Electromagnetics Wave and Fields
Lecture 5
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Engineering Electromagnetics
Chapter 5
Conductors, Dielectrics & Capacitance
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Chapter 5
Current and Conductors
Current and Current Density
◼ Electric charges in motion constitute a current.
◼ The unit of current is the ampere (A), defined as a rate of
movement of charge passing a given reference point (or
crossing a given reference plane).
dQ
I=
dt
◼ Current is defined as the motion of positive charges, although
conduction in metals takes place through the motion of
electrons.
◼ Current density J is defined, measured in amperes per square
meter (A/m2).
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Chapter 5
Current and Conductors
Current and Current Density
◼ The increment of current ΔI crossing an incremental surface
ΔS normal to the current density is:
I = J N S
◼ If the current density is not perpendicular to the surface,
I = J  S
◼ Through integration, the total current is obtained:
I =  J  dS
S
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Chapter 5
Current and Conductors
Current and Current Density
◼ Current density may be related to the velocity of volume
charge density at a point.
• An element of charge ΔQ = ρvΔSΔL moves
along the x axis
• In the time interval Δt, the element of charge
has moved a distance Δx
• The charge moving through a reference
plane perpendicular to the direction of
motion is ΔQ = ρvΔSΔx
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Dr. M. Tanseer Ali
x
Q
= v S
I =
t
t
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Chapter 5
Current and Conductors
Current and Current Density
◼ The limit of the moving charge with respect to time is:
I = v Svx
◼ In terms of current density, we find:
J x =  v vx
J = v v
◼ This last result shows clearly that charge in motion constitutes
a current. We name it here convection current.
◼ J = ρvv is then called convection current density.
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Chapter 5
Current and Conductors
Continuity of Current
◼ The principle of conservation of charge:
“Charges can be neither created nor destroyed.”
◼ But, equal amounts of positive and negative charge (pair of
charges) may be simultaneously created, obtained by
separation, destroyed, or lost by recombination.
I =  J  dS
S
• The Continuity Equation in Closed Surface
◼ Any outward flow of positive charge must be balanced by a
decrease of positive charge (or perhaps an increase of
negative charge) within the closed surface.
◼ If the charge inside the closed surface is denoted by Qi, then
the rate of decrease is –dQi/dt and the principle of
conservation of charge requires:
dQi
I =  J  dS = −
S
dt
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• The Integral Form of the Continuity Equation
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Chapter 5
Current and Conductors
Continuity of Current
◼ The differential form (or point form) of the continuity equation is
obtained by using the divergence theorem:
 J  dS =  (  J )dv
S
vol
◼ We next represent Qi by the volume integral of ρv:
d
vol (  J)dv = − dt vol v dv
◼ If we keep the surface constant, the derivative becomes a
partial derivative. Writing it within the integral,
v
vol (  J)dv = vol − t dv
v
(  J )v = −
v
t
v
J = −
• The Differential Form (Point Form)
of the Continuity Equation
t
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Chapter 5
Current and Conductors
Continuity of Current
◼ Example
1 −t
The current density is given by J = e a r A m 2 .
r
• Total outward current at time instant t = 1 s and r = 5 m.
I = J r S r = ( 15 e −1a r )(4 52 a r ) = 23.11 A
• Total outward current at time instant t = 1 s and r = 6 m.
I = J r Sr = ( 16 e −1a r )(4 62 a r ) = 27.74 A
• Finding volume charge density:
v
1 −t
1  2 1 −t
−
t
= J =
r r
2
(r
r
e )=
r
2
e
1 −t
1 −t
v = −  2 e dt = 2 e + K (r )
r
r
t → ,  v → 0
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 K (r ) = 0  v =
Jr
1 −t
3

v
=
=rm s
e C m
r
2
v
r
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Chapter 6
Dielectrics and Capacitance
Capacitance
◼ Now let us consider two conductors
embedded in a homogenous dielectric.
◼ Conductor M2 carries a total positive
charge Q, and M2 carries an equal
negative charge –Q.
◼ No other charges present → the total
charge of the system is zero.
• The charge is carried on the surface as
a surface charge density.
• The electric field is normal to the
conductor surface.
• Each conductor is an equipotential
surface
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Chapter 6
Dielectrics and Capacitance
Capacitance
◼ The electric flux is directed from M2
to M1, thus M2 is at the more positive
potential.
◼ Works must be done to carry a
positive charge from M1 to M2.
◼ Let us assign V0 as the potential
difference between M2 and M1.
◼ We may now define the capacitance
of this two-conductor system as the
ratio of the magnitude of the total
charge on either conductor to the
magnitude of the potential difference
between the conductors.
Q
C=
V0
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 E  dS

C=
−  E  dL
S
+
−
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Chapter 6
Dielectrics and Capacitance
Capacitance
◼ The capacitance is independent of the potential
and total charge for their ratio is constant.
◼ If the charge density is increased by a factor,
Gauss's law indicates that the electric flux
density or electric field intensity also increases
by the same factor, as does the potential
difference.
 E  dS

C=
−  E  dL
S
+
−
◼ Capacitance is a function only of the physical dimensions of
the system of conductors and of the permittivity of the
homogenous dielectric.
◼ Capacitance is measured in farads (F), 1 F = 1 C/V.
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Chapter 6
Dielectrics and Capacitance
Capacitance
◼ We will now apply the definition of capacitance to a simple twoconductor system, where the conductors are identical, infinite
parallel planes, and separated a distance d to each other.
S
E=
az

D = S a z
◼ The charge on the lower plane is positive, since D is upward.
DN = Dz = +  S
◼ The charge on the upper plane is negative,
DN = − Dz = −  S
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Dr. M. Tanseer Ali
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Chapter 6
Dielectrics and Capacitance
Capacitance
◼ The potential difference between lower and upper planes is:
S
S
 dz =
d
V0 = − 
E  dL = − 
d 
upper

lower
0
◼ The total charge for an area S of either plane, both with linear
dimensions much greater than their separation d, is:
Q = S S
◼ The capacitance of a portion of the infinite-plane arrangement,
far from the edges, is:
Q S
C=
=
V0
d
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Dr. M. Tanseer Ali
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Chapter 6
Dielectrics and Capacitance
Capacitance
◼ Example
Calculate the capacitance of a parallel-plate capacitor having a
mica dielectric, εr = 6, a plate area of 10 in2, and a separation
of 0.01 in.
S = 10 in 2
= 10 in 2  (2.54 10−2 m in) 2
= 6.452 10−3 m 2
d = 0.01 in
= 0.01 in  (2.54 10−2 m in)
= 2.54 10−4 m
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C=
S
d
(6)(8.854 10−12 )(6.452 10−3 )
=
2.54 10−4
= 1.349 nF
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Chapter 6
Dielectrics and Capacitance
Capacitance
◼ The total energy stored in the capacitor is:
WE = 12   E 2 dv
vol
2
 S 
1
= 2     dv
vol
  
C=
 S2
=  
dzdS
0 0 
2

= 12 S Sd
 2
1  S S
2
=2
d
d 2
1
2
S
d
V0 =
d
S
d

Q
C=
V0
2
Q
WE = 12 CV02 = 12 QV0 = 12
C
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S
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