Solution Wentworth

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Wentworth solutions chapter 2.pdf
2- !1
Solutions for Chapter 2 Problems 
1. Vectors in the Cartesian Coordinate System 
P2.1: Given P(4,2,1) and APQ=2ax +4ay +6az, find the point Q. 
APQ = 2 ax + 4 ay + 6 az = (Qx-Px)ax + (Qy-Py)ay+(Qz-Pz)az 
Qx-Px=Qx-4=2; Qx=6 
Qy-Py=Qy-2=4; Qy=6 
Qz-Pz=Qz-1=6; Qz=7 
Ans: Q(6,6,7) 
P2.2: Given the points P(4,1,0)m and Q(1,3,0)m, fill in the table and make a sketch of 
the vectors found in (a) through (f). 
a. AOP = (4-0)ax + (1-0)ay + (0-0)az = 4 ax + 1 ay. 
! 
! 
(see Figure P2.2ab) 
b. BOQ =(1-0)ax + (3-0)ay + (0-0)az = 1 ax + 3 ay. 
! 
! 
Vector Mag Unit Vector
a. Find the vector A 
from the origin to P
AOP = 4 ax + 1 ay 4.12 AOP = 0.97 ax + 0.24 ay
b. Find the vector B 
from the origin to Q
BOQ = 1 ax + 3 ay 3.16 aOQ = 0.32 ax + 0.95 ay
c. Find the vector C 
from P to Q
CPQ = -3 ax + 2 ay 3.61 aPQ = -0.83 ax + 0.55 ay
d. Find A + B A + B = 5 ax + 4 ay 6.4 a = 0.78 ax + 0.62 ay
e. Find C – A C - A = -7 ax + 1 ay 7.07 a = -0.99 ax + 0.14 ay
f. Find B - A B - A = -3 ax + 2 ay 3.6 a = -0.83 ax + 0.55 ay
Fig. P2.2ab
2- !2
(see Figure P2.2ab) 
c. CPQ = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay. 
! 
! 
(see Figure P2.2cd) 
d. A + B = (4+1)ax + (1+3)ay + (0-0)az = 5 ax + 4 ay. 
! 
! 
(see Figure P2.2cd) 
e. C - A = (-3-4)ax + (2-1)ay + (0-0)az = -7 ax + 1 ay. 
! 
! 
(see Figure P2.2ef) 
f. B - A = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay. 
! 
! 
(see Figure P2.2ef) 
P2.3: MATLAB: Write a program that will find the vector between a pair of arbitrary 
points in the Cartesian Coordinate System. 
A program or function for this task is really overkill, as it is so easy to perform the task. 
Enter points P and Q (for example, P=[1 2 3]; Q=[6 5 4]). Then, the vector from P toQ is 
simply given by Q-P. 
As a function we could have: 
function PQ=vector(P,Q) 
% Given a pair of Cartesian points 
FigP2.2ef
Fig. P2.2cd
2- !3
% P and Q, the program determines the 
% vector from P to Q. 
PQ=Q-P; 
Running this function we have: 
>> P=[1 2 3]; 
>> Q=[6 5 4]; 
>> PQ=vector(P,Q) 
PQ = 
 5 3 1 
Alternatively, we could simply perform the math in the command line window: 
>> PQ=Q-P 
PQ = 
 5 3 1 
>> 
2. Coulomb’s Law, Electric Field Intensity, and Field Lines 
P2.4: Suppose Q1(0.0, -3.0m, 0.0) = 4.0nC, Q2(0.0, 3.0m, 0.0) = 4.0nC, and Q3(4.0m, 
0.0, 0.0) = 1.0nC. (a) Find the total force acting on the charge Q3. (b) Repeat the problem 
after changing the charge of Q2 to –4.0nC. (c) Find the electric field intensity for parts (a) 
and (b). 
(a) where R13 = 4 ax + 3 ay =, R13 = 5m, a13 = 0.8 ax + 0.6 ay. 
so 
! 
Similarly, ! so ! 
(b) with Q2 = -4 nC, F13 is unchanged but so 
! 
(c) ! 
Fig. P2.4
2- !4
Likewise, ! 
P2.5: Find the force exerted by Q1(3.0m, 3.0m, 3.0m) = 1.0 µC on Q2(6.0m, 9.0m, 3.0m) 
= 10. nC. 
! 
R12 = (6-3)ax + (9-3)ay + (3-3)az = 3 ax + 6 ay m 
! 
! , so ! 
 
P2.6: Suppose 10.0 nC point charges are located on the corners of a square of side 10.0 
cm. Locating the square in the x-y plane (at z = 0.00) with one corner at the origin and 
one corner at P(10.0, 10.0, 0.00) cm, find the total force acting at point P. 
We arbitrarily label the charges as shown in Figure P2.6. Then 
ROP = 0.1 ax + 0.1 ay 
ROP = 0.141 m 
Fig. P2.5
2- !5
aOP = 0.707 ax + 0.707 ay. 
! 
! 
! 
and then the total (adjusting to 2 significant digits) is: 
! 
P2.7: 1.00 nC point charges are located at (0.00, -2.00, 0.00)m, (0.00, 2.00, 0.00)m, (0.00, 
0.00, -2.00)m and (0.00, 0.00, +2.00)m. Find the total force acting on a 1.00 nC charge 
located at (2.00, 0.00, 0.00)m. 
Figure P2.7a shows the situation, but we need only find the x-directed force from one of 
the charges on Qt (Figure P2.7b) and multiply this result by 4. Because of the problem’s 
symmetry, the rest of the components cancel. 
Fig. P2.6
Fig. P2.7b
Fig. P2.7a
2- !6
! 
so! 
The force from all charges is then 
! 
 
 
P2.8: A 20.0 nC point charge exists at P(0.00,0.00,-3.00m). Where must a 10.0 nC 
charge be located such that the total field is zero at the origin? 
For zero field at the origin, we must cancel the +az directed field from QP by placing Q at 
the point Q(0,0,z) (see Figure P2.8). Then we have Etot = EP + EQ = 0. 
So, ! 
and 
! 
So then 
! 
Thus, Q(0,0,2.12m). 
3. The Spherical Coordinate System 
P2.9: Convert the following points from Cartesian to Spherical coordinates: 
a. P(6.0, 2.0, 6.0) 
b. P(0.0, -4.0, 3.0) 
c. P(-5.0,-1.0, -4.0) 
Fig. P2.8
2- !7
(a) ! 
(b) ! ! 
(c) ! 
P2.10: Convert the following points from Spherical to Cartesian coordinates: 
a. P(3.0, 30.°, 45.°) 
b. P(5.0, π/4, 3π/2) 
c. P(10., 135°, 180°) 
(a) 
! 
(b) 
! 
(c) 
! 
P2.11: Given a volume defined by 1.0m ≤ r ≤ 3.0m, 0 ≤ θ ≤ 0°, 90° ≤ φ ≤ 90°, (a) 
sketch the volume, (b) perform the integration to find the volume, and (c) perform the 
necessary integrations to find the total surface area. 
(a) 
(b) 
! 
2- !8
So volume V = 14 m3. 
(c) There are 5 surfaces: an inner, an outer, and 3 identical sides. 
! 
So Stotal = 35 m2. 
4. Line Charges and the Cylindrical Coordinate System 
P2.12: Convert the following points from Cartesian to cylindrical coordinates: 
a. P(0.0, 4.0, 3.0) 
b. P(-2.0, 3.0, 2.0) 
c. P(4.0, -3.0, -4.0) 
(a) ! 
(b) ! 
(c) ! 
P2.13: Convert the following points from cylindrical to Cartesian coordinates: 
a. P(2.83, 45.0°, 2.00) 
Fig. P2.11
2- !9
b. P(6.00, 120.°, -3.00) 
c. P(10.0, -90.0°, 6.00) 
(a) 
! 
(b) 
! 
(c) 
! 
P2.14: A 20.0 cm long section of copper pipe has a 1.00 cm thick wall and outer diame-
ter of 6.00 cm. 
a. Sketch the pipe conveniently overlaying the cylindrical coordinate system, lining 
up the length direction with the z-axis 
b. Determine the total surface area (this could actually be useful if, say, you needed 
to do an electroplating step on this piece of pipe) 
c. Determine the weight of the pipe given the density of copper is 8.96 g/cm3 
(a) See Figure P2.14 
(b) The top area, Stop, is equal to the bottom area. We must also find the inner area, Sinner, 
and the outer area, Souter. 
! 
! 
2- !10
The total area, then, is 210π cm2, or Stot = 660 cm2. 
(c) Determining the weight of the pipe requires the volume: 
! 
So Mpipe = 2820g. 
P2.15: A line charge with charge density 2.00 nC/m exists at y = -2.00 m, x = 0.00. (a) A 
charge Q = 8.00 nC exists somewhere along the y-axis. Where must you locate Q so that 
the total electric field is zero at the origin? (b) Suppose instead of the 8.00 nC charge of 
part (a) that you locate a charge Q at (0.00, 6.00m, 0.00). What value of Q will result in a 
total electric field intensity of zero at the origin? 
(a) The contributions to E from the line and point charge must cancel, or ! 
For the line: ! 
and for the point charge, where the point is located a distance y along the y-axis, we have: 
! 
Therefore: 
 ! 
(b) 
Fig. P2.14
2- !11
! 
P2.16: You are given two z-directed line charges of charge density +1 nC/m at x = 0, y = 
-1.0 m, and charge density –1.0 nC/m at x = 0, y = 1.0 m. Find E at P(1.0m,0,0). 
The situation is represented by Figure P2.16a. A better 2-dimensional view in Figure 
P2.16b
is useful for solving the problem. 
! 
! , and ! 
So ETOT = 18 ay V/m. 
P2.17: MATLAB: Suppose you have a segment of line charge of length 2L centered on 
the z-axis and having a charge distribution ρL. Compare the electric field intensity at a 
point on the y-axis a distance d from the origin with the electric field at that point assum-
ing the line charge is of infinite length. The ratio of E for the segment to E for the infinite 
line is to be plotted versus the ratio L/d using MATLAB. 
This is similar to MATLAB 2.3. We have for the ideal case 
! 
Fig. P2.16a
Fig. P2.16b
Fig. P2.15
2- !12
For the actual 2L case, we have an integration to perform (Equation (2.35) with different 
limits): 
! 
Now we manipulate these expressions to get the following ratio: 
! 
In the program, the actual to ideal field ratio is termed “Eratio” and the charged line half-
length L ratioed to the distance d is termed “Lod”. 
% M-File: MLP0217 
% 
% This program is similar to ML0203. 
% It compares the E-field from a finite length 
% segment of charge (from -L to +L on the z-axis) 
% to the E-field from an infinite length line 
% of charge. The ratio (E from segment to E from 
% infinite length line) is plotted versus the ratio 
% Lod=L/d, where d is the distance along the y axis. 
% 
% Wentworth, 12/19/02 
% 
% Variables: 
% Lod the ratio L/d 
% Eratio ratio of E from segment to E from line 
clc %clears the command window 
clear %clears variables 
% Initialize Lod array and calculate Eratio 
Lod=0.1:0.01:100; 
Eratio=Lod./(sqrt(1+Lod.^2)); 
% Plot Eratio versus Lod 
semilogx(Lod,Eratio) 
grid on 
xlabel('Lod=L/d') 
ylabel('E ratio: segment to line') 
2- !13
Executing the program gives Figure P2.17. 
So we see that the field from a line segment of charge appears equivalent to the field from 
an infinite length line if the test point is close to the line. 
P2.18: A segment of line charge ρL =10 nC/m exists on the y-axis from the origin to y = 
+3.0 m. Determine E at the point (3.0, 0, 0)m. 
It is clear from a sketch of the problem in Figure P2.18a that the resultant field will be 
directed in the x-y plane. The situation is redrawn in a temporary coordinate system in 
Figure P2.18b. 
We have from Eqn (2.34) ! 
For Eρ we have: 
! 
With ρ = 3, we then have Eρ = 21.2 V/m. 
For Ez: 
! 
Fig. P2.17
2- !14
Thus we have ETOT = 21 aρ – 8.8 az V/m. 
Converting back to the original coordinates, we have ETOT = 21 ax – 8.8 ay V/m. 
5. Surface and Volume Charge 
P2.19: In free space, there is a point charge Q = 8.0 nC at (-2.0,0,0)m, a line charge ρL = 
10 nC/m at y = -9.0m, x = 0m, and a sheet charge ρs = 12. nC/m2 at z = -2.0m. Deter-
mine E at the origin. 
The situation is represented by Figure P2.19, and the total field is ETOT = EQ + EL + ES. 
! 
! 
! 
So: Etot = 18 ax + 20 ay + 680 az V/m. 
P2.20: An infinitely long line charge (ρL = 21π nC/m) lies along the z-axis. An infinite 
area sheet charge (ρs = 3 nC/m2) lies in the x-z plane at y = 10 m. Find a point on the y-
axis where the electric field intensity is zero. 
We have ETOT = EL + ES. 
! 
Fig. P2.19
Fig. P2.20
2- !15
! 
so 
! 
Therefore, P(0, 7m, 0). 
P2.21: Sketch the following surfaces and find the total charge on each surface given a 
surface charge density of ρs = 1nC/m2. Units (other than degrees) are meters. 
(a) –3 ≤ x ≤ 3, 0 ≤ y ≤ 4, z = 0 
(b) 1 ≤ r ≤ 4, 180° ≤ φ ≤ 360°, θ = π/2 
(c) 1 ≤ ρ ≤ 4, 180° ≤ φ ≤ 360°, z = 0 
 
(a) ! 
(b) ! 
(c) ! 
P2.22: Consider a circular disk in the x-y plane of radius 5.0 cm. Suppose the charge 
density is a function of radius such that ρs = 12ρ nC/cm2 (when ρ is in cm). Find the 
electric field intensity a point 20.0 cm above the origin on the z-axis. 
Fig. P2.21b&cFig. P2.21a
2- !16
From section 4 for a ring of charge of radius a,! Now we have 
ρL=ρsdρ and where ρs = Aρ nC/cm2. Now the total field is given 
by the integral: 
! 
This can be solved using integration by parts, where u = ρ, du = dρ, 
! This leads to 
! 
Plugging in the appropriate values we arrive at E = 6.7 kV/cm az. 
P2.23: Suppose a ribbon of charge with density ρs exists in the y-z plane of infinite 
length in the z direction and extending from –a to +a in the y direction. Find a general 
expression for the electric field intensity at a point d along the x-axis. 
The problem is represented by Figure P2.23a. A better representation for solving the 
problem is shown in Figure P2.23b. 
We have where ρL = ρsdy. Then, since 
FigP2.23a FigP2.23b
2- !17
! 
the integral becomes 
! 
It may be noted that the ay component will cancel by symmetry. The ax integral is found 
from the appendix and we have 
! 
 
P2.24: Sketch the following volumes and find the total charge for each given a volume 
charge density of ρv = 1nC/m3. Units (other than degrees) are meters. 
(a) 0 ≤ x ≤ 4, 0 ≤ y ≤ 5, 0 ≤ z ≤ 6 
(b) 1 ≤ r ≤ 5, 0 ≤ θ ≤ 60° 
(c) 1 ≤ ρ ≤ 5, 0° ≤ φ ≤ 90°, 0 ≤ z ≤ 5 
(a) ! 
(b) 
! 
(c) 
 ! 
P2.25: You have a cylinder of 4.00 inch diam-
eter and 5.00 inch length (imagine a can of 
tomatoes) that has a charge distribution that 
varies with radius as ρv = (6 ρ) nC/in3 where ρ is in inches. (It may help you with the 
Fig. P2.24b
Fig. P2.24a
2- !18
units to think of this as ρv (nC/in3)= 6 (nC/in4) ρ(in)). Find the total charge contained in 
this cylinder. 
! 
P2.26: MATLAB: Consider a rectangular volume with 0.00 ≤ x ≤ 4.00 m, 0.00 ≤ y ≤ 
5.00 m and –6.00 m ≤ z ≤ 0.00 with charge density ρv = 40.0 nC/m3. Find the electric 
field intensity at the point P(0.00,0.00,20.0m). 
% MLP0226 
% calculate E from a rectangular volume of charge 
% variables 
% xstart,xstop limits on x for vol charge (m) 
% ystart,ystop 
% zstart,zstop 
% xt,yt,zt test point (m) 
% rhov vol charge density, nC/m^3 
% Nx,Ny,Nz discretization points 
% dx,dy,dz differential lengths 
% dQ differential charge, nC 
% eo free space permittivity (F/m) 
% dEi differential field vector 
% dEix,dEiy,dEiz x,y and z components of dEi 
% dEjx,dEjy,dEjz of dEj 
% dEkx,dEky,dEkz of dEk 
% Etot total field vector, V/m 
clc 
Fig. P2.24c
2- !19
clear 
% initialize variables 
xstart=0;xstop=4; 
ystart=0;ystop=5; 
zstart=-6;zstop=0; 
xt=0;yt=0;zt=20; 
rhov=40e-9; 
Nx=10;Ny=10;Nz=10; 
eo=8.854e-12; 
dx=(xstop-xstart)/Nx; 
dy=(ystop-ystart)/Ny; 
dz=(zstop-zstart)/Nz; 
dQ=rhov*dx*dy*dz; 
for k=1:Nz 
 for j=1:Ny 
 for i=1:Nx 
 xv=xstart+(i-0.5)*dx; 
 yv=ystart+(j-0.5)*dy; 
 zv=zstart+(k-0.5)*dz; 
 R=[xt-xv yt-yv zt-zv]; 
 magR=magvector(R); 
 uvR=unitvector(R); 
 dEi=(dQ/(4*pi*eo*magR^2))*uvR; 
 dEix(i)=dEi(1); 
 dEiy(i)=dEi(2); 
 dEiz(i)=dEi(3); 
 end 
 dEjx(j)=sum(dEix); 
 dEjy(j)=sum(dEiy); 
 dEjz(j)=sum(dEiz); 
 end 
 dEkx(k)=sum(dEjx); 
 dEky(k)=sum(dEjy); 
 dEkz(k)=sum(dEjz); 
end 
Etotx=sum(dEkx); 
Etoty=sum(dEky); 
Etotz=sum(dEkz); 
Etot=[Etotx Etoty Etotz] 
 
Now to run the program: 
Etot = 
2- !20
 -6.9983 -8.7104 79.7668 
>> 
So E = -7.0 ax -8.7 ay + 80. az V/m 
P2.27: MATLAB: Consider a sphere with charge
density ρv = 120 nC/m3 centered at the 
origin with a radius of 2.00 m. Now, remove the top half of the sphere, leaving a hemi-
sphere below the x-y plane. Find the electric field intensity at the point P(8.00m,
0.00,0.00). (Hint: see MATLAB 2.4, and consider that your answer will now have two 
field components.) 
% M-File: MLP0227 
% 
% This program modifies ML0204 to find the field 
% at point P(8m,0,0) from a hemispherical 
% distribution of charge given by 
% rhov=120 nC/m^3 from 0 < r < 2m and 
% pi/2 < theta < pi. 
% 
% Wentworth, 12/23/02 
% 
% Variables: 
% d y axis distance to test point (m) 
% a sphere radius (m) 
% dV differential charge volume where 
% dV=delta_r*delta_theta*delta_phi 
% eo free space permittivity (F/m) 
% r,theta,phi spherical coordinate location of 
% center of a differential charge element 
% x,y,z cartesian coord location of charge % 
 element 
% R vector from charge element to P 
% Rmag magnitude of R 
% aR unit vector of R 
% dr,dtheta,dphi differential spherical elements 
% dEi,dEj,dEk partial field values 
% Etot total field at P resulting from charge 
clc %clears the command window 
clear %clears variables 
% Initialize variables 
2- !21
eo=8.854e-12; 
d=8;a=2; 
delta_r=40;delta_theta=72;delta_phi=144; 
% Perform calculation 
for k=(1:delta_phi) 
 for j=(1:delta_theta) 
 for i=(1:delta_r) 
 r=i*a/delta_r; 
 theta=(pi/2)+j*pi/(2*delta_theta); 
 phi=k*2*pi/delta_phi; 
 x=r*sin(theta)*cos(phi); 
 y=r*sin(theta)*sin(phi); 
 z=r*cos(theta); 
 R=[d-x,-y,-z]; 
 Rmag=magvector(R); 
 aR=R/Rmag; 
 dr=a/delta_r; 
 dtheta=pi/delta_theta; 
 dphi=2*pi/delta_phi; 
 dV=r^2*sin(theta)*dr*dtheta*dphi; 
 dQ=120e-9*dV; 
 dEi=dQ*aR/(4*pi*eo*Rmag^2); 
 dEix(i)=dEi(1); 
 dEiy(i)=dEi(2); 
 dEiz(i)=dEi(3); 
 end 
 dEjx(j)=sum(dEix); 
 dEjy(j)=sum(dEiy); 
 dEjz(j)=sum(dEiz); 
 end 
 dEkx(k)=sum(dEjx); 
 dEky(k)=sum(dEjy); 
 dEkz(k)=sum(dEjz); 
end 
Etotx=sum(dEkx); 
Etoty=sum(dEky); 
Etotz=sum(dEkz); 
Etot=[Etotx Etoty Etotz] 
 
Now to run the program: 
Etot = 
 579.4623 0.0000 56.5317 
So E = 580 ax + 57 az V/m. 
2- !22
6. Electric Flux Density 
P2.28: Use the definition of dot product to find the three interior angles for the triangle 
bounded by the points P(-3.00, -4.00, 5.00), Q(2.00, 0.00, -4.00), and R(5.00, -1.00, 
0.00). 
Here we use ! 
! 
! 
! 
! 
P2.29: Given D = 2ρ aρ + sin φ az C/m2, find the electric flux passing through the surface 
defined by 2.0 ≤ ρ ≤ 4.0 m, 90.° ≤ φ ≤ 180°, and z = 4.0 m. 
! 
! 
P2.30: Suppose the electric flux density is given by D = 3r ar –cos φ aθ + sin2θ aφ C/m2. 
Find the electric flux through both surfaces of a hemisphere of radius 2.00 m and 0.00° ≤ 
θ ≤ 90.0˚. 
! 
Fig. P2.28
2- !23
! 
! 
! 
! 
7. Gauss’s Law and Applications 
P2.31: Given a 3.00 mm radius solid wire centered on the z-axis with an evenly dis-
tributed 2.00 coulombs of charge per meter length of wire, plot the electric flux density 
Dρ versus radial distance from the z-axis over the range 0 ≤ ρ ≤ 9 mm. 
For a 1 m length, ! 
! , where L is the length of the Gaussian sur-
face. Note that this expression for Qenc is valid for both Gaussian surfaces. 
GS1 (ρ < a): 
! 
so ! 
GS2 (ρ > a): 
! 
This is plotted with the following Matlab routine: 
% M-File: MLP0231 
% 
Fig. P2.30
2- !24
% Gauss's Law Problem 
% solid cylinder with even charge 
% 
% Variables 
% rhov charge density (C/m^3) 
% a radius of cylinder (m) 
% rho radial distance from z-axis 
% rhomm rho in mm 
% D electric flux density (C/m^3) 
% N number of data points 
% maxrad max radius for plot (m) 
clc;clear; 
% initialize variables 
rhov=70.7e3; 
a=0.003; 
maxrad=.009; 
N=100; 
bndy=round(N*a/maxrad); 
for i=1:bndy 
 rho(i)=i*maxrad/N; 
 rhomm(i)=rho(i)*1000; 
 D(i)=rhov*rho(i)/2; 
end 
for i=bndy+1:N 
 rho(i)=i*maxrad/N; 
Fig. P2.31
2- !25
 rhomm(i)=rho(i)*1000; 
 D(i)=(rhov*a^2)/(2*rho(i)); 
end 
plot(rhomm,D) 
xlabel('radial distance (mm)') 
ylabel('elect. flux density (C/m^2)') 
grid on 
P2.32: Given a 2.00 cm radius solid wire centered on the z-axis with a charge density ρv 
= 6ρ C/cm3 (when ρ is in cm), plot the electric flux density Dρ versus radial distance 
from the z-axis over the range 0 ≤ ρ ≤ 8 cm. 
Choose Gaussian surface length L, and as usual we have 
! valid for both Gaussian surfaces. 
In GS1 (ρ < a): ! 
so ! 
For GS2 (ρ > a): ! 
This is plotted for the problem values in the following Matlab routine. 
% M-File: MLP0232 
% 
% Gauss's Law Problem 
% solid cylinder with radially-dependent charge 
% 
% Variables 
% a radius of cylinder (cm) 
% rho radial distance from z-axis 
% D electric flux density (C/cm^3) 
% N number of data points 
% maxrad max radius for plot (cm) 
clc;clear; 
% initialize variables 
a=2; 
maxrad=8; 
N=100; 
bndy=round(N*a/maxrad); 
for i=1:bndy 
 rho(i)=i*maxrad/N; 
2- !26
 D(i)=2*rho(i)^2; 
end 
for i=bndy+1:N 
 rho(i)=i*maxrad/N; 
 D(i)=(2*a^3)/rho(i); 
end 
plot(rho,D) 
xlabel('radial distance (cm)') 
ylabel('elect. flux density (C/cm^2)') 
grid on 
 
 
P2.33: A cylindrical pipe with a 1.00 cm wall thickness and an inner radius of 4.00 cm is 
centered on the z-axis and has an evenly distributed 3.00 C of charge per meter length of 
pipe. Plot Dρ as a function of radial distance from the z-axis over the range 0 ≤ ρ ≤ 10 
cm. 
! this is true for all the Gaussian surfaces. 
GS1 (ρ < a): since Qenc = 0, Dρ = 0. 
GS2(a < ρ < b): 
! 
Fig. P2.32
2- !27
! 
So, 
! 
GS3(ρ > b): 
Qenc = 3h, 
A plot with the appropriate values is generated by the following Matlab routine: 
% M-File: MLP0233 
% Gauss's Law Problem 
% cylindrical pipe with even charge distribution 
% 
% Variables 
% a inner radius of pipe (m) 
% b outer radius of pipe (m) 
% rho radial distance from z-axis (m) 
% rhocm radial distance in cm 
% D electric flux density (C/cm^3) 
% N number of data points 
% maxrad max radius for plot (m) 
clc;clear; 
% initialize variables 
a=.04;b=.05;maxrad=0.10;N=100; 
bndya=round(N*a/maxrad); 
bndyb=round(N*b/maxrad); 
for i=1:bndya 
 rho(i)=i*maxrad/N; 
 rhocm(i)=rho(i)*100; 
 D(i)=0; 
end 
for i=bndya+1:bndyb 
Fig. P2.33a
2- !28
 rho(i)=i*maxrad/N; 
 rhocm(i)=rho(i)*100; 
 D(i)=(3/(2*pi*rho(i)))*((rho(i)^2-a^2)/(b^2-a^2)); 
end 
for i=bndyb+1:N 
 rho(i)=i*maxrad/N; 
 rhocm(i)=rho(i)*100; 
 D(i)=3/(2*pi*rho(i)); 
end 
plot(rhocm,D) 
xlabel('radial distance (cm)') 
ylabel('elect. flux density (C/m^2)') 
grid on 
 
P2.34: An infinitesimally thin metallic cylindrical shell of radius 4.00 cm is centered on 
the z-axis and has an evenly distributed charge of 100. nC per meter length of shell. (a) 
Determine the value of the surface charge density on the conductive shell and (b) plot Dρ 
as a function of radial distance from the z-axis over the range
0 ≤ ρ ≤ 12 cm. 
! 
For all Gaussian surfaces, 
of height h and radius ρ, we have: 
Fig. P2.33b
2- !29
! 
GS1 (ρ < a): Qenc = 0 so Dρ = 0 
GS2 (ρ > a): 
! 
% M-File: MLP0234 
% 
% Gauss's Law Problem 
% cylindrical shell of charge 
% 
% Variables 
% a radius of cylinder (m) 
% Qs surface charge density (nC/m^2) 
% rho radial distance from z-axis (m) 
% rhocm radial distance in cm 
% D electric flux density (nC/cm^3) 
% N number of data points 
% maxrad max radius for plot (cm) 
clc;clear; 
% initialize variables 
a=.04;Qs=398;maxrad=0.12;N=100; 
bndy=round(N*a/maxrad); 
for i=1:bndy 
 rho(i)=i*maxrad/N; 
 rhocm(i)=rho(i)*100; 
 D(i)=0; 
end 
for i=bndy+1:N 
 rho(i)=i*maxrad/N; 
 rhocm(i)=rho(i)*100; 
 D(i)=Qs*a/rho(i); 
end 
plot(rhocm,D) 
xlabel('radial distance (cm)') 
ylabel('elect. flux density (nC/m^2)') 
Fig. P2.34a
2- !30
grid on 
P2.35: A spherical charge density is given by ρv = ρo r/a for 0 ≤ r ≤ a, and ρv = 0 for r > 
a. Derive equations for the electric flux density for all r. 
! This is valid for each Gaussian sur-
face. 
GS1 (r < a): ! 
So ! 
GS2 (r > a): ! 
P2.36: A thick-walled spherical shell, with inner radius 2.00 cm and outer radius 4.00 
cm, has an evenly distributed 12.0 nC charge. Plot Dr as a function of radial distance 
from the origin over the range 0 ≤ r ≤ 10 cm. 
Here we’ll let a = inner radius and b = outer radius. Then 
! This is true for each Gaussian surface. 
The volume containing charge is 
Fig. P2.34b
2- !31
! 
So ! 
Now we can evaluate Qenc for each Gaussian surface. 
GS1 (r < a): Qenc = 0 so Dr = 0. 
GS2 (a < r < b): ! 
Inserting our value for ρv, we find 
! 
GS3 (r >b): Qenc = Q, 
This is plotted for appropriate values using the following Matlab routine: 
% M-File: MLP0236 
% Gauss's Law Problem 
% thick spherical shell with even charge 
% 
% Variables 
% a inner radius of sphere (m) 
% b outer radius of sphere (m) 
% r radial distance from origin (m) 
% rcm radial distance in cm 
% D electric flux density (nC/cm^3) 
% N number of data points 
% maxr max radius for plot (m) 
% Q charge (nC) 
clc;clear; 
% initialize variables 
a=.02;b=.04; 
Q=12; 
maxrad=0.10; 
N=100; 
bndya=round(N*a/maxrad); 
bndyb=round(N*b/maxrad); 
for i=1:bndya 
2- !32
 r(i)=i*maxrad/N; 
 rcm(i)=r(i)*100; 
 D(i)=0; 
end 
for i=bndya+1:bndyb 
 r(i)=i*maxrad/N; 
 rcm(i)=r(i)*100; 
 D(i)=(Q/
(4*pi*r(i)^2))*(r(i)^3-a^3)/
(b^3-a^3); 
end 
for i=bndyb+1:N 
 r(i)=i*maxrad/N; 
 rcm(i)=r(i)*100; 
 D(i)=Q/(4*pi*r(i)^2); 
end 
plot(rcm,D) 
xlabel('radial distance (cm)') 
ylabel('elect. flux density (nC/m^2)') 
grid on 
P2.37: Given a coaxial cable with solid inner conductor of radius a, an outer conductor 
that goes from radius b to c, (so c > b > a), a charge +Q that is evenly distributed 
throughout a meter length of the inner conductor and a charge –Q that is evenly dis-
tributed throughout a meter length of the outer conductor, derive equations for the electric 
flux density for all ρ. You may orient the cable in any way you wish. 
We conveniently center the cable on the z-axis. Then, for a Gaussian surface of length L, 
! valid for all Gaussian surfaces. 
GS1: (ρ < a): ! 
! 
Fig. P2.36
2- !33
GS2 (a < ρ < b): ! 
GS3 (b < ρ < c): ! 
! 
so ! 
GS4 (ρ > c): Qenc = 0, Dρ = 0. 
8. Divergence and the Point Form of Gauss’s Law 
P2.38: Determine the charge density at the point P(3.0m,4.0m,0.0) if the electric flux 
density is given as D = xyz az C/m2. 
! 
ρv(3,4,0)=(3)(4)=12 C/m3. 
P2.39: Given D = 3ax +2xyay +8x2y3az C/m2, (a) determine the charge density at the 
point P(1,1,1). Find the total flux through the surface of a cube with 0.0 ≤ x ≤ 2.0m, 0.0 
≤ y ≤ 2.0m and 0.0 ≤ z ≤ 2.0m by evaluating (b) the left side of the divergence theorem 
and (c) the right side of the divergence theorem. 
(a) ! 
(b) ! 
! 
! 
2- !34
! 
! 
! 
! 
! 
(c) ! 
P2.40: Suppose D = 6ρcosφ aφ C/m2. (a) Determine the charge density at the point (3m, 
90°, -2m). Find the total flux through the surface of a quartered-cylinder defined by 0 ≤ 
ρ ≤ 4m, 0 ≤ φ ≤ 90°, and -4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence 
theorem and (c) the right side of the divergence theorem. 
(a) ! 
! 
(b) ! 
note that the top, bottom and outside integrals yield zero since there is no component of 
D in the these dS directions. 
! 
! 
So, ! 
2- !35
(c) 
! 
P2.41: Suppose D = r2sinθ ar + sinθcosφ aφ C/m2. (a) Determine the charge density at 
the point (1.0m, 45°, 90°). Find the total flux through the surface of a volume defined by 
0.0 ≤ r ≤ 2.0 m, 0.0° ≤ θ ≤ 90.°, and 0.0 ≤ φ ≤ 180° by evaluating (b) the left side of 
the divergence theorem and (c) the right side of the divergence theorem. 
The volume is that of a quartered-sphere, as indicated in Figure P2.41. 
(a) 
! 
(b) ! 
! 
! 
! 
Summing these terms we have Q = 4(π2 – 1)C = 35.5C. 
(c) 
! 
 
2- !36
9. Electric Potential 
P2.42: A sheet of charge density ρs = 100 nC/m2 occupies the x-z plane at y = 0. (a) Find 
the work required to move a 2.0 nC charge from P(-5.0m, 10.m, 2.0m) to M(2.0m, 3.0m, 
0.0). (b)Find VMP. 
(a) ! 
! 
Notice that we are only concerned with movement in the y-direction. We then have: 
! 
(b) ! 
P2.43: A surface is defined by the function 2x + 4y2 –ln z = 12. Use the gradient equa-
tion to find a unit vector normal to the plane at the point (3.00m,2.00m,1.00m). 
Let ! then 
! 
Fig. P2.41
2- !37
At (3,2,1), 
 ! 
P2.44: For the following potential distributions, use the gradient equation to find E. 
(a) V = x+y2z (V) 
(b) V = ρ2sinφ(V) 
(c) V = r sinθ cosφ (V). 
(a) ! 
(b) ! 
(c) 
! 
P2.45: A 100 nC point charge is located at the origin. (a) Determine the potential differ-
ence VBA between the point A(0.0,0.0,-6.0)m and point B(0.0,2.0,0.0)m. (b) How much 
work would be done to move a 1.0 nC charge from point A to point B against the electric 
field generated by the 100 nC point charge? 
(a) ! 
The potential difference is only a function of radial distance from the origin. Letting ra = 
6m and rb = 2m, we then have 
! 
(b) ! 
P2.46: MATLAB: Suppose you have a pair of charges Q1(0.0, -5.0m, 0.0) = 1.0 nC and 
Q2(0.0, 5.0m, 0.0) = 2.0 nC. Write a MATLAB routine to calculate the potential VRO 
moving from the origin to the point R(5.0m, 0.0, 0.0). Your numerical integration will 
involve choosing a step size ΔL and finding the field at the center of the step. You should 
try several different step sizes to see how much this affects the solution. 
2- !38
% M-File: MLP0246 
% 
% Modify ML0207 to calculate the potential 
% difference going from the origin (O) to the point 
% R(5,0,0) given a pair of point charges 
% Q1(0,-5,0)=1nC and Q2(0,5,0)=2nC. 
% 
% The approach will be to break up the distance 
% from O to R into k sections. The total field E will 
% be found at the center of each section (located 
% at point P) and then dot(Ep,dLv) will give the 
% potential drop across the kth section. Total 
% potential is found by summing the potential drops. 
% 
% Wentworth, 1/7/03 
% 
% Variables: 
% Q1,Q2
the point charges, in nC 
% k number of numerical integration steps 
% dL magnitude of one step 
% dLv vector for a step 
% x(n) x location at center of section at P 
% R1,R2 vector from Q1,Q2 to P 
% E1,E2 electric fields from Q1 & Q2 at P 
% Etot total electric field at P 
% V(n) portion of dot(Etot,dL) at P 
clc %clears the command window 
clear %clears variables 
% Initialize variables 
k=64; 
Q1=1; 
Q2=2; 
dL=5/k; 
dLv=dL*[1 0 0]; 
% Perform calculation 
for n=1:k 
 x(n)=(n-1)*dL+dL/2; 
 R1=[x(n) 5 0]; 
 R2=[x(n) -5 0]; 
 Rmag1=magvector(R1); 
 Rmag2=magvector(R2); 
2- !39
 E1=9*Q1*R1/Rmag1^3; 
 E2=9*Q2*R2/Rmag2^3; 
 Etot=E1+E2; 
 V(n)=dot(Etot,dLv); 
end 
Vtot=sum(-V) 
 
Now running the program: 
Vtot = 
 -1.5817 
So VRO = -1.6 V. 
P2.47: For an infinite length line of charge density ρL = 20 nC/m on the z-axis, find the 
potential difference VBA between point B(0, 2m, 0) and point A(0, 1m, 0). 
! 
P2.48: Find the electric field at point P(0.0,0.0,8.0m) resulting from a surface charge 
density ρs = 5.0 nC/m2 existing on the z = 0 plane from ρ = 2.0 m to ρ = 6.0 m. Assume 
V = 0 at a point an infinite distance from the origin. 
(Method 1) 
For a ring of charge it was previously found that 
! 
We can then break up our disk into differential rings (see Figure P2.48), each contributing 
dE as: 
! 
So we then have 
! 
2- !40
This is easy to integrate if we let u = ρ2 + h2, then du = 2 ρ dρ, and we have 
! 
Solving, we arrive at 
! 
Upon inserting the appropriate values we find E = 48 V/m az. 
(method 2) 
Find an expression for potential and then evaluate the gradient at the point. 
! 
! 
Now we let h = z and ! 
! 
Plugging in the values we find E = 48 V/m az. 
Fig. P2.48
2- !41
P2.49: Suppose a 6.0 m diameter ring with charge density 5.0 nC/m lies in the x-y plane 
with the origin at its center. Determine the potential difference Vho between the point 
h(0.0,0.0,4.0)m and the origin. (Hint: first find an expression for E on the z-axis as a 
general function of z.) 
For the ring of charge, replacing h with z, we have 
! 
! 
Letting u = a2 + z2, du = 2z dz, we have 
! 
Replacing u and evaluating from 0 to h, 
! 
10. Conductivity and Current 
P2.50: A columnular beam of electrons from 0 ≤ ρ ≤ 1 mm has a charge density ρv =-0.1 
cos(πρ/2) nC/mm3 (where ρ is in mm) and a velocity of 6 x 106 m/sec in the +az direc-
tion. Find the current. 
Let’s let where ρo = -0.1 nC/mm3. Then we’ll let u = uoaz, where uo = 
6x109 mm/s. Notice we convert the units to mm. Now, 
! 
and with dS = ρ dρ dφ az we then have 
! 
This becomes 
 where A = 2πρouo. 
Fig. P2.49
2- !42
Now we can integrate by parts, or ! where u = Aρ, du = Adρ, 
! and ! 
We then have 
! 
To evaluate, we first find A = 2π(-0.1x10-9)(6x109)=3.77, and then 
I = 2.40-1.53=0.87A. 
I = 0.87A. 
P2.51: Two spherical conductive shells of radius a and b (b > a) are separated by a mate-
rial with conductivity σ. Find an expression for the resistance between the two spheres. 
First find E for a < r < b, assuming +Q at r = a and –Q at r = b. From Gauss’s law: 
! 
Now find Vab: 
! 
Now can find I: 
! 
Finally, ! 
P2.52: The typical length of each piece of jumper wire on a student’s protoboard is 5.0 
cm. Assuming AWG-20 (wire diameter 0.812 mm) copper wire, (a) determine the resis-
tance for this length of wire. (b) Determine the power dissipated in the wire for 10. mA 
of current. 
2- !43
(a) ! 
so R = 1.7 mΩ 
(b) ! 
P2.53: A densely wrapped coil of AWG-22 (0.644 mm diameter) copper magnet wire is 
150 m long. The wire has a very thin insulative sheath. Determine the resistance for this 
length of wire. 
! 
so R = 7.9Ω 
P2.54: Determine an expression for the power dissipated per unit length in coaxial cable 
of inner radius a, outer radius b, and conductivity between the conductors σ if a potential 
difference Vab is applied. 
From Eqn(2.84) we have ! 
Now for a given potential difference Vab we have 
! 
P2.55: Find the resistance per unit length of a stainless steel pipe of inner radius 2.5 cm 
and outer radius 3.0 cm. 
! 
so we have ! 
so R/L = 1.0 mΩ/m 
2- !44
P2.56: A nickel wire of diameter 5.0 mm is surrounded by a 0.50 mm thick layer of sil-
ver. What is the resistance per unit length for this wire? Assuming 1.0 m of this wire 
carries 1.0 A of current, determine the power dissipated in the nickel portion and in the 
silver portion of the wire. 
We can treat this wire as two resistors in parallel. We have 
! 
! 
! 
To find the power dissipated, we first find the potential difference: 
! 
then 
! 
11. Dielectrics 
P2.57: A material has 12.0 V/m ax field intensity with permittivity 194.5 pF/m. Deter-
mine the electric flux density. 
! 
P2.58: MATLAB: A 20 nC point charge at the origin is embedded in Teflon (εr = 2.1). 
Find and plot the magnitudes of the polarization vector, the electric field intensity and the 
electric flux density at a radial distance from 0.1 cm out to 10 cm. 
We use the following equations: 
! 
% M-File: MLP0258 
% 
% Plot E, P and D vs distance r from a point 
% charge Q at the origin with a dielectric. 
2- !45
% 
% Variables 
% Q charge (C) 
% eo free space permittivity (F/m) 
% r radial distance (m) 
% Chi electric susceptibility 
% E electric field intensity(V/m) 
% D electric flux density (C/m^2) 
% P polarization vector (C/m^2) 
% initialize variables 
Q=20e-9; 
er=2.1; 
eo=8.854e-12; 
Chi=er-1; 
% perform calculations 
r=0.001:.001:0.100; 
rcm=r.*100; 
E=Q./(4*pi*r.^2); 
P=Chi*eo*E; 
D=er*eo*E; 
% plot data 
subplot(2,1,1) 
loglog(rcm,P,'--k',rcm,D,'-k') 
legend('P','D') 
ylabel('C/m^2') 
grid on 
subplot(2,1,2) 
loglog(rcm,E) 
ylabel('V/m') 
xlabel('radial dis-
tance (cm)') 
grid on 
Fig. P2.58
2- !46
P2.59: Suppose the force is very carefully measured between a pair of point charges sep-
arated by a dielectric material and is found to be 20 nN. The dielectric material is re-
moved without changing the position of the point charges, and the force has increased to 
100 nN. What is the relative permittivity of the dielectric? 
! 
P2.60: The potential field in a material with εr = 10.2 is V = 12 xy2 (V). Find E, P and D. 
! 
! 
! 
! 
P2.61: In a mineral oil dielectric, with breakdown voltage of 15 MV/m, the potential 
function is V = x3 – 6x2 –3.1x (MV). Is the dielectric likely to breakdown, and if so, 
where? 
! 
! so from 6x – 12 = 0 we find the maximum electric field 
occurs at x = 2m. 
At x = 2m, we have E = -12+24+3.1 = 15.1 MV/m, exceeding the breakdown voltage. 
12. Boundary Conditions 
P2.62: For y < 0, εr1 = 4.0 and E1 = 3ax + 6πay + 4az V/m. At y = 0, ρs = 0.25 nC/m2. If 
εr2 = 5.0 for y > 0, find E2. 
E1 = 3ax + 6πay + 4az V/m (g) E2 = 3ax + 20.7ay + 4az V/m
(a) EN1 = 6πay (f) EN2 = DN2/5εo = 20.7ay
2- !47
(e) ! 
! 
P2.63: For z ≤ 0, εr1 = 9.0 and for z > 0, εr2 = 4.0. If E1 makes a 30° angle with a normal 
to the surface, what angle does E2 make with a normal to the surface? 
Refer to Figure P2.63. 
! 
also 
! 
Therefore 
! and after routine math we find ! 
Using this formula we obtain for this problem θ2 = 14°. 
! 
P2.64:
A plane defined by 3x + 2y + z = 6 separates two dielectrics. The first dielectric, 
on the side of the plane containing the origin, has εr1 = 3.0 and E1 = 4.0az V/m. The other 
dielectric has εr2 = 6.0. Find E2. 
We first use gradient to find a normal to the planar surface. 
Let F = 3x + 2y + z – 6 = 0. 
(b) ET1 = 3ax + 4az (c) ET2 = ET1 = 3ax + 4az
(d) DN1 = εr1εoEN1 = 24πεo ay (e) DN2 = 0.92 ay
Fig. P2.63
2- !48
! 
Now we can work the boundary condition problem. 
! 
! 
! 
! 
Finally we have ! . 
P2.65: MATLAB: Consider a dielectric-dielectric charge free boundary at the plane z = 
0. Construct a program that will allow the user to enter εr1 (for z < 0), εr2, and E1, and 
will then calculate E2. (Just for fun, you may want to have the program calculate the an-
gles that E1 and E2 make with a normal to the surface). 
% M-File: MLP0265 
% 
% Given E1 at boundary between a pair of 
% dielectrics with no charge at boundary, 
% calculate E2. Also calculates angles. 
% 
clc 
clear 
% enter variables 
disp('enter vector quantities in brackets,') 
disp('for example: [1 2 3]') 
er1=input('relative permittivity in material 1: '); 
er2=input('relative permittivity in material 2: '); 
a12=input('unit vector from mtrl 1 to mtrl 2: '); 
E1=input('electric field intensity vector in mtrl 1: '); 
% perform calculations 
En1=dot(E1,a12)*a12; 
Et1=E1-En1; 
Et2=Et1; 
Dn1=er1*En1; %ignores eo since it will factor out 
Dn2=Dn1; 
En2=Dn2/er2; 
2- !49
E2=Et2+En2 
% calculate the angles 
th1=atan(magvector(Et1)/magvector(En1)); 
th2=atan(magvector(Et2)/magvector(En2)); 
th1r=th1*180/pi 
th2r=th2*180/pi 
Now run the program: 
enter vector quantities in brackets, 
for example: [1 2 3] 
relative permittivity in material 1: 2 
relative permittivity in material 2: 5 
unit vector from mtrl 1 to mtrl 2: [0 0 1] 
electric field intensity vector in mtrl 1: [3 4 5] 
E2 = 
 3 4 2 
th1r = 
 45 
th2r = 
 68.1986 
P2.66: A 1.0 cm diameter conductor is sheathed with a 0.50 cm thickness of Teflon and 
then a 2.0 cm (inner) diameter outer conductor. (a) Use Laplace’s equations to find an 
expression for the potential as a function of ρ in the dielectric. (b) Find E as a function of 
ρ. (c) What is the maximum potential difference that can be applied across this coaxial 
cable without breaking down the dielectric? 
(a) Since V is only a function of ρ, 
2- !50
! 
where A and B are constants. 
Now we apply boundary conditions. 
BC1: 
! 
BC2: ! 
or 
! 
(b) ! 
(c) 
! 
P2.67: A 1.0 m long carbon pipe of inner diameter 3.0 cm and outer diameter 5.0 cm is 
cut in half lengthwise. Determine the resistance between the inner surface and the outer 
surface of one of the half sections of pipe. 
One approach is to consider the resistance for the half-section of pipe is twice the resis-
tance for a complete cylindrical section, given by Eqn. (2.84). But we’ll used the 
LaPlace equation approach instead. 
Laplace: ! ; here we see V only depends on ρ 
So: ! ; 
Fig. P2.66
2- !51
where A and B are constants. 
Now apply boundary conditions. 
BC1: 
! 
BC2: 
! 
! 
J=σE 
! 
! 
P2.68: For a coaxial cable of inner conductor radius a and outer conductor radius b and a 
dielectric εr in-between, assume a charge density is added in the dielectric re-
gion. Use Poisson’s equation to derive an expression for V and E. Calculate ρs on each 
plate. 
! 
so 
! , where A is a constant. 
! , where B is a constant. 
Now apply boundary conditions: ! 
Fig. P2.67
2- !52
Applying the second one gives us: 
! 
Applying the first one: 
! 
Therefore, 
! 
! 
where 
! , 
so ! 
! 
! 
P2.69: For the parallel plate capacitor given in Figure 2.51, suppose a charge density 
2- !53
 ! 
is added between the plates. Use Poisson’s equation to derive a new expression for V and 
E. Calculate ρs on each plate. 
! 
! 
! 
Now apply the boundary conditions: 
! 
! 
! 
! 
at z = 0, ! ! 
at z = d, ! ! 
13. Capacitors 
P2.70: A parallel plate capacitor is constructed such that the dielectric can be easily re-
moved. With the dielectric in place, the capacitance is 48 nF. With the dielectric re-
moved, the capacitance drops to 12 nF. Determine the relative permittivity of the dielec-
tric. 
! 
2- !54
P2.71: A parallel plate capacitor with a 1.0 m2 surface area for each plate, a 2.0 mm plate 
separation, and a dielectric with relative permittivity of 1200 has a 12. V potential differ-
ence across the plates. (a) What is the minimum allowed dielectric strength for this ca-
pacitor? Calculate (b) the capacitance, and (c) the magnitude of the charge density on one 
of the plates. 
(a) ! 
(b) ! 
(c) ! 
P2.72: A conical section of material extends from 2.0 cm ≤ r ≤ 9.0 cm for 0 ≤ θ ≤ 30° 
with εr = 9.0 and σ = 0.020 S/m. Conductive plates are placed at each radial end of the 
section. Determine the resistance and capacitance of the section. 
! , where A and B are constants. 
Boundary conditions: r = a, V = 0 and r = b, V = Vb 
! 
! 
! 
2- !55
! 
! 
P2.73: An inhomogeneous dielectric fills a parallel plate capacitor of surface area 50. 
cm2 and thickness 1.0 cm. You are given εr = 3(1 + z), where z is measured from the bot-
tom plate in cm. Determine the capacitance. 
Place +Q at z = d and –Q at z = 0. 
! 
! 
evaluating the integral: 
! 
! 
Fig. P2.72
2- !56
P2.74: Given E = 5xyax + 3zaz V/m, find the electrostatic potential energy stored in a 
volume defined by 0 ≤ x ≤ 2 m, 0 ≤ y ≤ 1 m, and 0 ≤ z ≤ 1 m. Assume ε = εo. 
! 
P2.75: Suppose a coaxial capacitor with inner radius 1.0 cm, outer radius 2.0 cm and 
length 1.0 m is constructed with 2 different dielectrics. When oriented along the z-axis, 
εr for 0° ≤ φ ≤ 180° is 9.0, and for 180° ≤ φ ≤ 360° is 4.0. (a) Calculate the capaci-
tance. (b) If 9.0 V is applied across the conductors, determine the electrostatic potential 
energy stored in each dielectric for this capacitor. 
(a) a coaxial line, 
! 
But for only half the line, 
! 
So 
! 
and 
! 
So 
! 
(b) !
Fig. P2.75
Wentworth solutions chapter 3.pdf
3- !1
Solutions for Chapter 3 Problems 
1. Magnetic Fields and Cross Products 
P3.1: Find AxB for the following: 
a. A = 2ax – 3ay + 4az, B = 5ay - 1az 
b. A = aρ + 2aφ + 4az, B = 2aρ + 6az 
c. A = 2ar + 5aθ + 1aφ, B = ar + 3aφ 
(a) 
! 
(b) 
! 
(c) 
! 
P3.2: If a parallelogram has a short side a, a long side b, and an interior angle θ (the 
smaller of the two interior angles), the area of the parallelogram is given by 
 
! 
Determine how you would use the cross product of a pair of vectors to find the area of a 
parallelogram defined by the points O(0,0,0), P(6,0,0), Q(8,12,0) and R(2,12,0). 
(Assume dimensions in meters) 
 
! 
A = 6ax, B = 2ax + 12ay 
A x B = 72az, 
Area = 72 m2 
( ) ( ) ( )2 3 4 3 20 2 10 17 2 10
0 5 1
x y z
x y z x y z× = − = − − − + = − + +
−
a a a
A B a a a a a a
( ) ( )1 2 4 12 6 8 0 4 12 2 4
2 0 6
z
z z
ρ φ
ρ φ ρ φ× = = − − + − = + −
a a a
A B a a a a a a
( ) ( )2 5 1 15 6 1 0 5 15 5 5
1 0 3
r
r r
θ φ
θ φ θ φ× = = − − + − = − −
a a a
A B a a a a a a
sin .area ab θ=
sin
sin
area ab θ
θ
=
= × =A B A B Fig. P3.2
3- !2
P3.3: Given the vertices of a triangle P(1,2,0), Q(2,5,0) and R(0,4,7), find (a) the interior 
angles, (b) a unit vector normal to the surface containing the triangle and (c) the area of 
the triangle. 
(a) ! 
! 
! 
! 
! 
! 
(b) ! 
(c) ! 
2. Biot-Savart’s Law 
P3.4: A segment of conductor on the z-axis extends from z = 0 to z = h. If this segment 
conducts current I in the +az direction, find H(0,y,0). Compare your answer to that of 
Example 3.2. 
 
We use Eqn. (3.7) and change the limits: 
1 3 ; 3.16x y= + =PQ a a PQ
1 2 7 ; 7.348x y z= − + + =PR a a a PR
21 7 5
22.69
x y z× = − +
× =
PQ PR a a a
PQ PR
( )( )
sin
22.69sin ;
3.16 7.348
78
ab
P
P
θ
θ
θ
× =
=
=
A B A B
�
1 3 ; 3.16
2 1 7 ; 7.348
x y
x y z
= − − =
= − − + =
QP a a QP
QR a a a QR
22.69; 78 ; 180 24Q R P Qθ θ θ θ× = = = − − =QP QR
� � �
0.93 0.31 0.22n x y z
×
= = − +
×
PQ PRa a a a
PQ PR
21 11.4
2
area m= × =PQ PR
Fig. P3.3
3- !3
! 
Note that if the line of current is semi-
infinite (goes from z = 0 to z = ∞), we’d 
have: 
! 
P3.5: An infinite length line with 2.0 A current in the +ax direction exists at y = -3.0 m, z 
= 4.0 m. A second infinite length line with 3.0 A current in the +az direction exists at x = 
0, y = 3.0 m. Find H(0, 0, 0). 
This situation is shown in Figure P3.5a. 
Ho = H1 + H2 
Referring to the figure, 
R = 3ay – 4az, R = 5, 
aR = 0.6ay – 0.8az 
aφ = ax x aR = 0.80ay + 0.60az 
! 
! 
Ho = 159ax + 51ay +38az mA/m 
Or with 2 significant digits 
( )
2 2 2
0
2 2 2
2 2
4
4
4
h
x
x
I z
z
Iy h
y h y
Ih
y h y
φρ
π ρ ρ
π
π
⎡ ⎤
= ⎢ ⎥
+⎢ ⎥⎣ ⎦
⎡ ⎤−
= ⎢ ⎥
⎢ + ⎥⎣ ⎦
−
=
+
a
H
a
aH
4
xI
yπ
−
=
aH
( )
1
1 2
0.80 0.602
2 5
51 38 
y z
y z
I
A
m
mA
m
φπρ
π
=
+
=
= +
H a
a a
a a
2
2 2
3 159
2 3
x
x
I
A mA
mm
φπρ
π
=
= =
H a
a a
Fig. P3.4
Fig. P3.5a
Fig. P3.5b
3- !4
Ho = 160ax + 51ay +38az mA/m 
P3.6: A conductive loop in the shape of an equilateral triangle of side 8.0 cm is centered 
in the x-y plane. It carries 20.0 mA current clockwise when viewed from the +az 
direction. Find H(0, 0, 16cm). 
The situation is illustrated in Figure P3.6a. The sketch in Figure P3.6b is used to find the 
+x-axis intercept for the triangle. By simple trigonometry we have: 
! 
Now for one segment we adapt Eqn. (3.7): 
! 
with ρaρ = RaR, 
! 
! 
H1 = -4.7ax – 0.68az mA/m 
Now by symmetry the total H contains only the az component: 
Htot = -2.0 mA/m az. 
 
tan 30 ;and since = 4 cm we find = 2.31 cm.b a b
a
= �
1 2 24
a
a
I z
z
φ
πρ ρ
+
−
⎡ ⎤
= ⎢ ⎥
+⎢ ⎥⎣ ⎦
a
H
2.31 16 , 16.17 , 0.99 0.143x z y R x zcm φ= − + = = − × = − −R a a R a a a a a
( )( )
( )
( )3
1 2 2 2
20 10 0.99 0.143 2 4
4 16.17 10 4 16.17
x zx A
x mπ
−
−
− − ⎡ ⎤
= ⎢ ⎥
+⎣ ⎦
a a
H
Fig. P3.6a Fig. P3.6b
3- !5
P3.7: A square conductive loop of side 10.0 cm is centered in the x-y plane. It carries 
10.0 mA current clockwise when viewed from the +az direction. Find H(0, 0, 10cm). 
We find H for one section of the square 
by adopting Eqn. (3.7): 
! 
! 
With ρaρ = RaR, we have 
R = -5ax + 10az, |R| = 11.18x10-2 m 
aR = -0.447ax + 0.894az, 
aφ = -ay x aR = -0.894 ax – 0.447 az 
! 
Now by symmetry the total H contains only the az component: 
HTOT = -10.4az mA/m 
P3.8: A conductive loop on the x-y plane is bounded by ρ = 2.0 cm, ρ = 6.0 cm, φ = 0° 
and φ = 90°. 1.0 A of current flows in the loop, going in the aφ direction on the ρ = 2.0 
cm arm. Determine H at the origin. 
By inspection of the figure, we see 
that only the arc portions of the 
loop contribute to H. 
From a ring example we have: 
! 
For the ρ = a segment of the loop: 
! 
2 24
a
a
I z
z
φ
πρ ρ
+
−
⎡ ⎤
= ⎢ ⎥
+⎢ ⎥⎣ ⎦
a
H
2 22
Ia
a
φ
πρ ρ
=
+
a
H
( )( )( )
( )
3
1 2 2 2
10 10 5 0.894 0.447
5.2 2.6
2 11.18 10 5 11.18
x z
x z
x mA
mxπ
−
−
− −
= = − −
+
a a
H a a
( )
22
3
2 2 2 04
zIa d
h a
π
φ
π
=
+
∫
aH
/ 22
a 3
04 8
z
z
Ia Id
a a
π
φ
π
= =∫
aH a
Fig. P3.7
Fig. P3.8
3- !6
At ρ = b: ! ; 
So ! 
P3.9: MATLAB: How close do you have to be to the middle of a finite length of current-
carrying line before it appears infinite in length? Consider Hf(0, a, 0) is the field for the 
finite line of length 2h centered on the z-axis, and that Hi(0, a, 0) is the field for an 
infinite length line of current on the z-axis. In both cases consider current I in the +az 
direction. Plot Hf/Hi vs h/a. 
Adapting Eqn. (3.7), for the finite length line we have: 
! 
For the infinite length of line: 
! 
The ratio we wish to plot is: 
! 
The MATLAB routine follows. 
% M-File: MLP0309 
%Consider the field for a finite line of length 2h 
%oriented on z-axis with current I in +z direction. 
%The field is to be found a distance a away from 
%the current on the y axis (point (0,a,0)). 
%We want to compare this field with that of an infinite 
%length line of current. 
%Plot Hf/Hi versus h/a. We expect that as h/a grows large, 
02
b 3
/ 24 8
z
z
Ib Id
b bπ
φ
π
−
= =∫
aH a
1 1 1 1 1 4.2
8 8 0.02 0.06TOT z z z
I A
a b m
⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
H a a a
2 22f
a
I h
h
φ
ρ
πρ ρ
=
⎡ ⎤
= ⎢ ⎥
+⎢ ⎥⎣ ⎦
a
H
2i
I φ
πρ
=
a
H
( )
( )
2 2 2
1
f
i
hH h a
H h a h
a
= =
+ +
3- !7
%the line will appear more 'infinite' to an observation 
%point at (0,a,0). 
hova=0.01:.01:100; 
HfovHi=hova./sqrt(1+(hova).^2); 
semilogx(hova,HfovHi) 
xlabel('h/a') 
ylabel('Hf/Hi') 
grid on 
P3.10: MATLAB: For the ring of current described in MATLAB 3.2, find H at the 
following points (a) (0, 0, 1m), (b) (0, 2m, 0), and (c) (1m, 1m, 0). 
%M-File: MLP0310 
%Find the magnetic field intensity at any observation point 
%resulting from a ring of radius a and current I, 
%in the aphi direction centered in the x-y plane. 
df=1; %increment in degrees 
a=1; %ring radius in m 
I=1; %current in A 
Ro=input('vector location of observation point: '); 
for j=1:df:360; 
 Fr=j*pi/180; 
 Rs=[a*cos(Fr) a*sin(Fr) 0]; 
 as=unitvector(Rs); 
Fig. P3.9a
Fig. P3.9b
3- !8
 dL=a*df*(pi/180)*cross([0 0 1],as); 
 Rso=Ro-Rs; 
 aso=unitvector(Rso); 
 dH=I*cross(dL,aso)/(4*pi*(magvector(Rso))^2); 
 dHx(j)=dH(1); 
 dHy(j)=dH(2); 
 dHz(j)=dH(3); 
end 
H=[sum(dHx) sum(dHy) sum(dHz)] 
Now to run the program: 
>> MLP0310 
vector location of observation point: [0 0 1] 
H = -0.0000 -0.0000 0.1768 
>> MLP0310 
vector location of observation point: [0 2 0] 
H = 0 0 -0.0431 
>> MLP0310 
vector location of observation point: [1 1 0] 
H = 0 0 -0.1907 
>> MLP0310 
vector location of observation point: [0 1 1] 
H = 0.0000 0.0910 0.0768 
>> 
So we see: 
(a)
H = 0.18 az A/m 
(b) H = -0.043 az A/m 
(c) H = -0.19 az A/m 
(extra) H = 9.1ay + 7.7 az mA/m 
3- !9
P3.11: A solenoid has 200 turns, is 10.0 cm long, and has a radius of 1.0 cm. Assuming 
1.0 A of current, determine the magnetic field intensity at the very center of the solenoid. 
How does this compare with your solution if you make the assumption that 10 cm >> 1 
cm? 
Eqn. (3.10): 
 ! 
Or H = 1960 A/m az 
The approximate solution, assuming 10cm >> 1cm, is 
! 
P3.12: MATLAB: For the solenoid of the previous problem, plot the magnitude of the 
field versus position along the axis of the solenoid. Include the axis 2 cm beyond each 
end of the solenoid. 
% M-File: MLP0312 
% 
% Plot H vs length thru center of a solenoid 
% 
clc 
clear 
% initialize variables 
N=200; %number of turns 
h=0.10; %height of solenoid 
a=0.01; %radius of solenoid 
I=1; %current 
dz=0.001; %step change in z 
z=-.02:dz:h+.02; 
zcm=z.*100; 
( )
( )
( ) ( )
2 2 22
2 2 22
2
200 1 0.1 0.05 0.05 1961
2 0.1 0.05 0.010.1 0.05 0.01
z
z z
NI h z z
h z ah z a
A A
m m
⎡ ⎤
−⎢ ⎥= +
⎢ ⎥+− +⎣ ⎦
⎡ ⎤
−⎢ ⎥= + =
⎢ ⎥+− +⎣ ⎦
H a
H a a
( )200 1 2000
0.1z z z
ANI A
h m m
= = =H a a a
3- !10
A1=(h-z)./sqrt((h-z).^2+a^2); 
A2=z./sqrt(z.^2+a^2); 
Atot=A1+A2; 
H=N*I.*Atot/(2*h); 
% generate plot 
plot(zcm,H) 
xlabel('z(cm)') 
ylabel('H (A/m)') 
grid on 
P3.13: A 4.0 cm wide ribbon of current is centered about the y-axis on the x-y plane and 
has a surface current density K = 2π ay A/m. Determine the magnetic field intensity at 
the point (a) P(0, 0, 2cm), (b) Q(2cm, 2cm, 2cm). 
(a) Because of the symmetry (Figure P3.13a), 
we can use a modified Eqn. (3.14): 
! 
(b) Referring to Figure P3.13b; 
! 
! 
! 
1
1
tan
2 2tan 1.57
2
y
x
x x
K d
a
A
m
π
π
π
−
−
⎛ ⎞= ⎜ ⎟
⎝ ⎠
⎛ ⎞= =⎜ ⎟
⎝ ⎠
H a
a a
( )
; 
2
; r x z
Id
R where d x a
φ
ρ
πρ
ρ
=
= = − +
H a
a a R a a
( )
( )
( )
( )
( )
2 2
2 2
2 2
 , and
. Sox zR
x z
y R
so d x a
d x a
d x a
a x d
d x a
φ
ρ = − +
− +
=
− +
+ −
= × =
− +
a a
a
a a
a a a
( )
( )2 22
y x z zK a x d dx
d x aπ
+ −
=
− +∫
a a a
H
Fig. P3.12
Fig. P3.13a
Fig. P3.13b
3- !11
This is separated into 3 integrals, each 
one solved via numerical integration, 
resulting in: 
H=1.1083ax +0.3032az – 1.1083az; or 
H = 1.1ax – 0.80az A/m 
3. Ampere’s Circuit Law 
P3.14: A pair of infinite extent current sheets exists at z = -2.0 m and at z = +2.0 m. The 
top sheet has a uniform current density K = 3.0 ay A/m and the bottom one has K = -3.0 
ay A/m. Find H at (a) (0,0,4m), (b) (0,0,0) and (c) (0,0,-4m). 
 
We apply ! 
(a) ! 
(b) 
! 
(c) H = 0 
P3.15: An infinite extent current sheet with K = 6.0 ay A/m exists at z = 0. A conductive 
loop of radius 1.0 m, in the y-z plane centered at z = 2.0 m, has zero magnetic field 
intensity measured at its center. Determine the magnitude of the current in the loop and 
show its direction with a sketch. 
Htot = HS + HL 
! 
For the loop, we use Eqn. (3.10): 
! 
where here 
1 ,
2 N
= ×H K a
( ) ( )1 13 -3 0
2 2y z y z
= × + × =H a a a a
( ) ( )1 13 -3
2 2
3
y z y z
x
A
m
= ×− + ×
= −
H a a a a
a
( )1 1 6 3
2 2S N y z x
A
m
= × = × =H K a a a a
2 z
I
a
=H a
Fig. P3.14
Fig. P3.15
3- !12
! 
(sign is chosen opposite HS). 
So, I/2 = 3 and I = 6A. 
P3.16: Given the field H = 3y2 ax, find the current passing through a square in the x-y 
plane that has one corner at the origin and the opposite corner at (2, 2, 0). 
Referring to Figure P3.6, we evaluate the circulation of H around the square path. 
! 
! 
! 
! 
! 
So we have Ienc = 24 A. The negative 
Sign indicates current is going in the 
-az direction. 
P3.17: Given a 3.0 mm radius solid wire centered on the z-axis with an evenly distributed 
2.0 amps of current in the +az direction, plot the magnetic field intensity H versus radial 
distance from the z-axis over the range 0 ≤ ρ ≤ 9 mm. 
Figure P3.17 shows the situation along with the Amperian Paths. We have: 
! 
This will be true for each Amperian path. 
( )
2 2L x x
I I
a
−
= − =H a a
b c d a
enc
a b c d
d I= = + + +∫ ∫ ∫ ∫ ∫H L��
( )23 0 0
b
x x
a
dx =∫ a a�
23 0
c
x y
b
y dy =∫ a a�
( )
0
2
2
3 2 12 24
d
x x
c
dx dx= = −∫ ∫a a�
0
a
d
=∫
, ; 2enc encd I where H and d d H Iφ φ φ φρ φ πρ= = = =∫ H L H a L a� ��
Fig. P3.16
3- !13
AP1: ! 
So: ! 
AP2: Ienc = I, 
% MLP0317 
% generate plot for ACL problem 
a=3e-3; %radius of solid wire (m) 
I=2; %current (A) 
N=30; %number of data points to plot 
rmax=9e-3; %max radius for plot (m) 
dr=rmax/N; 
for i=1:round(a/dr) 
 r(i)=i*dr; 
 H(i)=(I/(2*pi*a^2))*r(i); 
end 
for i=round(a/dr)+1:N 
 r(i)=i*dr; 
 H(i)=I/(2*pi*r(i)); 
end 
plot(r,H) 
xlabel('rho(m)') 
ylabel('H (A/m)') 
grid on 
2 2
2 2 2
0 0
I I I, = ,
a a aenc z enc
I d I d d
ρ π ρ
ρ ρ φ
π π
= = =∫ ∫ ∫J S J a�
2 for 2
I a
a φ
ρ ρ
π
= ≤H a
 for 
2
I aφ ρπρ
= ≥H a
Fig. P3.17a Fig. P3.17b
3- !14
P3.18: Given a 2.0 cm radius solid wire centered on the z-axis with a current density J = 
3ρ A/cm2 az (for ρ in cm) plot the magnetic field intensity H versus radial distance from 
the z-axis over the range 0 ≤ ρ ≤ 8 cm. 
We’ll let a = 2 cm. 
! 
AP1 (ρ < a): ! 
and ! 
AP2 (ρ > a): Ienc = 2πa3, so 
The MATLAB plotting routine is as follows: 
% MLP0318 
% generate plot for ACL problem 
a=2; %radius of solid wire (cm) 
N=40; %number of data points to plot 
rmax=8; %max radius for plot (cm) 
dr=rmax/N; 
for i=1:round(a/dr) 
 r(i)=i*dr; 
 H(i)=r(i)^2; 
end 
for i=round(a/dr)+1:N 
 r(i)=i*dr; 
 H(i)=a^3/r(i); 
end 
plot(r,H) 
xlabel('rho(cm)') 
ylabel('H (A/cm)') 
grid on 
P3.19: An infinitesimally thin metallic cylindrical shell of radius 4.0 cm is centered on 
the z-axis and carries an evenly distributed current of 10.0 mA in the +az direction. (a) 
, ; 2enc encd I where H and d d H Iφ φ φ φρ φ πρ= = = =∫ H L H a L a� ��
33 2enc zI d d dρρ ρ ρ φ πρ= = =∫ ∫J S a a� �
2 for aφρ ρ= ≤H a
3
 for a aφ ρρ
= ≥H a
Fig. P3.18
3- !15
Determine the value of the surface current density on the conductive shell and (b) plot H 
as a function of radial distance from the z-axis over the range 0 ≤ ρ ≤ 12 cm. 
(a) ! 
(b) for ρ < a, H = 0. For ρ > a we have: 
! 
The MATLAB routine to generate the plot is as follows: 
% MLP0319 
% generate plot for ACL problem 
a=4; %radius of solid wire (cm) 
N=120; %number of data points to plot 
I=10e-3; %current (A) 
rmax=12; %max plot radius(cm) 
dr=rmax/N; 
for i=1:round(a/dr) 
 r(i)=i*dr; 
 H(i)=0; 
end 
for i=round(a/dr)+1:N 
 r(i)=i*dr; 
 H(i)=100*I/(2*pi*r(i)); 
end 
plot(r,H) 
xlabel('rho(cm)') 
ylabel('H (A/m)') 
grid on 
( )
10 39.8 ; so 40
2 2 0.04s z
I mA mA mAK
a m m mπ π
= = = =K a
2
I
φπρ
=H a
Fig. P3.19a
Fig. P3.19b
3- !16
 
P3.20: A cylindrical pipe with a 1.0 cm
wall thickness and an inner radius of 4.0 cm is 
centered on the z-axis and has an evenly distributed 3.0 amps of current in the +az 
direction. Plot the magnetic field intensity H versus radial distance from the z-axis over 
the range 0 ≤ ρ ≤ 10 cm. 
For each Amperian Path: 
! 
Now, for ρ < a, Ienc = 0 so H = 0. 
For a < ρ < b, ! 
! 
 
% MLP0320 
% generate plot for ACL problem 
a=4; %inner radius of pipe (cm) 
b=5; %outer radius of pipe(cm) 
N=120; %number of data points to plot 
I=3; %current (A) 
rmax=10; %max radius for plot (cm) 
dr=rmax/N; 
aoverdr=a/dr 
boverdr=b/dr 
, ; 2enc encd I where H and d d H Iφ φ φ φρ φ πρ= = = =∫ H L H a L a� ��
( )2 2
, where and =enc z z
II d d d d
b a
ρ ρ φ
π
= =
−∫ J S J a S a�
( ) ( )
( )2 22 2 2
2 22 2 2 2
0
,
2enc a
aI a II d d I
b ab a b a
ρ π
φ
ρρ
ρ ρ φ
ρπ π
−−
= = =
−− −∫ ∫ H a
Fig. P3.20bFig. P3.20a
3- !17
for i=1:round(a/dr) 
 r(i)=i*dr; 
 H(i)=0; 
end 
for i=round(a/dr)+1:round(b/dr) 
 r(i)=i*dr; 
 num(i)=I*(r(i)^2-a^2); 
 den(i)=2*pi*(b^2-a^2)*r(i); 
 H(i)=100*num(i)/den(i); 
end 
for i=round(b/dr)+1:N 
 r(i)=i*dr; 
 H(i)=100*I/(2*pi*r(i)); 
end 
 
plot(r,H) 
xlabel('rho(cm)') 
ylabel('H (A/m)') 
grid on 
P3.21: An infinite length line carries current I in the +az direction on the z-axis, and this 
is surrounded by an infinite length cylindrical shell (centered about the z-axis) of radius a 
carrying the return current I in the –az direction as a surface current. Find expressions for 
the magnetic field intensity everywhere. If the current is 1.0 A and the radius a is 2.0 cm, 
plot the magnitude of H versus radial distance from the z-axis from 0.1 cm to 4 cm. 
Fig. P3.21a
3- !18
! 
The MATLAB routine used to generate 
Figure P3.21b is as follows: 
% MLP0321 
% generate plot for ACL 
problem 
clc 
clear 
a=2; %inner radius of 
cylinder(cm) 
N=80; %number of data 
points to plot 
I=1; %current (A) 
rmax=4; %max radius for plot (cm) 
dr=rmax/N; 
for i=1:40 
 r(i)=.1+(i-1)*dr; 
 H(i)=100*I/(2*pi*r(i)); 
end 
for i=40:N 
 r(i)=i*dr; 
 H(i)=0; 
end 
plot(r,H) 
xlabel('rho(cm)') 
ylabel('H (A/m)') 
grid on 
P3.22: Consider a pair of collinear cylindrical shells centered on the z-axis. The inner 
shell has radius a and carries a sheet current totaling I amps in the +az direction while the 
outer shell of radius b carries the return current I in the –az direction. Find expressions 
for the magnetic field intensity everywhere. If a = 2cm, b = 4cm and I = 4A, plot the 
magnitude of H versus radial distance from the z-axis from 0 to 8 cm. 
; for 0< , and for , 0.
2enc
Id I a aφρ ρπρ
= < = > =∫H L H a H�
Fig. P3.21b
3- !19
! 
The MATLAB routine used to generate Figure P3.22b is as follows: 
% MLP0322 
% generate plot for ACL problem 
a=2; %inner radius of coax (cm) 
b=4; %outer radius of coax(cm) 
N=160; %number of data points to plot 
I=4; %current (A) 
rmax=8; %max radius for plot (cm) 
dr=rmax/N; 
aoverdr=a/dr 
boverdr=b/dr 
for i=1:round(a/dr) 
 r(i)=i*dr; 
 H(i)=0; 
end 
for i=round(a/dr)+1:round(b/dr) 
 r(i)=i*dr; 
 H(i)=100*I/(2*pi*r(i)); 
end 
for i=round(b/dr)+1:N 
; for 0< , = 0;
for a< , ;
2
 and for , 0.
encd I a
Ib
b
φ
ρ
ρ
πρ
ρ
= <
< =
> =
∫H L H
H a
H
�
Fig. P3.22a Fig. P3.22b
3- !20
 r(i)=i*dr; 
 H(i)=0; 
end 
plot(r,H) 
xlabel('rho(cm)') 
ylabel('H (A/m)') 
grid on 
P3.23: Consider the toroid in Figure 3.55 that is tightly wrapped with N turns of 
conductive wire. For an Amperian path with radius less than a, no current is enclosed 
and therefore the field is zero. Likewise, for radius greater than c, the net current 
enclosed is zero and again the field is zero. Use Ampere’s Circuital Law to find an 
expression for the magnetic field at radius b, the center of the toroid. 
! 
Within the toroid, H = Hφ aφ, so 
! 
Then, Ienc by the Amperian path is: Ienc = NI. 
! 
4. Curl and the Point Form of Ampere’s Circuital Law 
P3.24: Find ! for the following fields: 
a. A = 3xy2/z ax 
 b. A = ρsin2φ aρ – ρ2 z cosφ aφ 
c. A = r2sinθ ar + r/cos φ aθ 
(a) ! 
(b) ! 
! 
;encd I=∫H L�
2 .
b
d H d bHφ φ φ φρρ φ π== ⋅ =∫ ∫H L a a�
.
2
NI
b φπ
∴ =H a
∇×A
( )
2
2
2
3 63 x xx y z y z
A A xy xyxy z
z y z z
∂ ∂ −
∇× = − = −
∂ ∂
a a a a a
( ) ( )2 2 1sin cos z
A A A
z A
z z
φ ρ ρ
ρ φ ρ φ φρ φ ρ φ ρρ ρ φ
∂ ∂ ∂⎡ ⎤∂
∇× − = − + + −⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦
a a a a a
2
2 3 cos 1cos 0 2sin cosz z
z
ρ φ
ρ φ
ρ φ ρ φ φ
ρ ρ
= + − −a a a a
3- !21
! 
 (c) 
 ! 
! 
P3.25: Find J at (3m, 60°, 4m) for H = (z/sinφ) aρ – (ρ2/cosφ) az A/m. 
! 
! 
Now find J by evaluating ! at the given point: 
! 
P3.26: Suppose H = y2ax + x2ay A/m. 
a. Calculate ! around the path ! , where A(2m,0,0), 
B(2m,4m,0), C(0,4m,0) and D(0,0,0). 
b. Divide this by the area S (2m*4m = 8m2). 
c. Evaluate ! at the center point. 
d. Comment on your results for (b) & (c). 
(a) Referring to the figure, we evaluate ! 
[ ]2 cos 3 cos 2sin cos zzρρ φ ρ φ φ φ= − +a a
( )
2 sin
cos
1 1 1 1
sin sin
r
r r
r
rr
rAA A A
r r r r
θ
θθ
θ φ
θ
φ
θ φ θ φ θ
⎛ ⎞
∇× +⎜ ⎟
⎝ ⎠
∂⎡ ⎤⎡∂ ⎤ ⎡ ⎤∂ ∂−
= + + −⎢ ⎥⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦ ⎣ ⎦
a a
a a a
( ) ( )
1 2
2
2
cos 1 1 sin
sin cos
sin 2 cos
sin cos cos
r
r
r r r
r r r r
r
φ φ
φ
φ
θ
θ φ φ θ
φ
θ
θ φ φ
−∂ ⎛ ⎞− ∂ ∂
= + −⎜ ⎟∂ ∂ ∂⎝ ⎠
⎡ ⎤−
= + −⎢ ⎥
⎣ ⎦
a a a
a a
1 1z z
z
A AA A
z
ρ ρ
ρ φρ φ ρ ρ φ
∂ ∂⎡ ⎤∂ ∂ −
∇× = + − +⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦
H a a a
2 2
sin 1 2 cos
cos sin cos sin z
z
ρ φ
ρ φ ρ φ
φ φ φ ρ φ
⎛ ⎞−
= + + +⎜ ⎟
⎝ ⎠
a a a
∇×H
210 13 0.89 z
A
mρ φ
= − + +J a a a
d∫ H L�� A B C D A→ → → →
d∫ H L��
∇×H
B C D A
enc
A B C D
d I= = + + +∫ ∫ ∫ ∫ ∫H L�
3- !22
! 
! 
! 
! 
So we have ! 
(b) dividing by S = 8m2, we have -2 C/m2 
(c) Evaluating the curl of H: 
! , and at the center point (x = 1 and y = 2) we have 
! 
(d) In this particular case, ! even though S is of appreciable size. 
P3.27: For the coaxial cable example 3.8, we found: 
! 
a. Evaluate the curl in all 4 regions. 
b. Calculate the current density in the conductive regions by dividing the current by 
the area. Are these results the same as what you found in (a)? 
(a) ! 
! 
( )
4
2
0 2
4 4 16
B
A x
x dy C
=
= = =∫ ∫
0
2
2 4
32
C
B y
y dx C
=
= = −∫ ∫
0
2
4 0
0
D
C x
x dy
=
= =∫ ∫
2
2
0 0
0
A
D y
y dy
=
= =∫ ∫
16d C= −∫H L�
( )2 2y x z z
A A x y
x y
∂⎛ ⎞∂
∇× = − = −⎜ ⎟∂ ∂⎝ ⎠
H a a
22 zcenter
C
m
∇× = −H a
,d S∇× = ∫H H L��
( )
2
2 2
2 2
for a, ,
2
for a< b, ,
2
Ifor b< c, = ,
2 c
and for c< , =0.
I
a
I
c
b
φ
φ
φ
ρ
ρ
π
ρ
πρ
ρ
ρ
ρπ
ρ
≤ =
≤ =
⎛ ⎞−
≤ ⎜ ⎟
− ⎝ ⎠
H a
H a
H a
H
2
2 2
1 for 
2 z z
I I a
a a
ρ
ρ
ρ ρ π π
⎛ ⎞∂
∇× = = ≤⎜ ⎟∂ ⎝ ⎠
H a a
1 0 for a
2 z
I bρ
ρ ρ π
∂ ⎛ ⎞∇× = = < <⎜ ⎟∂ ⎝ ⎠
H a
Fig. P3.26
3- !23
! 
! 
(b) ! 
! 
Comment: ! is confirmed. 
P3.28: Suppose you have the field H = r cos θ aφ A/m. Now consider the cone specified 
by θ = π/4, with a height a as shown in Figure 3.56. The circular top of the cone has a 
radius a. 
a. Evaluate the right side of Stoke’s theorem through the dS = dSaθ surface. 
b. Evaluate the left side of Stoke’s theorem by integrating around the loop. 
(a) ! 
ar derivative: 
! 
aθ derivative: 
( )
( ) ( )
2 2
2 2 2 2
1 for b<
2 z z
I c I c
c b c b
ρ
ρ
ρ ρ π π
⎛ ⎞−∂ −
⎜ ⎟∇× = = ≤
⎜ ⎟∂ − −⎝ ⎠
H a a
0 for cρ∇× >H =
( )2 2for , , z zIIa S a S aρ π π≤ = = =J a a
( ) ( )
2 2
2 2
for b< , , z
Ic S c b
c b
ρ π
π
−
≤ = − =
−
J a
∇×H = J
( ) ( )1 1sinsin r
rH
H
r r r
φ
φ θθθ θ
∂∂⎡ ⎤∇× = −⎢ ⎥∂ ∂⎣ ⎦
H a a
( ) ( ) ( ) ( )
2 2
2 2
cos sin1sin cos cos sin ; sin
sin sinr r
r r H
r φ
θ θ
θ θ θ θ θ
θ θ θ θ
−∂ ∂⎡ ⎤= − =⎢ ⎥∂ ∂⎣ ⎦
a a
Fig. P3.28
3- !24
! 
So, ! 
Now we must integrate this over the aθ surface: 
! 
(b) ! 
Clearly in this case the circulation of H is the easiest approach. 
5. Magnetic Flux Density 
P3.29: An infinite length line of 3.0 A current in the +ay direction lies on the y-axis. Find 
the magnetic flux density at P(7.0m,0,0) in (a) Teslas, (b) Wb/m2, and (c) Gauss. 
! 
! 
! 
P3.30: Suppose an infinite extent sheet of current with K = 12ax A/m lies on the x-y 
plane at z = 0. Find B for any point above the sheet. Find the magnetic flux passing 
through a 2m2 area in the x-z plane for z > 0. 
! 
This is valid at any point above the sheet. 
( )2 cos1 2cosr
r r θ θ
θ
θ
∂−
= −
∂
a a
( )2 2cos sin
2cos
sin r θ
θ θ
θ
θ
−
∇× = −H a a
( )
( )
( )
2 2
2 2
2
4 0 0
cos sin
2cos sin
sin
2 sin cos 2 sin cos 2
r
a
d r drd
r drd rdr d a
θ θ
π
π
θ
θ θ
θ θ φ
θ
θ θ φ θ θ φ π
=
⎡ ⎤−
⎢ ⎥∇× = − −
⎢ ⎥⎣ ⎦
= = =
∫ ∫
∫ ∫ ∫
H S a a a� �
2
2 2
2
4 0
cos cos 2
r a
d r ad a d a
π
πφ φ θ
θ φ θ φ π
= =
= = =∫ ∫ ∫H L a a� ��
( )
( )3 68
2 2 7 z z
I A mA
m mφπρ π
= = − = −H a a a
7 9 9
24 10 68 86 10 86 10o z z z
H A Wb Wbx x x T
m m HA m
µ π − − −⎛ ⎞⎛ ⎞ ⎛ ⎞= = − = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
B H a a a
( )9 610,00086 10 860 10z zGx T x GT
− −= − = −B a a
( )7 6
2
1 ;
2
4 10
12 7.54 10
2 2
N
o
N x z y
x H m A Wbx
m m
πµ
−
−
= ×
⎛ ⎞= × = × = −⎜ ⎟
⎝ ⎠
H K a
B K a a a a
3- !25
Now, ! 
P3.31: An infinite length coaxial cable exists along the z-axis, with an inner shell of 
radius a carrying current I in the +az direction and outer shell of radius b carrying the 
return current. Find the magnetic flux passing through an area of length h along the z-
axis bounded by radius between a and b. 
For a < ρ < b, ! 
6. Magnetic Forces 
P3.32: A 1.0 nC charge with velocity 100. m/sec in the y direction enters a region where 
the electric field intensity is 100. V/m az and the magnetic flux density is 5.0 Wb/m2 ax. 
Determine the force vector acting on the charge. 
! 
! 
P3.33: A 10. nC charge with velocity 100. m/sec in the z direction enters a region where 
the electric field intensity is 800. V/m ax and the magnetic flux density 12.0 Wb/m2 ay. 
Determine the force vector acting on the charge. 
! 
( )( )-6 22-7.54x10 2 15y yWbd m Wbmφ µ
⎛ ⎞= = − =⎜ ⎟
⎝ ⎠∫
B S B S = a a� �
( )
, ,
2 2
ln
2 2
ln Wb
2
o
bo o
a
o
II
I Id d dz h
Ih b
a
φ φ
φ
φ
µ
πρ πρ
µ µ
φ ρ ρ
π ρ π
µ
φ
π
= =
= = =
⎛ ⎞= ⎜ ⎟
⎝ ⎠
∫ ∫
H a B a
a
B S a� �
( ) 2; 100 5 500y z
x
m Wb Wbq
s m sm
= + × × = × = −F E u B u B a a
a
910 100 500 400z z z
V Wb Vs mNC nN
m sm Wb VC
− ⎛ ⎞= + − = −⎜ ⎟
⎝ ⎠
F a a a
( ) 9 210 10 800 100 12 4x z x
y
V m Wbq x C N
m s m
µ−
⎛ ⎞
= + × = + × = −⎜ ⎟⎜ ⎟
⎝ ⎠
F E u B a a a
a
3- !26
P3.34: A 10. nC charged particle has a velocity v = 3.0ax + 4.0ay + 5.0az m/sec as it 
enters a magnetic field B = 1000. T ay (recall that a tesla T = Wb/m2). Calculate the force 
vector on the charge. 
! 
The cross-product: 
! 
Evaluating we find: F = -50ax + 30az µN 
P3.35: What electric field is required so that the velocity of the charged particle in the 
previous problem remains constant? 
! 
! 
P3.36: An electron (with rest mass Me= 9.11x10-31kg and charge q = -1.6 x 10-19 C) has a 
velocity of 1.0 km/sec as it enters a 1.0 nT magnetic field. The field is oriented normal to 
the velocity of the electron. Determine the magnitude of the acceleration on the electron 
caused by its encounter with the magnetic field. 
! 
! 
! 
( ) ( )9 210 10 3 4 5 1000x y z y
m Wbq x C
s m
− ⎡ ⎤= × = + + ×⎢ ⎥⎣ ⎦
F u B a a a a
3 4 5 5000 3000
0 1000 0
x y z
x z
⎡ ⎤
⎢ ⎥ = − +⎢ ⎥
⎢ ⎥⎣ ⎦
a a a
a a
0 (constant velocity)dm m
dt
= = =F va
( )
( ) 2
0;
4 5 1000 3000 5000
5 3 
x y z y z x
x z
q
m Wb V V
s m m m
kV
m
= + × =
= − × + + × = − +
= −
F E u B
E u B = - 3a a a a a a
E a a
( );m q= = ×F u Ba
( );q
m
= ×u Ba
( )( )( )
( )
19 9
2
3
231
1.6 10 1000 10
175 10
9.11 10
m Wbx CquB s m ma x sm x kg
− −
−
−
= = =
3- !27
P3.37: Suppose you have a surface current K = 20. ax A/m along the z = 0 plane. About a 
meter or so above this plane, a 5.0 nC charged particle is moving along with velocity v = 
-10.ax m/sec. Determine the force vector on this particle. 
! 
! 
! 
P3.38: A meter or so above the surface current of the previous problem there is an infinite 
length line conducting 1.0 A of current in the –ax direction. Determine the force per unit 
length acting on this line of current. 
! 
! 
P3.39: Recall that the gravitational force on a mass m is ! where, at the earth’s 
surface, g = 9.8 m/s2 (-az). A line of 2.0 A current with 100. g mass per meter length is 
horizontal with the earth’s surface and is directed from west to east. What magnitude and 
direction of uniform magnetic flux density would be required to levitate this line? 
! 
! 
By inspection, B = Bo(-ax) 
1 1 20 10
2 2N x z y
A
m
= × = × = −H K a a a a
( )7 72 210 4 10 40 10o y yWb Wbx xm mµ π π
− −= = − = −B H a a
( )9 7 25 10 10 40 10 0.63x y z
m Wbq x C x pN
s m
π− −⎛ ⎞= × = − ×− =⎜ ⎟
⎝ ⎠
F a a au b
0
7
12 2 2 1 2 240 10x y
L
WbI d I dx x
m
π −= × = ×−∫ ∫ a aF L B
( )( )712 2
712
40 10 ;
40 10 12.6
z
z z
I L x
Nx
L m
π
µ
π
−
−
= − −
= =
a
a a
F
F
,m=F g
( ) ( )
2
2; for 1 m 100 9.8 0.98 1000g g z z
m Ns kgm g N
s kg m g
⎛ ⎞⎛ ⎞= = − = −⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
a aF g F
1
0
0.98
m
g zId N= × = − = +∫ aF L B F
3- !28
! 
The unit conversion to arrive at 
 Newtons is as follows: 
! 
So we have Bo = 0.490 Wb/m2, and 
B = 0.490 Wb/m2 (-ax) (directed north) 
P3.40: Suppose you have a pair of parallel lines each with a mass per unit length of 0.10 
kg/m. One line