Download Chapter 3-ELECTROSTATICS(Dielectrics

Dielectrics
• Conductor has free electrons.
• Dielectric electrons are strongly bounded to the
atom.
• In a dielectric, an externally applied electric field,
Eext cannot cause mass migration of charges
since none are able to move freely.
• But, Eext can polarize the atoms or molecules in
the material.
• The polarization is represented by an electric
dipole.
• Note the field will apply a force on both the positively charged
nucleus and the negatively charged electron. However, these forces
will move these particles in opposite directions
• Note, an electric dipole has been created !
D, flux density is proportionally
increase as polarization increase
through induction of permittivity, ε of
the material relating the E and D
permittivity, ε = proportional
to the permittivity of free
space, ε0
ε = εr ε0
• However, the electron may be break free from the atom, creating
a positive ion and a free electron.
• We call these free charges, and the electric field will cause them to
move in opposite directions :
• Moving charge is electric current J(r ) .
Electric Boundary Conditions
• Electric field maybe continuous in each of two
dissimilar media
Dielectric
and dielectric
• But, the E-field maybe
discontinuous
at the
boundary between them
• Boundary
Boundary conditions specify how the tangential
Dielectric
and
conductor
and
normal
components
of
the
field
in one
condition
medium are related to the components in other
medium across the boundary
Conductor
• Two dissimilar media
could be:and
two conductor
different
dielectrics, or a conductor and a dielectric, or
two conductors
Dielectric- dielectric boundary
• Interface between two dielectric media
Dielectric- dielectric boundary
• Based on the figure in previous slide:
• First boundary condition related to the tangential
components of the electric field E is:
E1t  E2t
V/m 
• Second boundary condition related to the normal
components of the electric field E is:
D1n  D2n   S
• OR
 1 E1n   2 E2n   S
z
ε1
E2
Exy1
ε2
Ez1
E1
Exy2
Ez2
xy -plane
xy
• Solution:
ε1=2ε0
ε2=5ε0
1) Exy1=Exy2 thus,Exy2 = 3ax+4ay
2) Ez1 = 5az,
but, Ez2 = ??
E2
Ө2
Ө1
z
E1
Example 1:
1) Find E2 in the dielectric,
when E1 = 3ax+4ay+5az,
2) And find Ө1 and Ө2.
2ε0(5az) = 5ε0(Ez2)
Ez2 = 2az
thus, E2 =3az+4ay+2az
z
ε1 =2 ε0
Ө2
ε2=8ε0
E1
E2
xy
Ө1
Find E1 if E2 = 2x -3y +3z with s = 3.54 x 10-11(C/m2)
And find Ө1 and Ө2
Conductor- conductor boundary
• Boundary between two conducting media:
• Using the 1st and 2nd boundary conditions:
E1t  E2t V/m  and  1 E1n   2 E2 n   S
xy
J2
Jz1
Jxy1
Jz2
J1
Jxy2
z
Conductor- conductor boundary
• In conducting media, electric fields give rise to
current densities.
• From J  E, we have:
J 1t
1

J 2t
2
and
1
J1n
1
 2
J 2n
2
 S
• The normal component of J has be continuous
across the boundary between two different
media under electrostatic conditions.
Conductor- conductor boundary
• Hence, upon setting J 1n  J 2 n , we found the
boundary condition for conductor- conductor
boundary:
 1  2 
J1n     ρs
 1  2 
electrosta tics 
Dielectric-conductor boundary
• Assume medium 1 is a dielectric
• Medium 2 is a perfect conductor
Perfect conductor
• When a conducting slab is placed in an external
electric field, E0
• Charges that accumulate on the conductor
surfaces induces an internal electric field Ei  E0
• Hence, total field inside conductor is zero.
Dielectric-conductor boundary
• The fields in the dielectric medium, at the
boundary with the conductor is E1t  E2t .
• Since E2t  0 , it follows that E1t  D1t  0 .
• Using the equation, D1n   s ,
• we get: D1n  1E1n   s
• Hence, boundary condition at conductor surface:
D1   1 E1  nˆ s
at conductor surface 
where n̂ = normal vector pointing outward
Dielectric-conductor boundary
• Based on the figure in previous slide:
• In a perfect conductor,
E  D  0 everywhere in the conductor
• Hence, E2  D2  0
• This requires the tangential and normal
components of E2 and D2 to be zero.
Capacitance
• Capacitor – two conducting bodies
separated by a dielectric medium
ρs = Q / A
ε1
ρs = surface charge density
Q = charge (+ve / -ve)
A = surface Area
E
E
E
E = ρs/ε
ε2
E=0/V=0 on the surface
Capacitance
• Capacitance is defined as:
Q
C
V
C/V
or F
where: V = potential difference (V)
Q = charge (C)
C = capacitance (F)
Example 7
Obtain an expression for the capacitance C of a
parallel-plate capacitor comprised of two parallel
plates each of surface area A and separated by
a distance d. The capacitor is filled with a
dielectric material with permittivity ε.
Solution to Example 7
• expression for the capacitance C = Q/V
ρs = Q / A
and
•
the voltage difference is
d
   zˆE   zˆdz  Ed

0
Q
Q
A
C 

• Hence, the capacitance is:
V Ed
d
Example 8
Use image theory to determine E at an arbitrary
point P (x, y, z) in the region z > 0 due to a
charge Q in free space at a distance d above a
grounded conducting plane.
Solution to Example 8
• Charge Q is at (0, 0, d) and its image −Q is at
(0,0,−d) in Cartesian coordinates. Using
Coulomb’s law, E at point P(x,y,z) due to two
point charges:
Q  xˆ  yˆ  zˆ;
R1  ( x  0)  ( y  0)  ( z  d );
R2  ( x  0)  ( y  0)  ( z  d )
 QR


QR
 1
2
E
3 
40  R 3
R
2 
 1
1


Q  xˆx  yˆy  zˆz  d 
xˆx  yˆy  zˆz  d  
E 

3
/
2
3/ 2 

40
2
2
2
2
2
2
x  y  z  d 
 x  y  z  d 





Electrostatic potential energy
• Assume a capacitor with plates of good
conductors – zero resistance,
• Dielectric between two conductors has negligible
conductivity, σ ≈ 0 – no current can flow through
dielectric
• No ohmic losses occur anywhere in capacitor
• When a source is connected to a capacitor,
energy is stored in capacitor
• Charging-up energy is stored in the form of
electrostatic potential energy in the dielectric
medium
Electrostatic potential energy
1
2
W

CV
• Electrostatic potential energy, e
2
Q
Q
A

• The capacitance: C  
V Ed
d
• Hence, We for a parallel plate capacitor:
1 A
1 2
1 2
2
Ed   E ( Ad )  E v
We 
2 d
2
2
where V  Ed (voltage across capacitor)
v  Ad (volume of the capacitor)
Image Method
• Image theory states that a charge Q above a
grounded perfectly conducting plane is equal to
Q and its image –Q with ground plane removed.