Gauss’ Law
EE 311 - Lecture 20
• The first fundamental law of electrostatics in integral form is
Gauss’ Law:
Z
I
ρV
Qtot
dV
E · dS =
=
(1)
ǫ
ǫ0
0
V
S
• This equation tells us that the “electric field flux” out of any
closed surface is proportional to the total charge enclosed by
the surface
1. Gauss’ law
2. Divergence
3. Using Gauss’ law to calculate fields
4. Examples
• The total charge can be expressed as a volume integral of the
volume charge density
• The surface considered here doesn’t necessarily physically exist;
any closed surface that we can imagine will do
• Essentially this is telling us that charges are “sources” or
“sinks” for fields; if there are no charges in a volume, there
cannot be fields created or terminated in that volume
1
2
Divergence
Using Gauss’ Law
• Although Gauss’ Law is a fundamental law of electrostatics, it
is only of limited use for finding fields produced by sources
• This is because in integral form it describes an integration of
fields; after a function is integrated, a lot of information is lost!
• Example: Suppose the electric field flux out of a surface is zero
- is the field on that surface then necessarily zero?
• It is only possible to use the integral form of Gauss’ Law for
finding fields produced by particular sources if the sources have
a very high degree of symmetry
• Consider Gauss’ Law where the surface used is a tiny cube,
aligned with the xyz coordinates in a Cartesian system
• There are 6 faces of the cube to keep track of: three sets of 2
parallel planes. The dimensions are small enough that we can
neglect field variations in the cube in most cases.
• Notice for each pair of parallel planes, the normals are pointing
in opposite directions; also, the two planes are displaced by a
distance 2∆
• Take the faces centered at x = +∆ and x = −∆ for example,
each with area (2∆)2 :
Face 1 flux : x̂ · E(∆, y, z)(2∆)2
• We can also try to find a “differential form” of Gauss’ Law: one
that applies at a point instead of over some integrated region of
space
Face 2 flux : −x̂ · E(−∆, y, z)(2∆)2
dsz = z^ dx dy
• To do this, simply shrink the surface S to be infinitesimally
small - leads to the definition of “divergence”
dy
dx
^ dx dz
dsy = y
dz
d = dx dy dz
^ dy dz
dsx = x
3
dy
4
Divergence Continued
• The total for these two faces is then
(2∆)2 (Ex (∆, y, z) − Ex (−∆, y, z))
Gauss’ Law : differential form
• If we shrink the surface considered to zero size in Gauss’ Law,
then we obtain:
∇ · E = ρV /ǫ0
which starts to look like a derivative
• We get a similar result for the other two pairs of faces, except
they involve the y and z components of the fields
• Adding up these three, we find the flux out of the tiny cube is
Ψ
=
(2∆)2 (Ex (∆, y, z) − Ex (−∆, y, z) + Ey (x, ∆, z) − Ey (x, −∆, z)
+Ez (x, y, ∆) − Ez (x, y, −∆))
• The divergence is defined as this result divided by the volume
of the cube as the cube shrinks to zero size:
∂Ey
∂Ez
∂Ex
+
+
divE = Ψ/(2∆)3 =
∂x
∂y
∂z
• This is often notated as ∇ · E; the divergence is a “derivative”
that operates on a vector function to produce a scalar
5
• The ρV appears on the right hand side because we divide by
the volume in defining the divergence
• The above is the “differential form” of Gauss’ Law; it applies at
every point in space
• The divergence essentially measures how much a field is flowing
“into” or “out of” a point in space; it will be zero unless
charges are located at this point
• The “divergence theorem” allows us to connect the differential
and integral forms directly; this is a general mathematical
relationship:
I
Z
E · dS =
S
dV ∇ · E
(2)
V
6
Using Gauss’ law to find fields
• We can use the integral form of Gauss’ law to find the fields
produced by a specific set of charges if the charges possess a
high degree of symmetry
• Based on the symmetry, we can find “Gaussian surfaces” over
which the field has only a normal component, and on which
that component is constant
• If this is true, the total flux is just the field component times
the surface area; setting this equal to the charge enclosed
determines the amplitude of the field component
• Types of sources for which this will work:
– infinite line charge (cylindrical coords)
– infinite cylinder surface/volume charges (cylindrical coords)
– spherical surface/volume charges (spherical coords)
– surface charges on an infinite plane (Cartesian coords)
• In all these cases, the line, volume, or surface charge densities
have to be constant
7
Example
Use Gauss’ Law to find the electric field produced by an infinitely
long line charge on the z axis; the line charge density ρl is constant
on the line.
• Line charge: use cylindrical coordinates and a cylindrical
Gaussian surface
• Think about symmetry with respect to each cylindrical
coordinate:
– Because the source is infinitely long, there can be no z
dependence in the resulting fields.
– Because the source is rotationally symmetric, there can be
no φ dependence in the resulting fields
– Fields will depend on r because this involves distance from
the line
8
Example
• Now think about symmetry with respect to field direction:
– Imagine the line charge as made up of many small positive
charges. At a given observation point, we add up all the
radially outward pointing contributions from each charge
– Because the line charge is infinite, there are always an equal
number of charges above and below a given observation
point. Thus, all z components of fields cancel out
– Because the source is rotationally symmetric, there can be
no φ components of the field
– A r component makes sense, and will not cancel out.
• Resulting form for field is then E(r, φ, z) = r̂Er (r)
• This will be constant over the surface area of a cylinder with
radius r, and is normal to the cylinder body. Total flux is then
2πrLEr (r) for a cylinder of radius r and length L
Finish Example
• According to Gauss’ law in integral form, the total flux out of
this cylinder equals to the charge enclosed divided by ǫ0
• The cylinder of length L encloses a total charge of ρl L, since
the line charge holds ρl coulombs per meter.
• Setting these two equal, we find
Er (r) =
ρl
2πǫ0 r
• For this highly symmetric problem, Gauss’ law made it simple
to find the field produced by a charge distribution
• Notice the field of the line charge falls off as one over the
distance, instead of one over distance squared for a point
charge.
• The infinite size of the line charge makes this slower fall-off
possible
9
10
Electric potential
• Because electric fields can exert forces on charged objects, they
have the ability to do work; i.e. to transfer energy
EE 311 - Lecture 21
• Just like moving an object to a higher point in a gravitational
field increases its “potential energy”, moving a charged object
in an electric field modifies its “electric potential”
• Consider a test charge qtest that is moved through an electric
field (presumably created by other charges somewhere)
1. Electric potential
2. Path independence: KVL
• The work obtained in doing this is
Z
Z
F · dl = qtest
E · dl
W =
3. Examples
C
C
where C is the path that qtest is moved through
• We obtain work here because the field is exerting the force.
R
The work required by us is −qtest C E · dl
11
12
Electric potential cont’d
• An electric potential difference between points P1 and P2 is
defined as the work per unit charge required to move a test
charge from P1 to P2 in the presence of the electric field E that
exists between P1 and P2
• Mathematically this is:
V (P2 ) − V (P1 ) = −
Z
P2
E · dl
P1
with V representing the electric potential; note qtest vanished
due to “work per unit charge”
• The units of V are Newton-meters/Coulomb or Joules per
coulomb. This unit is defined to be the Volt
• Notice V is a scalar that results from the underlying vector
electric field; in many cases it can be much easier to work with
V than E
• This electric potential is exactly the voltage of circuit theory,
but now we have a connection to the underlying fields
13
Path independence
• Notice we originally talked about a path C connecting points
P1 and P2 when defining the potential, but finally we just used
integration starting and stopping points at P1 and P2
• It turns out the particular path chosen doesn’t matter because
line integrals of electric fields are path independent
• Recall our second fundamental law of electrostatics:
I
E · dl = 0
(3)
C
• This looks like the computation of an electric potential, but the
path is a closed loop (i.e. starts and stops at the same point)
• Consider two alternate paths from P1 to P2 . By reversing the
direction of one of these paths, we can create a closed loop
path by combining the two. Because the integral over the
closed loop is zero, we find the integral over either path from
P1 to P2 must be the same
• Thus the potential difference between two points is unique
14
Absolute potential
KVL
• The path-independent property in defining the electric scalar
potential also results in “Kirchhoff’s voltage law” (KVL)
• If we break a closed loop path into pieces, the line integral of E
over each piece will produce a potential difference.
• However all of these will have to add to zero; thus the sum of
the voltage rises equals the sum of the voltage drops
• KVL is one of the fundamental circuit equations, but again we
now have a connection to the fields that underly this circuit
statement
• Remember we’re considering static fields here; this means there
are no time variations in the sources producing the fields. KVL
will need to be modified when we have time varying sources in
EE 312
15
• Notice so far we’ve only defined a potential difference between
two points, but not the potential at a particular point
• We can get an “absolute potential” by choosing the point P1 to
be a reference point (i.e. “ground”), and defining the potential
at all other points in terms of this point.
• In many cases, the point at distance infinity is taken to be zero
absolute potential.
• The absolute potential difference at another point then
becomes:
Z P2
E · dl
V (P2 ) = −
∞
This is the work per unit charge required to move a test charge
from ∞ to P2
• This is not always the reference point chosen though so it is
important to keep track of the “zero volts” reference point in
all problems
16
Absolute potential of a point charge
• Absolute potential of a point charge at the origin is:
Z R
q
R̂
V (R) = −
· dl
4πǫ0 R2
∞
• Take our path as radially outward from the point charge to
infinity; the integral can be performed to get
Z ∞
q
(R̂ · R̂dR)
V (R) =
4πǫ0 R2
R
q
V (R) =
4πǫ0 R
• We can evaluate the amount of work needed to move a test
charge qtest in this field simply by taking qtest (V (R2 ) − V (R1 ))
′
• For a point charge at position R , the result is
q
¯
¯
V (R) =
′¯
¯
4πǫ0 ¯R − R ¯
• The law of superposition still applies; simply add these for
multiple charges or integrate for charge distributions
Finding fields from V
• We’ve defined the electric potential in terms of a line integral of
the electric field
• Notice if we take a very small path, the differential voltage
obtained is:
dV = −E · dl
• However, for a scalar function we can also use the gradient to
find
dV = ∇V · dl
• Comparing these two,
E = −∇V
i.e. the electric field is the gradient of the potential
• Using our gradient properties, this tells us that the electric field
points in the direction of maximum decrease of the potential
• Also, the electric field is everywhere normal to surfaces of
constant potential. These surfaces are called “equipotentials”
18
17
An example
• Consider a capacitor connected to a battery; the battery
supplies positive charges to the top plate of the cap and
negative charges to the bottom plate
Another example
• Consider equipotentials in the x-y plane as shown. What is the
electric field?
• This results in a downward pointing electric field
• The potential difference from the point P2 to P1 is
Z P2
E · dl
V (P2 ) − V (P1 ) = −
y
V= 2 Volts
P1
• Notice E · dl will be negative here; this means the potential
difference is positive, and P2 is at a higher potential than P1
• Electric potentials generally increase as we move closer to
positive charges; they decrease as we move closer to negative
charges
P
+ + + +
y =2m
V= 1 Volt
y= 0m
V= 0 Volts
2
y=−2 m
+
V−
dl
E
− − − −
P
V=−1 Volts
1
19
20
x
Curl
EE 311 - Lecture 22
• We have seen that the computation of V from E is path
independent, due to our second law of electrostatics:
I
E · dl = 0
(4)
C
• Just as we derived the divergence operator by thinking about
Gauss’ law evaluated for a small box, we can derive the “curl”
operator by thinking about the above equation on a small path.
H
• Consider C E · dl for a small square path in the xy plane:
1. Curl
2. Types of media
3. Conductors
y
4. Examples
2∆
x
2∆
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22
Curl continued
Curl properties
• If we add up the four contributions from the sides of this path
we will get:
2∆(Ey (∆, y) − Ey (−∆, y) + Ex (x, ∆) − Ex (x, ∆))
• If we divide this by the area bounded by the path, it becomes
∂Ex
∂Ey
−
∂x
∂y
• We could also choose similar paths in the yz or xz planes;
separate these as different components of a vector. To do so,
multiply each result by a unit vector normal to the area
• These are open surfaces: choose direction of normal based on
right hand rule with dl
• Sum of all three components is the curl; measures the tendency
of a vector function for rotation about a point in space
• The curl of a vector field produces a vector!
23
• Performing the mathematical analysis shows that the curl
operator in Cartesian coordinates is:
¯
¯
¯ x̂
ŷ
ẑ ¯¯
¯
¯
¯
∂
∂
∂ ¯
curl A = ¯¯ ∂x
∂y
∂z ¯
¯
¯
¯ Ax Ay Az ¯
• The curl of A is also written as ∇ × A
• Electrostatic fields have ∇ × E = 0 due to path independence;
differential form of second fundamental law
• As the divergence theorem connects the divergence operator to
a flux integral, Stokes’ Theorem connects the curl operator to a
line integral:
I
Z
¢
¡
A · dl
∇ × A · dS =
C
S
Here the contour C bounds the area S; dl and dS satsify a
r-h-rule relationship
24
Perfect conductors
Types of media
• So far we’ve been talking about electric fields in free space, i.e.
nothing around but vacuum. Things change when fields exist in
the presence of other objects or media
• We can classify material media into two basic types:
– Conductors: contain some loosely held electrons that are
free to move (metals)
– Insulators: electrons are tightly held, cannot move easily
• The conductivity of a material σ in S/m is a measure of this
property: high conductivity = good conductor, low
conductivity = good insulator
• We will focus on either “perfect electric conductors” (PEC)
with σ = ∞ or “perfect insulators” with σ = 0 in many cases
• A perfect conductor with σ = ∞ is capable of relocating an
arbitrary amount of charge from within the conductor
• This is not really possible, but a reasonable approximation for
most metals
• Assume we have a PEC material, and put a charge inside; the
inside charge produces an electric field, so that the “free”
charges in the PEC start moving around
• These “free” charges create their own electric fields as they
move around; they keep moving around until all the electric
fields cancel out
• The end result is that the field is zero inside the PEC; a surface
charge density can be left on the surface
PEC Material
+
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Point charge inside PEC
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26
Perfect conductor “boundary conditions”
Potentials and PEC materials
• We can now start to talk about problems involving both free
space and PEC regions
• We know that there is no volume charge density inside a PEC,
that there is no E field inside a PEC, and that there can be
surface charge densities ρS on the surface of a PEC
• The presence of the surface charges means that there cannot be
any tangential E field on the surface of a PEC; if there were,
the surface charges would move around to cancel it out
• However, there can be a normal E field on the surface of a
PEC, because the surface charges can’t move off the PEC
• Using Gauss’ law with a small box on the PEC gives us a
relationship between the normal component of E on the PEC
surface and the “induced” surface charge density ρS
• The final “boundary conditions” on a PEC surface are:
Etan = 0
Enorm = ρS /ǫ0
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Volume charge density=0 everywhere
inside PEC; surface charge density can exist
on surface; fields inside are zero
• The fact that Etan = 0 on the surface of a PEC means that the
surface of a PEC is an equipotential surface; i.e. a conductor is
everywhere at the same potential in electrostatics
• The above statement disagrees with the v(z, t) we studied in
transmission line theory; however those were dynamic fields,
not static as we are covering here
R
• PEC’s are equipotentials because evaluating E · dl along the
surface of the PEC will always produce zero
• We already know that E is normal to equipotential surfaces;
our PEC boundary conditions confirm this fact
• Also the fact that E vanishes in a PEC means that a solid PEC
material has the same potential even everywhere inside the
PEC
• Placing a PEC in a static field causes surface charges to be
induced on the conductor so that Etan vanishes
28
An example
Solution
• Gauss’ law can still be used to find E fields produced with
charges in the presence of PEC materials; just remember
properties of PEC’s!
• The high degree of spherical symmetry allows Gauss’ Law to be
used to find fields here. Fields are of the form E = R̂ER (R)
Problem: A point charge is located inside a (non-charged) hollowed
PEC sphere. Sphere inner radius is Ri , outer radius is R0 . Find E
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Point charge q
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• For R < Ri , only the point charge is enclosed and the resulting
field is ER (R) = 4πǫq0 R2
• For Ri < R < Ro , we are inside the PEC material, where the
field must vanish: ER (R) = 0. This is consistent with Gauss’
law because a surface charge density of total charge −q is
induced on the inner PEC surface to cancel the total charge
enclose; ρS on inner surface is −q/(4πRi2 )
• For R > Ro what happens? We said the PEC material is
uncharged; this means the induced charge on the inner surface
had to come from somewhere.
• Thus a surface charge density of total charge +q is left on the
outer sphere surface; field is then ER (R) = 4πǫq0 R2 for R > Ro
29
30
Another example: Method of Images
Method of Images
• The method of images is a trick that is useful for
understanding fields produced by charges in the presence of an
infinite PEC plane (i.e “ground plane”)
• Consider a point charge above an infinite PEC plane; surface
charge density will be induced on plane to cancel tangential
electric field
• The total field above the PEC is then determined both by the
point charge and the surface charge distribution on the PEC
surface; complicated!
Fields ignoring ground plane
Effect of ground plane charges
q
Infinite PEC "ground plane"
q
Method of Images
q
− − − − − − − −
Infinite PEC "ground plane"
−q
31
• The method of images says that the total fields above the PEC
are the same as those in an “image” problem
• In the image problem, we get rid of the PEC plane so that only
free space is left
• In the image problem, we consider the original source above the
PEC plane, plus an “image” of the source. The image is the
“reflection”of the original source through the PEC plane
• The image source has opposite charge to the original, and is
the same normal distance away from the PEC plane
• It is much easier to find the fields in the “image” problem: just
a sum of two point charges. The case with the PEC is much
harder due to the distribution of charges on the PEC surface
• The “image” sources in this particular case are known as an
“electric dipole”: equal but opposite charges separated by a
distance
32
Dielectrics
• A perfect dielectric material is a perfect insulator; i.e. σ = 0
• There are no “free” charges in a perfect dielectric, so you might
think that these materials produce no electric effects
• However, the “bound” charges in the atoms of a dielectric still
cause electric effects. These electrons are not free to move
among atoms, but an electric field still exerts a force on them
EE 311 - Lecture 23
• The effect of this is to distort the orbit on the electrons; result
is effectively an induced bound-charge dipole, which does
produce electric effects
1. Dielectrics
2. Dielectric boundary conditions
Distorted electron orbit
3. Examples
Negative electron orbits
Model of distorted orbit effects
−
−
Positive nucleus
+
E
Positive nucleus
+
E
E
+
dipole
−
33
34
Electric Flux Density
Other Dielectric Properties
• Notice that the fields of the induced dipoles oppose those of the
applied field; total field inside is smaller
• We have to account for the effect of these induced dipoles if we
want to describe fields in dielectrics. To handle this, we
introduce a new field, known as D, the “electric flux density”
• The effect of the induced dipoles is then included by modifying
Gauss’ Law to:
I
Z
D · dS =
dV ρV = Qtot
(5)
S
V
It is clear from this equation that the units of D are C/m2
• This is pretty similar to the old Gauss’ law, but the connection
between D and E is now important; this connection models
how the dipoles in a medium respond to an applied electric field
• It turns out in most cases that the induced dipoles are in the
same direction as the applied field, with an amplitude directly
proportional to the applied field
35
• The end result of this is that D and E are related through a
constant ǫ, the “permittivity” of the medium:
D = ǫE
In free space, D = ǫ0 E and the new and old forms of Gauss’
law are identical
• We used the permittivities of media in our study of
transmission lines, but hopefully now the meaning is clearer
• The induced dipoles don’t remain “bound” charges forever if
we keep increasing the applied field; eventually these charges
are “ripped” from the atoms to become free charges
• This effect is called “dielectric breakdown”; produces sparks in
air (at 3 MV/m)
• The “dielectric strength” of a material is the maximum E
amplitude the material can tolerate before breakdown occurs
• Limits the power handling capabilities of devices; table in book
36
Dielectric boundary conditions
• We talked about how fields behave at an interface between free
space and a PEC; now consider an interface of dielectrics
• Consider fields E 1 , D1 above an interface and E 2 , D2 below;
apply laws of electrostatics on small regions at the interface
R
• The law C E · dl = 0 provides information about tangential
electric fields E1t and E2t on the interface. We can show that
E1t = E2t , i.e. tangential electric fields are continuous
H
• The law S D · dS = Q provides information about normal flux
densities D1n and D2n . We can show that D1n − D2n = ρS .
Typically ρS will exist only on a PEC surface
D1n
^2
n
Medium 1
ε1
E1
E1n
} ∆h2
} ∆h2
E1t
E2n
E2t
E2
a
∆s
c
d
b
∆h
2
∆h
2
∆l
Medium 2
ε2
{
{
ρs
^
D2n n1
Boundary conditions
• Note that normal D will be discontinuous if there is a surface
charge density present
• Knowledge of normal D can tangential E allows us to
determine the total fields D and E both above and below the
interface
• Remember that these boundary conditions apply only at the
interface! Fields away from the interface in the two regions are
not necessarily related in any way
• A few examples: Are BC’s satisfied? (region 1 is z > 0, region
2 is z < 0):
– Region 1 free space, Region 2 PEC: E 1 = x̂2 V/m, E 2 = 0
– Region 1 free space, Region 2 ǫ = 2ǫ0 : E 1 = ŷ3 V/m,
E 2 = ŷ(z + 3)
– Region 1 ǫ = 3ǫ0 , Region 2 ǫ = 2ǫ0 : E 1 = (x̂2 + ŷ3 + ẑ)
V/m, E 2 = (x̂2 + ŷ3 + 1.5ẑ) V/m
37
38
Example: Gauss’ Law with dielectrics
Solution
• It is easy to use the new Gauss’ Law with dielectrics, just
replace previous E with D and forget ǫ0 factors
Problem: A point charge q is located at the center of a hollowed
dielectric sphere (inner radius Ri , outer radius Ro , permittivity ǫ).
Find the electric field everywhere.
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Permittivity ε
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Permittivity ε
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Point charge q
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39
• The high degree of spherical symmetry allows Gauss’ Law to be
used to find fields here. Fields are of the form D = R̂DR (R)
• For R < Ri , only the point charge is enclosed and the resulting
q
field is DR (R) = 4πR
2
• For Ri < R < Ro , we are inside the dielectric material.
However using D automatically takes care of this for us. There
are no more charges enclosed so DR is the same as for R < Ri
• For R > Ro , no new charges again, so DR stays the same
• However, the electric field E is varying in the three regions due
to the changing permittivities.
• The final E field satisfies the bc’s, and has E = R̂ER (R):
q
for R < Ri
ER =
4πǫ0 R2
q
for Ri < R < Ro
ER =
4πǫR2
q
for Ro < R
ER =
4πǫ0 R2
40
Capacitance
• Consider two PEC objects of arbitrary shape. Place a total
charge of +Q on one of them and −Q on the other; this charge
is distributed over the surface of the conductors in a
complicated way
EE 311 - Lecture 24
• These charges result in an electric field between the two
conductors (from + conductor to − conductor). Relationship
between charges and E is complicated
• If we know E, we can find the potential difference V between
the two conductors by line integrating the electric field. The +
conductor will be at high potential.
1. Capacitance
2. Calculating capacitance
3. Examples
Surface S
+
+
+
+
Q
+
Conductor 1
+
+ + +
V
+
-
+
+
ρs
E
- - -Q
Conductor 2 - - -
41
Capacitors
• This structure is called a capacitor. The capacitance of the
capacitor is
C = |Q|/|V |
and has units of Coulombs/V olt or Farads (F)
• Because the voltage V that results between the two conductors
is directly proportional to the applied charge Q, the
capacitance is independent of the amount of charge applied
• Capacitance does depend on the geometry of the conductors
and the materials surrounding them
• We can find the capacitance of a capacitor if we know the
electric field everywhere. Normal D boundary condition on
PEC surfaces allows ρS , then total charge Q to be determined.
Line integral of E determine V . Ratio is capacitance
43
42
Calculating capacitance
The following steps are useful for computing the capacitance of
highly symmetric caps:
1. Choose an appropriate coordinate system based on symmetry
properties of conductors
2. Assume total charges +Q and −Q on the two conductors; for
highly symmetric caps, these also determine ρS
3. Find E from Q using Gauss’ law
4. Find the potential difference between the conductors using
R1
V = − 2 E · dl, where points 1 and 2 are on the + and −
conductors respectively. Choose dl to be as simple as possible
5. The capacitance is then C = Q/V
44
Example: Parallel plate capacitor
Example: Follow steps
• A parallel plate capacitor (as in our parallel plate T-line) is
made from two PEC plates of area A separated by distance d
• Cartesian coordinates are best for this structure, due to planar
symmetry
• Because the plates are of finite size, there are not many
symmetry properties of this device, and our procedure doesn’t
work. However for “large” plates, we can approximate that the
E field for finite plate area is the same as if the plates were
infinite.
• Assume +Q on top plate, −Q on bottom. ρS = ±Q/A on
top/bottom plates
• This is called the “no fringing” approximation, because
“fringing” fields at the edge of the finite plates are neglected
Conducting plate
z
Area A
ρs
z=d
+
V
-
z=0
+Q
+
+ + + + + + + + + +
ds
E E E
- - - - - - - - - -
-
-ρs
+
+
+
E E -
+
-
+
-
+
Fringing
field lines
Dielectric ε
Conducting plate
• Symmetry properties of fields: no x or y variations, only a ẑ
component for infinite plates. Thus D = ẑDz (z). Choose a
rectangular box for the Gaussian surface, top of box inside
upper PEC plate, bottom of box above bottom plate.
• The box encloses charge Q on the top plate, flux out of bottom
is −Dz (z)A. Gauss’ Law gives Dz (z) = −Q/A = −ρS . Then
E = −ẑρS /ǫ
• Note this field has constant amplitude and does not decay with
distance from the charges; due to assumed infinite size of plane.
Also note that PEC boundary condition is satisfied
R z=d
• Line integral of electric field is V = − z=0 E · ẑdz = Qd/(ǫA)
• Capacitance is then C = Q/V = Q/(Qd)ǫA = ǫA/d as expected
45
46
Example: Coaxial capacitor
Example: Follow steps
• A coaxial capacitor (as in our coaxial T-line) is made from two
coaxial cylinders
Q
on
• Assume −Q on inner cylinder, +Q on outer. ρS = − 2πal
Q
inner cylinder, 2πbl on outer
• Because the line is of finite length, there are not many
symmetry properties of this device. However for “large”
lengths, we can approximate that the E field for finite length is
the same as if the line were infinitely long.
• Symmetry properties of fields: no φ or z variations, only a r̂
component for infinite length. Thus D = r̂Dr (r). Choose a
cylinder for the Gaussian surface, with the cylinder radius
between a and b.
• This is called the “no fringing” approximation, because
“fringing” fields at the ends of the line are neglected
• The box encloses charge −Q on the inner cylinder, flux out of
cylinder is 2πrDr (r)l. Gauss’ Law gives Dr (r) = −Q/(2πrl).
Then E = −r̂Q/(2πrlǫ)
• Cylindrical coordinates are best for this structure
l
+
V -
b
a
+ + + + + + + + + + + + +
E
E
E
- - - - - - - - - - - - - - - - - - - - - - - - ε
E
E
E
+ + + + + + + + + + + + +
ρl
-ρl
Inner conductor
Dielectric material
Outer conductor
47
• Line integral of electric field is
Z r=b
Z r=b
Q
Q
ln(b/a)
1/r =
E · r̂dr =
V =−
2πlǫ
2πlǫ
r=a
r=a
• Capacitance is then
C = Q/V = Q/(Q ln(b/a))2πlǫ = 2πǫl/ ln(b/a)
• We previously used this result for our coaxial transmission
lines!
48
Capacitor properties
• Because a capacitor can hold separated positive and negative
charges, it can store electric potential energy; we could get a
lot of work out of these charges by letting them move toward
each other under their own force
• We know from circuit theory the energy stored in a capacitor is
1
2
2 CV ; this is an electrostatic potential energy
• Notice both cap types had capacitance directly proportional to
ǫ. Larger ǫ values result in a smaller voltage for a fixed amount
of charge. This is due to the opposition effect of induced dipole
moments in a dielectric
• Geometry of conductors matters too: larger individual
conductor areas increase capacitance; C increases as conductors
brought closer together
• Dielectric breakdown limits the maximum voltage that can be
applied. Occurs first on inner conductor in coaxial cap
49
Other capacitor properties
• Multiple capacitors connected in parallel share the same
voltage but combine charges; effective capacitance is sum of
individual capacitances
• Multiple capacitors in series produce a total voltage that is the
sum of the voltage across each. Effective capacitance is
“parallel” combination of individual capacitances
• We have been able to compute capacitances of a few highly
symmetric caps here through the use of Gauss’ law
• However procedure is much more difficult for non-symmetric
capacitors; need computers to handle this case
• Most familiar capacitors: coaxial, “chip” capacitors on PCB’s
( parallel plate)
• Any object capable of holding electric charge has a
capacitance; second conductor may be the Earth surface
50
Imperfect conductors
EE 311 - Lecture 25
1. Imperfect conductors
• We’ve now considered both perfect conductors (σ = ∞) and
perfect insulators σ = 0; a capacitor combined these two
materials. No currents flow in either of these material types
• However real materials lie somewhere between these two. In
this case, it is possible for currents to flow inside the medium
• Current is a measure of the number of coulombs per unit time
flowing through a specified area. Total current flowing through
a surface is then a scalar quantity, measured in Amps (C/sec)
2. Resistance
3. Example
• The current ∆I flowing through a small piece of area dS in a
given amount of time can be written as
4. Charge conservation
∆I = J · dS
where J is the “current density”, and is a vector having units
of A/m2
51
52
Current density
Point form of Ohm’s Law
• J is a vector point function of space, and captures currents
flowing in any possible direction
• Example: A volume charge density ρV moving with velocity u
produces a current density J = ρV u
• While J essentially is a current spread over and flowing through
an area, is also possible to have a surface current density J S
spread over and flowing through a line. Units of J S are A/m
• For example, J S can exist on the surface of a PEC if the
surface charges on PEC move along the surface
Volume Current density
Surface current density
J
Js
l
S
• Given our definition of J, the total current flowing out of a
closed surface S is then
I
I=
J · dS
S
• An imperfect conductor contains free charges that can move in
response to an applied field
• In many media, the current density that results is in the same
direction as the applied field, and directly proportional to the
applied field magnitude: J = σE
• In these materials, the parameter σ characterizes the current
density induced by applied fields; the units of σ are
Amps/V olts/m or Siemens/m
• The above equation is the “point form of Ohm’s Law”
• Note we can have currents flowing in a medium even if ρV = 0
everywhere in the medium; the currents must be “steady” so
that equal amounts of charge move in and out together
53
54
Resistance
• In an ohmic medium, an applied electric field results in a
current flowing through the medium; the electric field also
produces a potential difference in the medium
• The ratio of the voltage to the current between two points in
the medium is the resistance: R = V /I
• We can write this ratio mathematically as:
R1
R1
− E · dl
− E · dl
= R2
R= R 2
J · dS
σ S E · dS
S
Simple Resistance Example
• Consider our parallel plate capacitor again, but in this case the
dielectric medium inside has conductivity σ
• For steady currents, there are no charges in this problem that
we didn’t include before when analyzing the capacitor. Thus,
fields are identical to the previous solution: E = −ẑQ/(Aǫ)
• However a current density now results between the plates:
J = σE = −ẑσQ/(Aǫ)
• Recall we found C = Q/V or 1/C = V /Q. Write this as
R1
R1
− 2 E · dl
− 2 E · dl
1/C = R
= R
D · dS
ǫ S E · dS
S
The denominator uses a Gaussian surface enclosing one of the
conductors.
• Combine them: RC = ǫ/σ; i.e. it is easy to find R or C from
knowing the other. We used this in our T-lines study as well
55
Conducting plate
z
Area A
ρs
+
z=d
V
z=0
+Q
+ + + + + + + + + +
ds
E E E
- - - - - - - - - -ρs
+
-
+
+
+
E
E - -
+
-
+
-
+
Fringing
field lines
Dielectric ε
Conducting plate
56
Continue Example
• The voltage between the plates is Qd/(Aǫ); to find the total
R
current flowing between the plates, use I = S J · dS
• Choose S to be the same size as the plates, but located in
between the two plates. Resulting I is σQ/(ǫ)
• The resistance is then R = V /I = Qd/(Aǫ)ǫ/(σQ) = d/(σA)
• The resistance between two conductors typically increases with
the distance between the two conductors, decreases as σ
increases, and decreases as the area through which the currents
flow increases
• Our answer satisfies RC = d/(σA)ǫA/d = ǫ/σ.
• In our T-line studies, this R modeled the current flow between
conductors. We represented this as a conductance G which for
the parallel plate line is σA/d
57
Equation of charge conservation
• One of the fundamental properties of electrostatics is that
charge is conserved; it can be moved around, but can neither
be created or destroyed
• Since moving charges are represented by currents, we should be
able to write an equation relating the total current flowing out
of a volume to the rate of change of the total charge enclosed
• The integral form of the equation of charge conservation is:
I
Z
∂Qenc
∂
dV ρV = −
J · dS = −
∂t V
∂t
• This equation is saying that current flowing out of a closed
surface results in a decrease of the amount of charge in the
volume enclosed by the surface
58
Power dissipation
Steady currents and KCL
• Using the divergence theorem, we can convert the left hand
side to a volume integral. Canceling the volume integrals then
yields a differential form:
∇·J =−
∂ρV
∂t
• For steady currents, the rate of change of the volume charge
density is zero, so the net current flux through any closed
surface is zero
• This is “Kirchhoff’s current law” for circuits: the sum of the
currents out of any closed surface is zero; at circuit junctions
the sum of the incoming currents must equal the sum of the
outgoing currents
• Now that we’ve got both voltages and currents, we can talk
about power dissipation
• Circuit theory tells us that dc power = VI. How is this related
to fields?
• Consider a current density that results from moving charges,
i.e. J = ρV u
• If these charges are moving through an electric field, the
electric force on the charges results in the performance of work
• Power=work per unit time: P = F ·
charge
= qE · u for a point
• Here we’ve got a volume charge density, so the power density in
W atts/m3 is ∆P = E · (ρV u) = E · J
• The total power dissipated in a volume is then
Z
Pdiss =
E·J
V
This is “Joule’s Law” in point form
59
dl
dt
60