Lecture 11
Perfect Conductors, Boundary
Conditions, Method of Images
Sections: 5.4, 5.5
Homework: See homework file
LECTURE 11
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1
Perfect Conductors – 1
• metals such as Cu, Ag, Al are closely approximated by the
concept of a perfect electric conductor (PEC)   ,  0
• charge relaxation is instantaneous
• charge is distributed in an infinitesimally thin layer on the
surface of the conductor
• inside the conductor there is no charge (i.e., excess of electrons)
due to Coulombic repulsion
biased (charged) conductor
(Q  0)
s  0
isolated conductor in external field
(Q  0)
s  0
E
ext
s  0
v  0
E0
v  0
E0
V0
s  0
2
Perfect Conductors – 2
underlying principle: charge arranges on surface so that resultant E
field inside is zero; otherwise charge will keep moving until E = 0
remember: E field is zero in the volume of a perfect conductor
consequence: perfect conductors are equipotential and so are their
surfaces
B
proof: VAB   E
  dL  0  VA  VB
A 0
(A and B belong to the volume or the surface of the conductor)
LECTURE 11
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3
Shielding by Conducting Shell (Faraday Cage)
in a cavity of a conductor E = 0 (if there are no free charges inside)
isolated conducting shell in external field
(Q  0)
v  0
E0
E
ext
biased (charged) conducting shell
(Q  0)
v  0 E  0
E0
E0
V0


proof: take any closed contour Γ passing part through the cavity
and part through the shell, apply the conservative property of E
 E  dL   E  dL   E  dL  0  E  0

shell 0
cavity
Is there any charge on the inner surface of the cavity?
LECTURE 11
Hint: Gauss law
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Faraday Cage: Illustrations
• every computer is enclosed in a metallic box, which protects it
from EMI
• workers in the power industry are protected by Faraday cages
when dealing with high voltage
LECTURE 11
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Boundary Conditions at PEC Surfaces
• tangential components
E  EN a N
PEC is equipotential
E t  0 Dt  0
PEC
interface surface
 0 on PEC surfaces
• normal components
lim
h 0

 D  ds  DN  S  Qs   s  S
S
D  DN a N
 sa N
DN   s , C/m 2 EN   s / 
compare with D of planar charge: D  a N  s / 2
LECTURE 11
D0
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Boundary Conditions at PEC Surface – Examples
Voltage V0 = 100 V is applied to a coaxial cable whose inner wire
has radius a = 1 mm and whose outer shield has radius b = 4 mm.
The insulator is air (ε = ε0). (a) Find E as a function of the distance
ρ from the center of the inner wire. (b) Find the surface charge
densities at the inner wire and at the outer shield.
Homework:
Voltage V0 = 100 V is applied to a pair of concentric spheres. The
smaller sphere has radius a = 1 mm and the larger sphere has radius
b = 4 mm. The insulator between the two is air (ε = ε0). (a) Find E
as a function of the distance r from the center of the spheres. (b)
Find the charge densities at the surfaces of the two spheres.
LECTURE 11
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The Equation of Electrostatics
  D  v    ( E)  v    V   v
• in a uniform medium with sources (Poisson equation)
 2V   v / 
• in a uniform source-free medium (Laplace equation)
 2V  0
uniqueness theorem for the Poisson/Laplace equation
if the following is given
• either V or ∂V/∂n at the boundary of the analyzed region
(boundary conditions), and
• all charge densities in the analyzed region (source conditions)
then the Poisson equation has one and only one solution
LECTURE 11
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Method of Images – 1
consider the two electrostatic problems
Q
Q


ground
h
h
Vb  0
Vb  0
h

Q  Q
• source and BC are the same if analyzed region is upper hemi-sphere
• the field solution in the upper hemi-sphere identical in both cases
• we say that the two problems are equivalent
LECTURE 11
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Method of Images – 2
point charge at two orthogonal ground planes
h2
Q
h1
h1
Vb  0
Q
Vb  0
Q
Q
h2
LECTURE 11
Vb  0
Vb  0
h2
h1
Q
h2
h1
h1
h2
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Method of Images – 3
• point charge at two ground planes at an angle /n, n is integer
• a total of (2n − 1) images required to build the equivalent problem
Q
Example: n = 3
0
Q
Vb
Vb
0
Q
 /3
 /3
Vb  0
Vb  0
• if n is not integer, exact equivalent problem does not exist
LECTURE 11
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You have learned:
that the electric field is zero inside a perfect conductor
the potential is constant everywhere inside and on a perfect
conductor, i.e., the voltage between any two points is zero
a closed conducting shell shields the inside volume from external
electric fields
the tangential E field is zero on the surface of a conductor
the normal D component is equal to the surface charge density at
the conductor’s surface
how the method of images can simplify solutions involving
infinite PEC planes at an angle of π/n where n is integer
LECTURE 11
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