University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
1 Part a
Jackson Problem 2.15
B. J. Mattson
In general, the Green function must satisfy V2GD : 47r6(fa; For Dirichlet boundary
conditions, the Green function has boundary conditions: G(b0undaries) = 0. So I need
to solve:
V'QGD = 4776(:c $)(5(y y)
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 5 Solutions Kimel 2. 2.23 The system is pictured in the following figure:
a) As suggested in the text and in class, we will superpose solutions of the form (2.56) for the two sides with V x, y, z V. 1) First consider the side V x, y, a z :
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 5 Solutions Kimel 1. 2.13 The system is pictured in the following figure:
a) Notice from the figure, , , ; thus from Eq. (2.71) in the text,
, a 0
n1
a n n cos n a 0 V1 V2 2
/2 b,
3/2
2a 0 V 1 V 2
Using
/2 cos m cos nd nm
3/2
Applyi
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 4 Solutions Kimel 4. 2.11 The system is pictured in the following figure:
a) The potential for a line charge is (see problem 2.3) ln rr r 2 0 Thus for this system r 2 ln
0
r R r
2
ln
0
r R r
To determine and R , we need two conditions:
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 4 Solutions Kimel 3. 2.10 As done in class we simulate the electric field E 0 by two charges at
3 0 E 0 cos a) This charge distribution simulates the given system for cos 0. We have treated this probem in class. The potential is given by
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 4 Solutions Kimel 2. 2.9 The system is pictured below
a) We have treated this probem in class. We found the charge density induced was 3 0 E 0 cos We also note the radial force/unit area outward from the surface is 2 /2 0 . Thus the force
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 4 Solutions Kimel 1. 2.8 The system is pictured below
a) Using the known potential for a line charge, the two line charges above give the potential 1 ln r V, a constant. Let us define V 4 0 V r r 2 0 Then the above equation can be written
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 3 Solutions Kimel 5. 2.7 The system is described by
a) The Greens function, which vanishes on the surface is obviously G , xx where x y * z k, I x y * z k x x b) There is no free charge distribution, so the potential everywhere is determin
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 3 Solutions Kimel 4. 2.6 We are considering two conducting spheres of radii r a and r b respectively. The charges on the spheres are Q a and Q b . a) The process is that you start with q a 1 and q b 1 at the centers of the spheres, and sph
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 3 Solutions Kimel 3. 2.5 a) W
r Fdy
qiqj
q2a 4 0
r
dy y3 1
a2 y2 2
q2a 8 0 r 2 a 2
Let us compare this to disassemble the charges W 1 8
0 x x i j 
i j
1 4
0
aq 2 r
1 r 1
a2 r2
q2a 4 0 r 2 a 2
W
The reason for this difference is that
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 3 Solutions Kimel 2. 2.3 The system is described by
a) Given the potenial for a line charge in the problem, we write down the solution from the figure, R2 R2 R2 R2 ln ln ln 2 2 2 0 o o1 o2 o3 2 x x x x x x x x 2 2 2 x x x x x x Looking at
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 2 Solutions Kimel 5. 2.2 The system is described by
a) Using the method of images x with y b)
a2 y 0
1 4
0
q q y y x  x 
, and q q xa 0 x xa n
a y 0
1 4
0
x
x y
2
2
q 2xy cos
1/2
y2 a2
x y
2
2
q 2xy cos
1/2
q 1 4 y 2 a 2 Note q i
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 2 Solutions Kimel 4. 2.1 We will work in cylidrical coordinate, (, z, , with the charge q located at the point dz, and the conducting plane is in the z 0 plane. d Then we know from class the potential is given by x q 4 1 4 q
0
xd 1
q xd 1
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
Jackson Problem 1.1
B. J. Mattson
1 Part a
Im assuming that we are dealing with a conductor at equilibrium. Then when we place
excess charge on it. Consider a Gaussian surface with an enclosed charge of q. In this
case:
. 1
/E ' x da : _Qenclosed
60
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 2 Solutions Kimel 3. 1.9 I will be using the principle of virtual work. In the figure below, Fl is the work done by an external force. If F is along l (ie. is positive), then the force between the plates is attractive. This work goes into
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 2 Solutions Kimel 2. 1.8 We will be using Gauss's law E da a) 1) Parallel plate capacitor
Q enc
0
From Gauss's law E W
0
0
Q A 0 0
12 d Q A 0 2
Q Ad
A 0 12 d A 0 12 d 2
2 2 2) Spherical capacitor
E
2 3
d x
0E
2
Ad
2
Q2 1 d 1 2 0 A 2 0
A
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 1 Solutions Kimel 4. 1.5 1 r 2 4 0 r 2 2 r , 1 r2 0 r r2 r r q 1 r e 2 r2 r 2e 4 0 r 2 r r r qe
r
Using
2 1 r
4 r
2
q 4
0
1 r2
r
r
re
r
r
e
r 2
r
r 4 r
1 r
2 r2e 2 2e r
r
r r
q 4
0
e r2
2 e r
r e 2 r
3 e 2
1
0
q r
q 3 e 8
r
q 3 r e 8 Tha
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 1 Solutions Kimel 3. 1.4 Gauss's Law: E da Q enclosed
0 Q a) Conducting sphere: all of the charge is on the surface 4a 2 Q E4r 2 0, r a E4r 2 0 , r a
E 0, r a
E
Q r, r a 4 0 r 2
b) Uniform charge density: E4r 2 E4r 2
4 3
Q a 3
, r a, 0,
University of Arkansas Community College at Morrilton
Electrodynamis
PHYSIS 506

Spring 2008
PHY 5346 HW Set 1 Solutions Kimel 2 1.3 In general we take the charge density to be of the form f , where f is determined by r r 3 physical constraints, such as d x Q. a) variables: r, , . d 3 x dd cos r 2 dr fr R fr r R fRr R r
d 3 x f R r 2 drdd r
r
R