Superposition principles

If we had more than one charge and each charge were at a different lo-cation in a vacuum, the total electric field in the space external to the lolo-cation of these charges would be the vector summation of the electric field originating from each individual charge. The vacuum is a linear media. In fact, a vacuum has the greatest number of linear properties that can be found in any media. The introduction of an additional material into the vacuum may cause the region to become nonlinear. The principles of superposition apply in a vacuum. The only caveat that we will encounter will be that we must be careful to apply vector superposition principles and just not scalar superposition principles. Both the magnitude and the direction of the individual electric fields from each charge must be included in the addition. For N separate charges in the region of interest, this vector summation can be written as

^{  ˜˜˜}

¦

^N vectors and this is illustrated in Figure 2-3 for the particular case of two charges Q1 and Q2. Remember that we have to include the correct sign of the charge in the vector addition operation. For the case depicted in Figure 2-3, the total

u indicates the unit vector associated with each individual charge Qj to the point where the electric field is to be computed. The distance between the

charge and this point is given by Rj. This vector addition is best illustrated with two examples.

Figure 2-3. The total electric field is the vector sum of individual components.

Example 2-4. Two charges Q1 = +4C and Q2 = -2C are located at the points indicated on the graph. The units of the graph are in meters. Find the electric field at the origin (0, 0) of the coordinate system.

Answer: The electric field E is computed from (2.11)

R2

2 2 o

2 2 R

1 o

1

1 4 R

Q R

4

Q u u

E E

ET 2 1  SH

 SH

The vector direction of the electric field is directed from the charge #1 to the charge #2 at the origin. Using the numerical values specified for the charges and the distances as determined from the graph, we write

y

x

Q

1

Q

2

E

1

E

2

E

T

Q

1

Q

2

R

1

R

2

36

Note that the electric fields from the two charges add up at the origin.

Example 2-5. Three charges Q1 = + 1 C, Q2 = + 2 C, and Q3 = 3 C are placed at the indicated points on the graph. Find the electric field at the point P.

Answer: The electric field at the point P is computed from a linear superposition on the individual electric field components due to the individual charges. We

The further evaluation of this electric field is straightforward.

x y

Q

1

Q

2

Q

3

P

Up to this point, our discussion of electrostatic fields has assumed that it was possible to calculate the electric field by merely summing the vector contributions from each individual charge. In theory, this is the correct procedure that should always be followed. However, in many cases that one encounters in practice, we would quickly run out of steam in following such a procedure when describing realistic situations where the number of charged particles in a confined volume may be of the order of a power of ten, say fifteen or twenty.

Numerical tools would soon be required to perform this summation.

If we can make certain assumptions concerning the distribution of the charges in a region and realize that an integration of the distributed charges over the region follows directly from a summation if we let a certain parameter become extremely small, then it is possible to obtain analytical solutions for a particular problem. We will include some of these solutions in the following discussion.

The assumption that we will employ is that if a total charge ¨Q is dis-tributed within a volume ¨v and we take the limit as this volume ¨v ^o 0, then we can define a volume charge density U

v as

¸¹

¨ ·

©

§ '

U_v ' ₃

m C v

Q (2.12)

This charge density may be inhomogeneous such that it depends on the local position r, and we write this charge density as Uv^{ U}v(r). There may also be a uniform volume charge distribution.

If the charge is distributed on a surface whose area is ¨s and it is independent of the distance normal to the surface, then we can define this as a surface charge density Us as

¸¹

¨ ·

©

§ '

U_s ' ₂

m C s

Q (2.13)

We take the limit of ¨s ^o 0. This surface charge density could depend on its location on the surface r and we write Us = ^Us(r). It could also be uniformly

distributed on the surface and the charge distribution would be a constant in this case.

The charge could also be distributed along a line whose length is ¨L. The charge would have a uniform distribution in the two transverse coordinates of the line. This would yield a linear charge density UL where

¸¹

¨ ·

©

§ ' U '

m C L Q

L (2.14)

We again take the limit of ¨L ^o 0. Once again, the charge could be distributed nonuniformly or uniformly along the line.

We will find it advantageous to use all three definitions in later derivations.

We will later encounter "infinite sheets" or "infinite lines" that have charge densities given by (2.13) and (2.14). This merely implies that an infinite amount of charge is distributed over these infinite surfaces or lines but the ratios given in these two equations (2.13) and (2.14) are finite.

Figure 2-4. Distributed charge densities: a) The charge is distributed in a volume 'v creating a volume charge density U_v, b) The charge is distributed on a surface 's creating a surface charge density U_s, c) The charge is distributed along a line 'L creating a linear charge density U_L.

If we want to calculate the electric field that is created by either of the dis-tributed charge density distributions, we will make use of the principle of superposition that was stated in equation (2.10) and shown in Figure 2-5. A quick

(a) (b) (c)

glimpse at Figure 2-5 should convince us that numerical techniques may have to be employed for most charge distributions in order to calculate the electric field.

Fortunately for us, there are a few examples that can be treated analytically and some of them will appear in this text.

Figure 2-5. In order to calculate the electric field at the point P, the differential electric fields ¨Ej caused by the charges in the differential volumes ¨vj are added together vectorially.

If we let the differential volumes ¨vj become very small and the number of the small volumes to become very large, then the summation of the distinct electric fields caused by the discrete charges within these volumes will eventually cause the summation to become an integral that must be performed over the entire volume v where the distributed charge density is located. This integration is written as

³

v R v o

R u dv

E ₂

4

1 U

SH ^(2.15)

Equation (2.15) implies that there exists a differential electric field that is directed radially from each of the differential charges that is enclosed within each of the differential volumes. The total electric field that will emanate from the

'E

₁

'E

3

Q

₁

Q

₂

Q

₃

R

₁

R

3

R

2

P

'E

₂

entire volume v is calculated by integrating the charge density over the entire volume. Each of the incremental electric fields will have their individual unit vectors and the integration must incorporate this fact.

If we are given a particular charged object and wish to analytically calculate the electric field caused by it, the first thing that we must do is to select the proper coordinate system in which the integration must be performed. This choice is usually predicated on any possible symmetry that can be found in the problem. For example, if the charged body were a sphere that was centered on the origin of a coordinate system, we should attempt the solution in spherical coordinates. If the charged body were a long cylindrical rod that was centered at the origin, we should use cylindrical coordinates.

The variables that appear in this integral are defined as follows. The vari-able R is the distance between the point of observation and the location of a particular charge element U_vr')dv that is within the volume of integration v. In Cartesian coordinates, we write



x x'



² (y y')² (z z')²

R      (2.16)

where x', y', and z' specify the location of the differential charge element and x, y, and z specify the location where the electric field is to be determined at the point P. The unit vector u_R is directed from this charge element to the point P. The unit vector u_R will change as the integration is performed. This will be noted when ac-tually performing the integration. In this general equation (2.15), we have to be careful since vectors are present and we would have to perform the integration separately over the three components in the differential volume dv.

In order to illustrate the procedure involved in setting up the integral and identifying each term in the integral, we calculate the electric field from a finite amount of charge that is uniformly distributed on a finite line. The linear charge density on this line will be UL. This linear charge density is depicted in Figure 2-6.

From this figure, we find that the unit vector from a differential charge that is localized on a section of the line whose length is dz to the point of observation is given by

2 2

z

r z

r z





 _r

R

u

u u (2.17)

The variation of the unit vector upon the variable z alluded to earlier is clearly displayed in (2.17). In the calculation, we will assume that there is symmetry in that the point of observation is taken to be at the midpoint of the line. Therefore, for every charge segment at a distance +z, there will be an equivalent charge element located at -z. This is an example of symmetry and it is shown in Figure 2-6. Because of this symmetry, the components of the electric fields polarized in the ±z directions will cancel ('E_z+ = 'E_z-). If the line of charge were infinite in length, the center of symmetry could be placed anywhere along the line.

Figure 2-6. Calculating the electric field from a uniformly distributed finite line of charge. The radial axis is at the center of the charged line. Because of this symmetry, the tangential components 'Ez of the electric field cancel.

2a

z' dz

z r

'E

_z

'E 'E

_r

R

T

The term polarization means that the field is directed in that particular direction.

Therefore, the radial component of the electric field is given in terms of the dif-ferential electric field dE by

2

where the magnitude of differential electric field dE is calculated from the charge that is contained in the length dz. This charge is equal to U_L dz. Therefore,

The total radial electric field is given by the summation of all of the infinitesimal components dEr since this is a linear media and superposition applies. This summation becomes an integration of the linear charge density over the length of the line and it can be performed analytically.

 

^{Lo 2 2}

This integral can be performed with the substitution of z = r tan T or by using an integral table. As the length of the line is made extremely long (2aĺ^f), the electric field decreases as this distance increases.

r

Example 2-6. Calculate the electric field from an infinite charged plane. Assume that the plane consists of an infinite number of parallel charged lines as shown in the figure.

Answer: It is possible to consider the infinite plane as a parallel array of juxtaposed infinite charged lines. Hence, we can use (2.21) as our point of embarkation, where the distance R x²y² . The linear charge density U_L of a particular line whose width is dy is just equal to U_L Us dx. Due to symmetry, the components of the electric field that are tangent to the plane will cancel.

Therefore, we need only find the component of the electric field that is normal to the plane assuming that the differential surface areas are concentric circular washers.

Due to the symmetry found in these two examples, we have been able to obtain analytical solutions for the electric field from two different charge configu-rations using the integral given in (2.15). We already know the electric field due

D

x

z

to a point charge in (2.8). The field varies respectively in distance from the charge region as R^-2, R^-1 and being independent of R for the electric field from a point charge, an infinite line charge and an infinite surface charge. We would expect a difference since the infinite line charge and the infinite surface charge each contain an increasing order of infinity more charge than the point charge.

The assumption of symmetry has made these two examples problems that can easily be solved. There are, however, many more examples in which one cannot invoke these arguments of symmetry. The resulting integration may have to be performed numerically and we’ll discuss this topic with reference to MATLAB later after we encounter the subject of the electric potential. This will permit us to neglect any vector notation and this will simplify our discussion of that topic. In the material that immediately follows, we’ll continue to make the symmetry assumption.

In document Parte 1 Fundamentos Electromagneticos Con Matlab_Lonngren & Savov (Page 110-121)