Maxwell’s Electromagnetic Field Equation No. 1
By George J. Spix
1.0 Statement of Equation
The following Electrostatic Field equations will be developed in this section:
Maxwell’s first equation is based on Gauss’ law of electrostatics published in 1832, wherein Gauss established the relationship between static electric charges and their accompanying static fields.
The above integral equation states that the electric flux through a closed surface area is equal to the total charge enclosed.
The differential form of the equation states that the divergence or outward flow of electric flux from a point is equal to the volume charge density at that point.
1.1. Maxwell’s Equation No.1; Area Integral
We will derive the integral equation by considering the summation of electric flux density on a surface area, and then as a summation of volume containing electric charge. The two integrals are shown to be equal when they are based on the same charge. Two examples using the equations are shown.
1.1.1 Gauss’ Law
Gauss’ electrostatics law states that lines of electric flux,fE, emanate from a positive charge, q, and terminate, if they terminate, on a negative charge. The space within which the charges exert their influence is called the electrostatic field.
The sketch in Figure 1.1 represents the charges and the three dimensional field. The field is visualized as being made up of lines of flux. For an isolated charge, the lines of flux do not terminate and are considered to extend to infinity.
To obtain the equation relating an electric charge q, and its fluxf E, assume that the charge is centered in a sphere of radius r meters. The electric flux density, D, is then equal to the electric flux emanating from the charge, q, divided by the area of the sphere.
coulombs per square meter; where the area is perpendicular to the lines of flux. (One coulomb is equal to the magnitude of charge of 6.25X 1018 electrons.) The charge enclosed in the sphere is then equal to the electric flux density on its surface times the area enclosing the charge.
q (coulombs enclosed) =D x 4 r2.
The lines of flux contributing to the flux density are those that leave the sphere perpendicular to the surface of the sphere. This leads to the integral statement of this portion of Gauss’ law;
The integral sign indicates the summation of infinitesimal areas, da, in order to obtain the entire surface area.
The circle on the integral sign indicates that the integral or summation of area is taken of a closed continuous surface.
Bold face letters indicate that the letter represents a vector, i.e., this quantity has magnitude and direction. Distance, velocity, acceleration and force are common examples of vectors.
Dis the electric flux density vector in coulombs per square meter.
The (said dot) following D shows that the vector dot product must be used when multiplying the two vectors, D and da.
The dot product (discussed below) indicates that the magnitudes of the two vectors are multiplied together and then that product is multiplied by the cosine of the angle between the two vectors. The dot product here enables us to determine the effective lines of flux flowing through the surface.
1.1.2 Vector Dot Product
A vector dot product application can be illustrated by calculating work in the following physics problem.
Recall that work = force times distance. Work is equal to the product of the force, that is in the direction of force movement, times the distance the force moves.
In the following example, assume a person is pushing a mop across a floor with the mop handle at an angle of 60 degrees to the floor as in Figure 1.2. Arrows are used in these diagrams to represent vectors.
A force of 20 pounds is applied through the mop handle. As shown in the diagram, only that component of the force in the direction parallel to the floor is used in the calculation of work. We see that the force parallel to the floor is;
20 pounds x cosine 600 = 10 pounds.
What is the work done when pushing the mop 8 feet across the floor?
10 lbs x 8 ft = 80 ft. lbs.
Using the dot product, the equation for work is:
Work = Force distance.
The dot product indicates that work equals the magnitude of the force times the magnitude of the distance moved, times the cosine of the angle between the two vectors.
Or, Work = Force on the mop handle, times the distance the force moves, times the cosine of the angle between the force and the floor.
Work = 20 pounds times 8 feet times 1/2 = 80 foot pounds.
Only that component of the total force in the direction parallel to the floor, as obtained through the dot product, is used in the calculation of work.
This shows that the dot product is defined as the method of vector multiplication in which the vector magnitudes are multiplied together and then that product is multiplied by the cosine of the vector’s included angle. There must always be a vector on each side of the dot in the dot product.
Therefore, when the dot product is used in Gauss’ Law, only that component of flux parallel to the vector representing area will contribute to the total enclosed charge.
1.2 Gauss’ Law; Area Integral Examples
The method of determining charge by using the dot product of is similar to finding Work as the dot product between applied Force and Distance. Through this method, only those components of the vector lines of flux in the same direction as the vector representing area will be summed in the calculation of charge. Or, stated in another way: Only those flux lines perpendicular to the surface are incorporated in the result of the dot product to obtain the charge enclosed.
A scalar value is always the resultant of a dot product. In this case, the result is a number of coulombs. Examples of other scalar quantities are temperature, mass and power. A scalar quantity, in contrast to a vector, does not have direction.
The differential element of area isda. A vector representing an area is pointed normal, i.e. perpendicular, to that area. Using the dot product between the vector representing area da and the flux density D, results in obtaining the effective flux through the area. The summation of the entire area is in square meters.
The preliminary equation (Gauss’ Law) in our procedure to obtain Maxwell’s first equation is now;
This integral equation states that the amount of electric flux density normal to a surface is caused by a specific amount of charge, q, enclosed by the surface.
Consider the following examples of finding the electric flux density on a spherical surface and on a cylindrical surface.
1.2.1 Determine Flux Density on a Sphere
Assume a charge of one coulomb is centered in a sphere of radius r meters as in Figure 1.3. Calculate the electric flux densityD on the surface on the sphere. The integral or summation of area of the sphere is 4 p r2 square meters.
A vector representing an area is directed normal to that area. The vector representing the small area,da, is then directly in line with a line of electric flux leaving the sphere. D represents the density of those lines of electric flux leaving the sphere. The angle between the displacement density, D, and the arrow representing the infinitesimal area is zero degrees. The cosine of zero degrees is one.
Restating the area integral equation of Gauss’ Law:
q (coulombs enclosed) =
one coulomb = D x 4 pr2.
coulomb per square meter on the surface of the sphere.
1.2.2 Determine Flux Density on a Cylinder
Assume a long line of stationary charges of q coulombs per meter as shown in Figure 1.4. There is a cylinder of length "L" and radius "r" centered on the charges. What is the electric flux density on the surface of the cylinder?
Gauss’ equation is:
QT, the total charge enclosed, = q coulombs per meter x L meters.
= QT = D x 2pr x L .
Din coulombs per square meter =
1.3 Maxwell’s Equation No. 1; Volume Integral
Gauss’ electrostatics law is also written as a volume integral:
This equation states that the charge enclosed in a volume is equal to the volume charge density,r, (rho) summed for the entire volume.
q is the charge enclosed in the volume.
ris the volume charge density in coulombs per cubic meter.
is an infinitesimal element of volume.
The entire volume is in cubic meters.
The total charge enclosed in the volume is the volume in cubic meters times the charge density in coulombs per cubic meter. The average volume charge density summed for the entire volume is the charge enclosed. A discussion of r is found in Section 1.5.3.
1.4 Maxwell’s Equation No. 1; Integral Form Completed
To obtain the integral form of Maxwell’s Equation No.1, assume that an experiment is set up so that the same charge of q coulombs is contained in each of Gauss’ law equations. Then the integrals due to the same charge must be equal.
Thus we have obtained the integral form of Maxwell’s Equation No.1. This equation states that the effective electric field through a surface enclosing a volume is equal to the total charge within the volume. The equation shows that the area enclosed by the left hand integral must enclose the volume of the right integral. This is similar to stating that the surface area of a ball or box encloses the volume of the ball or box. The area and volume indicated by the equations need not be observable physical surfaces, often they will be mathematical limits.
To remember the integral form of Maxwell’s Equation No. 1, consider that a charge q, enclosed in a volume, must be equal to the volume charge density,r, times the volume. Also, the same charge q, will cause a certain area flux density, D, times a certain area. The area must enclose the volume. The integrals (summations) must be equal since the same number of coulombs must be obtained on each side of the equal sign.
1.5 Maxwell’s Equation No.1; Differential form
The differential form of Maxwell’s Equation No.1 is:
is a differential operator read "del" (discussed below).
is read "divergence".
Dis the electric flux density in coulombs per square meter.
r(rho) is the volume charge density in coulombs per cubic meter.
1.5.1 Discussion of (del)
is the mathematical extension of the ordinary single dimension calculus derivative into three dimensions.
We will begin the discussion of by reviewing ordinary derivatives.
As an example, a derivative is used as the notation for velocity. Velocity (v) is the increase in distance, s, for an increase in time, t.
Velocity (v) =.
As the change in time is made very small the differential calculus symbol is used for velocity.
Now consider the ordinary single dimension derivative for acceleration.
Recall the equation for obtaining the velocity of an object when it is dropped from a height. The velocity that the object attains is found by;
Velocity, v, = acceleration due to gravity, g, times the time during which the object is falling.
Velocity = v = g t. Or
In the integral form of Gauss’ Law we summed the infinitesimal values of area and volume, da and dv. Here we are using the differential, ds, dv and dt to find instantaneous rates of change of distance and velocity with respect to time.
There is a rule of differential calculus that we will point out here as we will use the rule below.
Notice that velocity is equal to acceleration (g) times time(t). So we can take the derivative of velocity in this manner:
The calculus rule is that the derivative of a variable times a constant is the constant.
We will now extend that concept of ordinary derivatives to partial derivatives .This will allow us to obtain the rate of change of a volume in three dimensions, which in turn leads to the definition ofÑ< FONT FACE="Arial" SIZE=4> (del). To illustrate the rate of change in three dimensions, assume a box is placed at the origin of a rectangular coordinate system, as shown in Figure 1.5.
Box volume, V, = length x width x height; = L x W x H.
What is the rate of change of volume when only the length increases by a small amount but the width and height remain constant?
This is where the symbol for the partial derivative is used. The symbol for partial derivatives is slightly different than the symbol for ordinary (one dimension) derivatives. The symbol indicates that only one of the independent variables is changing at the moment under consideration. The dependent variable, volume,v changes as determined by changes of the independent variables; L,W and H. When only the length changes;
The partial derivative symbol, shows that the change in volume is due to a change in length only. The width and height are being held constant. We also see that the derivative of a variable times a constant is the constant, W times H. It follows, by symmetry, that the rate of change of volume as a function of either width or height is expressed as a partial derivative.
When the change in volume is due to a simultaneous change in length, width and height, the changes will occur in the x, y and z directions and the partial derivatives are added to find the resultant rate of change of volume. This is accomplished in vector form by multiplying each partial derivative by unit vectors pointing in the x, y and z directions. Unit vectors are indicated here and discussed further in Section 1.5.2. Using unit vectors and simultaneous changes in three dimensions, the total change in volume is designated by:
For the more general case of a volume,V changing in the x, y and z directions;
This discussion of ordinary and partial derivatives was aimed at obtaining the group of three partial derivative terms in the above parenthesis.
means the gradient of . We will not use the gradient in this paper.
We will have need for Ñ · (del dot, discussed below) and Ñ 5 (del cross, discussed in Section 3).
1.5.2 Discussion of Vector Components in Relation to
Before continuing the part of the differential form of Maxwell’s equation No.1, we must consider the x, y and z components of a vector in rectangular coordinates.
In section 1.1.2 the dot product of two vectors, force and distance, was used to calculate work. We will here calculate the same work using the vector components of force and distance, and employ the dot product.
Notice that the vector components of the force in the rectangular coordinates are placed at either at zero or 90 degrees to the distance movement. The cosine of zero degrees is one and the cosine of 90 degrees is zero.
From Figure 1.3;
The vector components of force = 17.3 lbs. y + 10 lbs. x.
The vector components of distance = 0 ft. y + 8 ft. x.
The vector · (dot) multiplication procedure now is to multiply the vector component magnitudes and the cosine of the angle between them, term by term.
The zero y distance times the two components of the force is zero.
The 8 ft x distance times the 17.8 lb force in the y direction times the cosine of 90 degrees is zero. The 8 ft x directed distance times the 10 lb force in the x direction times the cosine of zero degrees is the same 80 ft lbs that we found in the previous example.
This procedure of multiplying vector x, y and z components is followed in performing theproduct below. The multiplication of vector components, that are always at zero or 90 degrees apart, greatly simplifies vector mathematics.
We determined the components of in Section 1.5.1. Above we discussed dot product multiplication of vectors using their components. These two concepts are now used to calculate
The components of vector D are its projections on the x, y and z axis. The vector directions of D components are designated by the unit vectors x, y and z. In Figure 1.6, the vector D starts at the origin, points up and to the right and is indicated as coming out of the paper. The magnitudes of projections of D along the axes are Dx, Dy, and Dz. Figure 1.7 shows the unit vectors in the x, y and z directions that give the components of D their vector relationship. The same unit vectors are designated in
The equation for vectorD as projected on the three coordinate axes is, D = Dx x + Dyy + Dzz. We will now do the indicated dot product of .
The dot product indicates that we must multiply the parentheses, term by term, times the cosine of the included angle between each pair of terms. This series of multiplications could result in nine terms, but notice that a unit vector dotted into the same unit vector:
The other six combinations of unit vector dot product multiplications contain the cosine of 90 degrees and are therefore, zero.
The final result of the operation is a scalar of only three terms:
This equation indicates the sum of a change in electric flux density,D, in each of the three orthogonal directions. The change is due to a small, (approaching zero), distance change in the same orthogonal directions.
The change of distance in the three orthogonal directions is a volume change, as shown in Section 1.5.1. Therefore, the electric density (D) change in the three directions, that we obtained by using the dot product with the unit vectors in del, is actually a per unit volume change. Since charge is measured in coulombs, the sum of charge is in coulombs. The result of adding the three electric density changes is coulombs per cubic meter. This definesr , the volume charge density, as noted in Section 1.3.
1.6 Equation No.1, Differential Form Completed
By doing the indicated operation we obtainedr, the volume charge density. This is the differential statement of Maxwell’s equation No.1.
The equation states that the divergence of the electric flux density at a point is equal to the charge per unit volume at that point. The dot product, as always, produces a scalar result. In this case, the result isr , the number of coulombs of charge per cubic meter.
1.7 Divergence Theorem
It is instructive at this point to continue using the integral and differential equations just developed for Maxwell’s Equation No.1 in order to illustrate a vector identity called, "Gauss’ Divergence Theorem". This identity equates a vector surface integral to a vector volume integral, and will be required later in Section 2.5 .
From Section 1.4,
From Section 1.5;
By substituting forr in the integral equation we obtain;
This is a typical illustration of Gauss’ divergence theorem, using vectorD as the example. The point here is that any time we have a vector surface integral of this type we can substitute the volume integral. If we have a vector volume integral of the above type, we can substitute the surface integral. The integral of the divergence of a vector summed throughout the volume is equal to the integral of the product of the vector times its effective area summed over the area. This is analogous to stating that the volume of a ball is contained within its surface area. The circle on the integral sign indicates that the integral is taken over a continuous area.
If we had simply used Gauss’ Divergence Theorem from a textbook list of vector identities, we could have immediately written down the differential form of Maxwell’s equation No.1 from the integral form. This more detailed way of obtaining the identity will be helpful in later derivations.
1.8 Relation ofD, E and e
The space in which electric charges exert their influence is called the field of the electric charge. Surrounding an electric charge q, there is an electric field of field strengthE. It is the electric field strength, E, that will cause an amount of flux density, D, depending on the permittivity, e of the surrounding medium.
Dis in coulombs per square meter.
Eis in newtons per coulomb or volts per meter.
eis in coulomb2 per newton meter2.
Eis the stress in space that causes D to be manifested. Due to this equation and for reasons discussed in Section 3.7, D is often designated as electric flux displacement density in addition to electric flux density. Also, it will be shown in Section 4, that a magnetically induced electric field is also designated E, with a dimension of volts per meter. That induced electric field is the same field as the static field strength discussed here but it is generated by a changing magnetic field.
The permittivitye, is the degree to which the surrounding medium will permit the electric flux density, D, to occur due to a given electric field strength, E. In the medium of air or free space,
e= 8.85x10-12 coulomb2 per newton meter 2.
These concepts and definitions will be used in Sections 6 and 7.
1.9 Coulomb’s Law
In Section 1.2.1 we found that the electric flux density, D, due to a charge q, located within the sphere is:
Then, using E and e as defined in Section 1.8;
When another charge q2, is placed r meters from q1, a force is experienced by q2. The force isE times q2 Newton’s. Or,
where q1 and q2 designate individual charges. This equation is Coulomb’s law. Recall that a force of one newton will accelerate a mass of one kilogram at one meter per second2.
Coulomb’s law states that the force between two charges is proportional to the product of the two charges over the distance between the two charges squared. The equation is the basis for experimentally determining the force between two charges and the permittivity of different mediums. This important electrical law is not included in Maxwell’s list as it is considered derivable from Gauss’ law, and is not used in these field equations.
This completes the discussion of Maxwell’s Equation No.1.