A buckingham is 1 × 10−26 statC⋅cm2 and is equivalent to 1 D⋅Å. An atomic unit of electric quadrupole moment is e⋅a02 and is about 4.486 551 484(28) × 10−40 C⋅m2.

## 2. - Continuous Charge Distributions

“Often it is convenient to ignore the fact that charges come in packages like electrons and protons, and think of them as being spread out in a continuous smear − or in a 'distribution,' as it is called. This is O.K. so long as we are not interested in what is happening on too small a scale. We describe a charge distribution by the 'charge density,' ρ(r) ≡ ρ(x,y,z). If the amount of charge in a small volume ΔV1 located at the point r1 is Δq1, then ρ is defined by”[L1“Coulomb’s law; superposition”
The Feynman Lectures on Physics, Online Edition,
Volume II, Chapter 4, Section 4.2
]

 Δq1 = ρ(r1) ΔV1 (2.1)

As always, the volume integral of the charge density ρ(r) over a region of space V in ℝ3 is the charge contained in that region.

 q = ∫ V ρ(r) d3r = ∫ V ρ(r) dV = ∭ V ρ(x,y,z) dx dy dz (2.2)

In the Euclidean space, the components of the position vector r are the cartesian coordinates x, y and z, and the volume element dV is given by the product of the differentials of the cartesian coordinates, dV = dx dy dz.

Electric charge is the zero-th rank moment of of the distribution. Higher rank electric moments are obtained by evaluation of volume integrals of the type

 m(n) = ∫ V ρ(r) r(n) d3r (2.3)

The discussion of equation (2.3) is postponed to the next page of the manual.

## 3. - Electrostatics

Electrostatics deals with the circumstance in which nothing depends on the time, i.e. the static case. All charges (and their density) are permanently fixed in space. The basic equation of electrostatics is Gauss' law, which in its differential form

 ∇ ⋅ E = 4 π ρ(r) (3.1)

states that the divergence of the electric field E is proportional to the total electric charge density ρ(r). Eq. (3.1) is written in the natural system of Hartree atomic units. In the SI the right hand side of eq. (3.1) is multiplied by the Coulomb's constant, ke.[2Definition of Hartree atomic units.] From this fundamental equation connecting the charge density with the electric field, the potential Φ of the field can be derived. Introducing the potential Φ in such a way its gradient equals −E,

 E = − ∇ Φ (3.2)

eq. (3.1) can be rewritten in the form of a Poisson equation[9Poisson's equation.]

 ∇2 Φ = − 4 π ρ(r) (3.3)

The nabla symbol ∇ in eqs (3.1-3) indicates a differential vector operator whose components are partial derivative operators. It is most commonly used as a convenient mathematical notation to simplify expressions for the gradient (as in eq. 3.2), divergence (as in eq. 3.1), curl, directional derivative, Laplacian (as in eq. 3.3), Jacobian and tensor derivative.[8The vector differential operator .]

3.1. - The electric potential.  A solution of Poisson's equation[9Solution of Poisson's equation.] is given by eq. (3.4)

 Φ(r) = ∫ ℝ3 ρ(r') / |r − r'| d3r' (3.4)

which defines the electric potential for a continous charge distribution of density ρ(r). In the case of a discrete charge distribution, where the charges are characterized by a strength qi and are located at certain positions ri, the charge density is given by eq. (1.5). Therefore, inserting eq. (1.5) into eq. (3.4) yields

 Φ(r) = N ∑ i=1 qi ∫ ℝ3 δ(r' − ri) / |r − r'| d3r' = N ∑ i=1 qi / |r − ri| = N ∑ i=1 Φi(r) (3.5)

where the second equality holds because of the sifting property of the Dirac delta function.[6The Dirac Delta Function.]

The SI unit of electric potential is the volt (V, in honor of Alessandro Volta), which corresponds to J ⋅ C−1 and in terms of SI base units is kg ⋅ m2 ⋅ s−3 ⋅ A−1. The atomic unit of electric potential (Eh/e) has a value of 27.111 386 02(17) V.

3.2. - The electric field.  The electric field is a vector quantity generated by a distribution of electric charges, as described by Gauss' law, either in its differential form (3.1) or in its integral form.[10Gauss' flux theorem.]  Using the definition of eq. (3.2), an expression of the electric field, E(r), is derived for both a continous charge distribution

 E(r) = ∫ ℝ3 ρ(r') r − r' / |r − r'|3 d3r' (3.6)

and a discrete charge distribution

 E(r) = N ∑ i=1 qi r − ri / |r − ri|3 = N ∑ i=1 Ei(r) (3.7)

The SI units of the electric field are newtons per coulomb (N ⋅ C−1) or, equivalently, volts per metre (V ⋅ m−1), which in terms of SI base units are kg ⋅ m ⋅ s−3 ⋅ A−1. The atomic unit of electric field (Eh/e a0) has a value of 5.142 206 707(32) × 1011 V ⋅ m−1.

3.3. - The electric field gradient.  The electric field gradient is a tensor quantity generated by a distribution of electric charges. Indeed, a given charge distribution generates an electric potential Φ according to equations (3.4) and (3.5). The gradient of this potential is the negative of the electric field generated, as defined by eq. (3.2). The first derivatives of the field, or the second derivatives of the potential, yields the electric field gradient (EFG) or the hessian of the electric potential.

H(r) = − E(r) = ∇ ∇ Φ(r) =
 ∂2 / ∂ x2 ∂2 / ∂ x ∂ y ∂2 / ∂ x ∂ z ∂2 / ∂ y ∂ x ∂2 / ∂ y2 ∂2 / ∂ y ∂ z ∂2 / ∂ z ∂ x ∂2 / ∂ z ∂ y ∂2 / ∂ z2
Φ(r)
(3.8)
 Hαβ(r) = − ∂ Eβ(r) / ∂ rα = ∂2 Φ(r) / ∂ rα ∂ rβ = N ∑ i=1 qi [ 3 (rα − ri,α) (rβ − ri,β) / |r − ri|5 − δαβ / |r − ri|3 ] (3.9)
 Tr H(r) = ∇2 Φ(r) = 0 (3.10)

The SI units of the electric field gradient are V ⋅ m−2, which in terms of SI base units are kg ⋅ s−3 ⋅ A−1. The atomic unit of electric field gradient (Eh/e a02) has a value of 9.717 362 356(60) × 1021 V ⋅ m−2.

3.4. - Coulomb's law.  The force exerted by one point charge on another acts along the line between the charges. It varies inversely as the square of the distance separating the charges and is proportional to the product of the charges. The force is repulsive if the charges have the same sign and attractive if the charges have opposite signs. If q1 is at position r1 and q2 is at r2, the force F12 exerted by q1 on q2 is the vector

 F12 = ke q1 q2 / r123 r12 (3.11)

where r12 = r2r1 is the vector pointing from q1 to q2, and ke is an experimentally determined positive constant called the Coulomb constant.[2ke in SI units = 8.987 551 787 3681 × 109 kg m3 s−2 C−2.]  The magnitude of vector (3.11), i.e. the magnitude of the electric force exerted by a point charge q1 on another point charge q2 a distance r12 away, is given by

 F12 = ke q1 q2 / r122 (3.12)

In the system of Hartree atomic units the Coulomb constant is unity by definition and therefore it will be omitted hereafter. In accord with Newton’s third law,[11Newton’s third law.] the electrostatic force F21 exerted by q2 on q1 is the negative of F12, i.e. F21 = − F12.

It is worth noting that, by combining eqs (3.7) and (3.11), the electrostatic force F12 exerted by q1 on q2 can be written as

 F12 = q2 E1(r2) (3.13)

where E1(r2) is the electric field generated by the point charge q1 at position r2. According to eq. (3.13), the field E1(r2) can be viewed as the force per unit charge on q2 (due to all other charges). In a system of charges, each charge exerts a force given by eq. (3.13) on every other charge. The net force on any charge q is the vector sum of the individual forces exerted on that charge by all the other charges in the system.

 F(r) = q E(r) (3.14)

where E(r), according to eq. (3.7), is the electric field generated by the system of charges at position r. This follows from the principle of superposition of forces, which states that the force acting on a point charge due to a system of point charges is simply the vector addition of the individual forces acting alone on that point charge due to each one of the charges. The resulting force vector is parallel to the electric field vector at that point, with that point charge removed. Therefore, the force F(r) on a small charge q at position r, due to a system of N discrete charges in vacuum is:

 F(r) = q N ∑ 1 qi r − ri / |r − ri|3 = q N ∑ 1 qi Ri / |Ri|3 (3.15)

where qi and ri are the magnitude and position respectively of the i-th charge, and the vector Ri = rri points from charges qi to q.   For a continous charge distribution where ρ(r′) gives the charge per unit volume at position r′, the principle of superposition is also used. In this case an integral over the region containing the charge is equivalent to an infinite summation over infinitesimal elements of space containing a point charge   dq' = ρ(r′) d3r'   (where   d V' = d3r'   is an infinitesimal element of volume):

 F(r) = q ∫ ℝ3 ρ(r') r − r' / |r − r'|3 d3r' (3.16)

There are three conditions to be fulfilled for the validity of Coulomb’s law:[L5Coulomb’s law
]

1. The charges must have a spherically symmetric distribution (e.g. be point charges, or a charged metal sphere).
2. The charges must not overlap (e.g. be distinct point charges).
3. The charges must be stationary with respect to each other.

The SI unit of force is the newton (N), which in terms of SI base units is kg ⋅ m ⋅ s−2. In atomic units the force is expressed in hartrees per Bohr radius (Eh/a0). The atomic unit of force has a value of 8.238 723 36(10) × 10−8 N. The CGS unit of force is the dyne, which in in terms of CGS base units is g ⋅ cm ⋅ s−2, so the SI unit of force, the newton, is equal to 105 dynes.

3.5. - Electrostatic potential energy and electrostatic potential energy density.  Electrostatic potential energy, UE, is the energy possessed by an electrically charged object because of its position relative to other electrically charged objects.

The electrostatic potential energy, UE, of one point charge q at position r in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force (3.14) to bring q from infinity to position r,

 UE = − W∞→r = − q r ∫ ∞ E(r') ⋅ d r' = q r ∫ ∞ ∇ Φ(r') ⋅ d r' = q [Φ(r) − Φ(∞)] = q Φ(r) (3.17)

where E is electrostatic field defined by eq. (3.2) and dr' is the displacement vector in a curve from infinity to the final position r. The third equality holds beacause Coulomb forces are conservative, so that the work they do in moving a particle between two points is independent of the taken path. In the limit as r → ∞, the electrostatic potential Φ(r) falls off like 1/r, and is consequently zero. This explains the last equality in eq. (3.17).

The electrostatic potential energy of a system of point N charges is the work needed to bring the N charges from an infinite separation to their final positions.
 UE =   ∑ all pairs qi qj / rij = 1 / 2 N ∑ i=1 qi Φi (3.18)
where the first summation runs over all of the N! / 2 (N − 2)! pairs of charges and Φi is the potential at the location of the i-th charge due to all of the other charges.[12Tipler, Paul A.; Mosca, Gene
"Physics for Scientists and Engineers", - 5th ed.
W. H. Freeman and Company, San Francisco, 2004; Vol. I, p. 749.
]
For a point charge q1 at point r1, the potential Φ2 at point r2 a distance r12 away is given by Φ2 = q1r12.   To bring a second point charge q2 from rest at infinity to rest at point r2 requires that the work   W2 = q2 Φ2 = q2 q1r12   is done. The potential at point r3, at distance r13 from q1 and r23 from q2, is given by   Φ3 = q1r13 + q2r23.   To bring in an additional point charge q3 from rest at infinity to rest at point r3 requires that the additional work   W3 = q3 Φ3 = q3 q1r13 + q3 q2r23   is done. The total work required to assemble the three charges is the electrostatic potential energy UE of the system of three point charges:   UE = q2 q1r12 + q3 q1r13 + q3 q2r23   This quantity of work is independent of the order in which the charges are brought to their final positions. The last equation can be rewritten as[12Tipler, Paul A.; Mosca, Gene
"Physics for Scientists and Engineers", - 5th ed.
W. H. Freeman and Company, San Francisco, 2004; Vol. I, p. 749.
]
 UE = 1 / 2 (UE + UE) = 1 / 2 ( q2 q1 / r12 + q3 q1 / r13 + q3 q2 / r23 + q2 q1 / r12 + q3 q1 / r13 + q3 q2 / r23 ) = 1 / 2 [ q1 ( q2 / r12 + q3 / r13 ) + q2 ( q3 / r23 + q1 / r12 ) + q3 ( q1 / r13 + q2 / r23 ) ] = 1 / 2 (q1 Φ1 + q2 Φ2 + q3 Φ3)

Equation (3.18) also describes the electrostatic potential energy of a continuous charge distribution. By considering that each volume element dV contains the element of charge dq = ρ dV, eq. (3.18) can be written

 UE = 1 / 2   ∫ all space ρ(r1) ρ(r2) / r12 dV1 dV2 = 1 / 2   ∫ V1 ρ(r1) Φ(r1) dV1 ≡ 1 / 2   ∫ V ρ(r) Φ(r) dV (3.19)

where the factor 12 is introduced because in the double integral over dV1 and dV2 all pairs of charge elements have been counted twice. In addition, the second equality in eq. (3.19) follows from the fact that the integral over dV2 is just the potential at r1 (cf eq. 3.4):

 Φ1 ≡ Φ(r1) =   ∫ V2 ρ(r2) / r12 dV2 (3.20)

The last identity in eq. (3.19) comes from the fact that the point r2 no longer appears in the second equality, so that the suffix 1 is reduntant and can be safely neglected.

Finally, it's worth to note that the electrostatic energy can be described not only in terms of the charges, but also in terms of the electric fields they produce:

 UE =   ∫ V uE dV = 1 / 8 π   ∫ V E2(r) d3r (3.21)

This formula shows that when an electric field E(r) is present, there is located in space an energy UE whose density (energy per unit volume) is proportional to the square of the electric field strength, i.e. uE = E2(r) ⁄ 8π.

Though the quantity UE defined in eq. (3.21) is usually known as the energy stored in the electric field generated by a charge distribution or more shortly as the electrostatic field energy, it can be shown that it is numerically equal to the electrostatic potential energy defined by equations (3.18) and (3.19).[L6“Energy in the electrostatic field”
The Feynman Lectures on Physics, Online Edition,
Volume II, Chapter 8, Section 8.5
]
By introducing into eq. (3.19) the expression (3.3) for the electric charge density ρ(r), the electrostatic potential energy UE can be written in terms of the electric potential Φ and its laplacian ∇2Φ:
 UE = − 1 / 8 π   ∫ V Φ ∇2Φ dV (3.21-1)
“ Writing out the components of the integrand, we see that
 Φ ∇2Φ =  Φ ( ∂2 Φ / ∂ x2 + ∂2 Φ / ∂ y2 + ∂2 Φ / ∂ z2 ) =  ∂ / ∂ x ( Φ ∂ Φ / ∂ x ) − ( ∂ Φ / ∂ x )2 + ∂ / ∂ y ( Φ ∂ Φ / ∂ y ) − ( ∂ Φ / ∂ y )2 + ∂ / ∂ z ( Φ ∂ Φ / ∂ z ) − ( ∂ Φ / ∂ z )2 =  ∇ ⋅ (Φ ∇Φ) − (∇Φ) ⋅ (∇Φ) (3.21-2)
Our energy integral (3.21-1) is then
 UE = 1 / 8 π   ∫ V (∇Φ) ⋅ (∇Φ) dV − 1 / 8 π   ∫ V ∇ ⋅ (Φ ∇Φ) dV (3.21-3)
We can use Gauss’ theorem to change the second integral of eq. (3.21-3) into a surface integral:
 ∫ V ∇ ⋅ (Φ ∇Φ) dV =   ∫ S (Φ ∇Φ) ⋅ n dS (3.21-4)
We evaluate the surface integral in the case that the surface goes to infinity (so the volume integrals become integrals over all space), supposing that all the charges are located within some finite distance. The simple way to proceed is to take a spherical surface of enormous radius R whose center is at the origin of coordinates. We know that when we are very far away from all charges, Φ varies as 1⁄R and Φ as 1⁄R2. (Both will decrease even faster with R if there the net charge in the distribution is zero.) Since the surface area of the large sphere increases as R2, we see that the surface integral falls off as (1⁄R)(1⁄R2)R2 = (1⁄R) as the radius of the sphere increases. So if we include all space in our integration (R → ∞), the surface integral goes to zero and we have that
 UE = 1 / 8 π   ∫ V (∇Φ) ⋅ (∇Φ) dV = 1 / 8 π   ∫ V E ⋅ E dV (3.21-5)
We see that it is possible for us to represent the energy of any charge distribution as being the integral over an energy density located in the field.[L6“Energy in the electrostatic field”
The Feynman Lectures on Physics, Online Edition,
Volume II, Chapter 8, Section 8.5
]

The SI unit of energy is the joule (J), which in terms of SI base units is kg ⋅ m2 ⋅ s−2. In atomic units the energy is expressed in hartree (Eh). The atomic unit of energy has a value of 4.359 744 17(75) × 10−18 J or, in more common non SI units, 27.211 385 eV and 627.509 kcal ⋅ mol−1.

## Unit Converter

Enter the value to be converted in the input box with the appropriate units, then press the "tab" or the "Enter/return" key, or click outside of the input box.

Electric Charge Electric Charge Density Electric Dipole Moment
 au (e) coulomb (C) esu (statC)
 au (e / a03) C m−3 C nm−3 C Å−3 e Å−3 e nm−3
 au (e ⋅ a0) debye (D) C⋅m esu⋅cm
 au (e ⋅ a02) D⋅Å C⋅m2 esu⋅cm2
 au (e ⋅ a03) D⋅Å2 C⋅m3 esu⋅cm3
 au (e ⋅ a04) D⋅Å3 C⋅m4 esu⋅cm4
Electric Potential Electric Field Electric Field Gradient
 au (Eh ⁄ e) V = J⋅C−1
 au (Eh ⁄ e a0) V⋅m−1
 au (Eh ⁄ e a02) V⋅m−2
Force Energy Energy Density
 au (Eh ⁄ a0) N dyne
 au (Eh) J eV kJ/mol kcal/mol cm-1 MHz nm
 au (Eh ⁄ a03) TJ ⁄ m3 MJ⋅mol-1⋅Å-3

## References and Notes

1. A point particle defines a particle which lacks spatial extension: being zero-dimensional, it does not take up space.
2. Hartree atomic units (au or a.u.) form a system of natural units, i.e. physical units of measurement based only on universal physical constants. In this system, the numerical values of the following four fundamental physical constants are all unity by definition:
• Electron mass, me;
• Elementary charge, e;
• Reduced Plank's constant, ħ = h / 2 π ;
• Coulomb's constant or electric force constant or electrostatic constant, ke = 1 / 4 π ε0 ;
The numerical values of these fundamental constants in SI units are[3(a) Mohr, P. J.; Newell, D. B.; Taylor, B. N.
Reviews of Modern Physics 2016, 88, 035009, 73 pages.
(b) Mohr, P. J.; Newell, D. B.; Taylor, B. N.
J. Phys. Chem. Ref. Data 2016, 45, 043102, 74 pages.
, L2CODATA recommended values of the fundamental physical constants: 2014*
at http://physics.nist.gov/constants
]   me = 9.109 383 56(11) × 10−31 kg;  e = 1.602 176 6208(98) × 10−19 C;  ħ = 1.054 571 800(13) × 10−34 J s;  ke = 8.987 551 787 3681 × 109 kg m3 s−2 C−2. The constant ε0 is the permittivity of free space, ε0 = 8.85418782 × 10−12 C2 ⋅ m−2 ⋅ N−1 or m−3 ⋅ kg−1 ⋅ s4 ⋅ A2.
3. (a) "CODATA recommended values of the fundamental physical constants: 2014*"
Mohr, P. J.; Newell, D. B.; Taylor, B. N.   Rev. Mod. Phys. 2016, 88, 035009, 73 pages,
DOI: 10.1103/RevModPhys.88.035009.
(b) "CODATA recommended values of the fundamental physical constants: 2014*"
Mohr, P. J.; Newell, D. B.; Taylor, B. N.   J. Phys. Chem. Ref. Data 2016, 45, 043102, 74 pages,
DOI: 10.1063/1.4954402.
4. Dirac, Paul “The Principles of Quantum Mechanics” (4th ed.), Oxford at the Clarendon Press: 1958, ISBN 978-0-19-852011-5.
5. Messiah, Albert “Quantum Mechanics”, Volume I, North-Holland Publishing Company, Amsterdam: 1970, ISBN 7204-0044-9, pp. 179-184 and 468-471.
6. Definition of the Dirac delta function in one dimension, δ(xx0), can be found in reference [5Messiah, A. “Quantum Mechanics”, Volume I
North-Holland Publishing Company, Amsterdam: 1970, pp. 179-184 and 468-471.
]. Here it is extended to three-dimensional Dirac delta function, δ3(rr0) or simply δ(rr0), because of its relevance to electrostatics. The easiest way to define a three-dimensional delta function is just to take the product of three one-dimensional functions, i.e. in cartesian coordinates:

 δ(r − r0) = δ(x − x0) δ(y − y0) δ(z − z0) (N6.1)

Expressions for cylindrical and spherical coordinates can be be found at link [L3Weisstein, Eric W. “Delta Function.”
From MathWorld − A Wolfram Web Resource.   http://mathworld.wolfram.com/DeltaFunction.html
].   The Dirac delta function δ(rr0) is defined by the property

 ∫   f(r) δ(r − r0) d3r = f(r0) (N6.2)

for any function f(r) that is continous at the point r = r0. The integral (N6.2) is over any volume that contains the point r = r0. At this point δ(rr0) is singular in such a manner that

 ∫   δ(r − r0) d3r = +∞ ∫ −∞ δ(x − x0) dx  +∞ ∫ −∞ δ(y − y0) dy  +∞ ∫ −∞ δ(z − z0) dz = 1 (N6.3)

In addition, δ(rr0) is zero everywhere except at r = r0,   i.e.

 δ(r − r0) = 0         ∀   r ≠ r0 (N6.4)

The discrete analog of the Dirac delta function is the Kronecker delta function, which is usually defined on a discrete domain and takes values 0 and 1.

Eq. (N6.2) is also known as the “sifting” property of the Dirac delta function since that, when the delta function appears in a product within an integrand, it has the property of “sifting” out the value of the integrand at the point where the delta function is not null.

7. When discussing the dipole moment of a non-neutral system, a dependence on the choice of reference point arises. In such cases it is conventional to choose the reference point to be the center of mass of the system, instead of some arbitrary origin. For a charged molecule the center of charge should be the reference point instead of the center of mass. For neutral systems the references point is not important.
8. Eqs. (3.1-3) are expressed in terms of the vector differential operator nabla (from the Hellenistic Greek word ναβλα that indicates a Phoenician harp). It is also known as del and, less frequently, as atled (i.e. delta spelled backwards, the symbol being an inverted delta). Its components are partial derivative operators
= / r =
 ∂ / ∂ x ∂ / ∂ y ∂ / ∂ z
(N8.1)

It is a convenient mathematical notation for different differential operators, like the gradient of a function f,

 ∂ f / ∂ x ∂ f / ∂ y ∂ f / ∂ z
(N8.2)

the divergence of a vector v,

 div v = ∇ ⋅ v = ∇† v = ∂ vx / ∂ x + ∂ vy / ∂ y + ∂ vz / ∂ z (N8.3)

the curl of a vector v,

curl v = × v = det
 i j k ∂ / ∂ x ∂ / ∂ y ∂ / ∂ z vx vy vz
=
 0 − ∂ / ∂ z ∂ / ∂ y ∂ / ∂ z 0 − ∂ / ∂ x − ∂ / ∂ y ∂ / ∂ x 0
v =
 ∂ vz / ∂ y − ∂ vy / ∂ z ∂ vx / ∂ z − ∂ vz / ∂ x ∂ vy / ∂ x − ∂ vx / ∂ y
(N8.4)

and the divergence of the gradient of a function f, i.e. the laplacian of f,

 div grad f = ∇† ∇ f = ∇ ⋅ ∇ f = ∇2 f = ∂2 f / ∂ x2 + ∂2 f / ∂ y2 + ∂2 f / ∂ z2 (N8.5)

It's worth noting that the gradient operator acts on a scalar differentiable function f = f(r), where r ∈ ℝn (with n = 3 in eq. N8.2), and returns a vector. Here, the simple notation is a useful mnemonic device for the product of a vector by a scalar. The divergence operator acts on a vector field v = v(r), where r, v ∈ ℝn (with n = 3 in eq. N8.3), and returns a scalar function. The notation  ⋅ is a useful mnemonic device for the dot product of two vectors. The curl operator acts on a vector field v = v(r), where r, v ∈ ℝ3 (notice that, unlike the gradient and divergence, the curl operator does not generalize in n dimensions), and returns a vector field. The notation  × is a useful mnemonic device for the cross product of two vectors. The Laplacian is a scalar operator that acts on a scalar function and generates a scalar function. The notation ∇2 is a shorthand for the dot product of a vector operator with itself, which gives the squared norm of the vector operator.

9. The Poisson's equation is a partial differential equation (PDE) of the form

 ∇2 u(r) = f(r) (N9.1)

which describes the response u(r) of a physical system to an arbitrary source f(r). The general solution of eq. (N9.1) is given by the sum of any harmonic function [which satisfies the homogeneous Laplace equation, 2 u(r) = 0] and a particular solution of the non-homogeneous Poisson equation (N9.1). Let's first consider the non-homogeneous problem associated with the presence of a point source located at r', mathematically represented by the Dirac delta function δ(rr'):

 ∇2 G(r, r') = δ(r − r') (N9.2)

where the Green's function G(r, r') describes the response of the system at the point r to a point source located at r'.

If G is the fundamental solution of eq. (N9.2), then the convolution

 G * f = ∫ ℝ3 G(r, r') f(r') d3r' (N9.3)

will be a solution of eq. (N9.1). Indeed, because the Laplacian ∇2 is a linear differential operator, it follows that

 ∇2(G * f )  =  (∇2 G) * f  =  δ * f  =  ∫ ℝ3 δ(r − r') f(r') d3r'  =  f(r) (N9.4)

where the first equality comes from linearity of the Laplacian, the second equality comes from eq. (N9.2), the third equality follows from the definition of convolution, and the last equality is a consequence of the fact that the Dirac delta function is an identity element for convolution, as stated by eq. (N6.2). Therefore, if G is the fundamental solution, the convolution u = G * f is one solution of 2 u(r) = f(r). This does not mean that it is the only solution. Several solutions for different initial conditions can be found.

The definition (N9.2) of the fundamental solution implies that, if the Laplacian of G is integrated over any volume that encloses the source point, then, according to eq. (N6.3),

 ∭ V ∇2 G dV = 1 (N9.5)

The Laplace equation is unchanged under a rotation of coordinates, and hence we can expect that a fundamental solution may be obtained among solutions that only depend upon the distance r from the source point. If we choose the volume to be a ball of radius a around the source point, then Gauss' divergence theorem[10The divergence theorem.] implies that

 ∭ V ∇2 G dV = ∬ S d G / d r d S = 4 π a2 d G / d r |r=a = 1 (N9.6)

Rearranging the last equality of eq. (N9.6) yields

 d G / d r |r=a = 1 / 4 π a2 (N9.7)

and hence

 G(r, r') = − 1 / 4 π a = − 1 / 4 π ⋅ 1 / |r − r'| (N9.8)

as   a2 = (xx')2 + (yy')2 + (zz')2 = |rr'|2   is the equation of a sphere of radius a centered at the source point r'. Finally, inserting eq. (N9.8) into eq. (N9.3) gives the solution of Poisson's equation (N9.1):

 u(r) = − 1 / 4 π ∫ ℝ3 f(r') / |r − r'| d3r' (N9.9)

Eq. (N9.9) is a solution of Poisson's equation of electrostatics (3.3), provided that u(r) = Φ(r) and f(r) = − 4 π ρ(r).

10. The divergence theorem states that the surface integral of the normal component of an arbitrary vector F over a closed surface S is equal to the volume integral of the divergence of the vector over the volume V interior to the surface S:

 ∬ S (F ⋅ n) dS = ∭ V (∇ ⋅ F) dV (N10.1)

where dS may be used as a shorthand for n dS. The left-hand side of the equation represents the total flow across the closed boundary S, and the right-hand side represents the total of the sources in the volume V. In other words, the divergence theorem is a mathematical statement of the physical fact that, in the absence of the creation or destruction of matter, the density within a region of space can change only by having it flow into or away from the region through its boundary. In the special case where the vector F is defined as the gradient of a scalar G, F = G, and the volume V is that is that of a ball of radius a, the divergence theorem yields eq. (N9.5).

11. Newton's third law: forces always occur in equal and opposite pairs. If object A exerts a force FA,B on object B, an equal but opposite force FB,A is exerted by object B on object A. Thus,

 FB,A = − FA,B (N11.1)

This is also known as the Action - Reaction Principle: "For every action there is always an opposite and equal reaction".

12. Tipler, Paul A.; Mosca, Gene   "Physics for Scientists and Engineers", - 5th ed., W. H. Freeman and Company, San Francisco, 2004, ISBN: 0-7167-4389-2; Vol. I, p. 749.