Download Unit - SVCE

UNIT II
Conductors and Dielectrics
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Capacitance
1. To store charge
2. To store energy
3. To control variation time
scales in a circuit
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When a capacitor is charged, its plates have charges of equal magnitudes but opposite signs: q+
and q-. However, we refer to the charge of a capacitor as being q, the absolute value of these
charges on the plates.
The charge q and the potential difference V for a capacitor are proportional to each other:
The proportionality constant C is called the capacitance of the capacitor. Its value depends only on
the geometry of the plates and not on their charge or potential difference.
The SI unit is called the farad (F): 1 farad (1 F)= 1 coulomb per volt =1 C/V.
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Charging a Capacitor
The circuit shown is incomplete because switch S is open; that is, the switch does
not electrically connect the wires attached to it. When the switch is closed,
electrically connecting those wires, the circuit is complete and charge can then
flow through the switch and the wires.
As the plates become oppositely charged, that potential difference increases until
it equals the potential difference V between the terminals of the battery. With
the electric field zero, there is no further drive of electrons. The capacitor is then
said to be fully charged, with a potential difference V and charge q.
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Calculating the Capacitance:
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Calculating the Capacitance; A Cylindrical Capacitor :
As a Gaussian surface, we choose a cylinder of length L and
radius r, closed by end caps and placed as is shown. It is
coaxial with the cylinders and encloses the central cylinder
and thus also the charge q on that cylinder.
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Calculating the Capacitance; A Spherical Capacitor:
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Calculating the Capacitance; An Isolated Sphere:
*We can assign a capacitance to a single isolated spherical conductor of radius R
by assuming that the “missing plate” is a conducting sphere of infinite radius.
*The field lines that leave the surface of a positively charged isolated conductor
must end somewhere; the walls of the room in which the conductor is housed can
serve effectively as our sphere of infinite radius.
*To find the capacitance of the conductor, we first rewrite the capacitance as:
Now letting b→∞, and substituting R for a,
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Example, Charging the Plates in a Parallel-Plate Capacitor:
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Capacitors in Parallel
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Capacitors in Series
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Example, Capacitors in Parallel and in Series:
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Example, Capacitors in Parallel and in Series:
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Example, One Capacitor Charging up Another Capacitor:
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Energy Stored in an Electric Field:
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Energy Density:
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Example
Potential Energy
&
Energy Density of an Electric Field:
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Example
Work and Energy when a Dielectric is inserted inside a Capacitor:
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Dielectrics, an Atomic View:
1. Polar dielectrics. The molecules of some dielectrics, like water, have permanent electric
dipole moments. In such materials (called polar dielectrics), the electric dipoles tend to
line up with an external electric field as in Fig. 25-14. Since the molecules are
continuously jostling each other as a result of their random thermal motion, this
alignment is not complete, but it becomes more complete as the magnitude of the
applied field is increased (or as the temperature, and thus the jostling, are
decreased).The alignment of the electric dipoles produces an electric field that is
directed opposite the applied field and is smaller in magnitude.
2. Nonpolar dielectrics. Regardless of whether they have permanent electric dipole
moments, molecules acquire dipole moments by induction when placed in an external
electric field. This occurs because the external field tends to “stretch” the molecules,
slightly separating the centers of negative and positive charge.
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Dielectrics and Gauss’ Law:
A dielectric, is an insulating material such as mineral oil or plastic,
and is characterized by a numerical factor , called the dielectric
constant of the material.
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Dielectrics and Gauss’ Law:
(
Q )
e0 Þ e = ke0
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Dielectrics and Gauss’ Law:
1. The flux integral now involves E, not just E. The vector (0 E) is
sometimes called the electric displacement, D. The above equation can
be written as:
1. The charge q enclosed by the Gaussian surface is now taken to be the
free charge only. The induced surface charge is deliberately ignored on
the right side of the above equation, having been taken fully into
account by introducing the dielectric constant  on the left side.
1. 0 gets replaced by 0. We keep  inside the integral of the above
equation to allow for cases in which  is not constant over the entire
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Capacitor with a Dielectric:
The introduction of a dielectric limits the
potential difference that can be applied
between the plates to a certain value Vmax,
called the breakdown potential. Every dielectric
material has a characteristic dielectric strength,
which is the maximum value of the electric
field that it can tolerate without breakdown.
It actually can increase the capacitance of the
device. Recall that
e0 Þ e = ke0
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Example, Dielectric Partially Filling a Gap in a Capacitor:
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Why are boundary conditions important ?
 When a free-space electromagnetic wave is incident upon a medium secondary
waves are
 transmitted wave
 reflected wave
 The transmitted wave is due to the E and H fields at the boundary as seen from the
incident side
 The reflected wave is due to the E and H fields at the boundary as seen from the
transmitted side
 To calculate the transmitted and reflected fields we need to know the fields at the
boundary
 These are determined by the boundary conditions
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m1,1,s1
m2,2,s2
 At a boundary between two media, mr,ers are different on either side.
 An abrupt change in these values changes the characteristic impedance
experienced by propagating waves
 Discontinuities results in partial reflection and transmission of EM
waves
 The characteristics of the reflected and transmitted waves can be
determined from a solution of Maxwells equations along the boundary
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Boundary conditions
 The tangential component of E is continuous at a surface
of discontinuity
E1t, H1t
 E1t,= E2t
 Except for a perfect conductor, the tangential
component of H is continuous at a surface of
discontinuity
E2t, H2t
m1,1,s1
m2,2,s2
 H1t,= H2t
•
•
The normal component of D is continuous at the
surface of a discontinuity if there is no surface charge
density. If there is surface charge density D is
discontinuous by an amount equal to the surface
charge density.
– D1n,= D2n+rs
The normal component of B is continuous at the
surface of discontinuity
– B1n,= B2n
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m1,1,s1
D2n, B2n
m2,2,s2
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Proof of boundary conditions - continuity of Et
x
y
E x1
E y1
E y3
E y2
E y4
m1,1,s1
m2,2,s2
Ex2
 Integral form of Faraday’s law:
0
E y2
0
B
.dA
A t
 E.ds  
0
B
y
y
y
y
 E y1
 E x1x  E y 3
 E y4
 E x 2 x   z xy
2
2
2
2
t
0
As y  0, Bz t xy  0
0
E x1x  E x 2 x  0

E x1  E x 2
That is, the tangential component of E is continuous
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x
y
H x1
H y1
H y4
m1,1,s1
m2,2,s2
0
0
H y3
H y2
H x2
 Ampere’s law
0
H y2
0

H.ds 
 D

 J .dA

A  t


y
y
y
y
 Dz

 H y1
 H x1x  H y 3
 H y4
 H x 2 x  
 J z xy
2
2
2
2
 t

0
As y  0, Dz t  J z xy  0 That is, the tangential
H x1x  H x 2 x  0

H x1  H x 2
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component of H is
continuous
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Proof of boundary conditions - Dn
Dn1
y
x
m1,1,s1
m2,2,s2
z
Dn 2
 The integral form of Gauss’ law for electrostatics is:
 D.dA  V dV
applied to the box gives
Dn1xy  Dn2 xy  edge   s xy
As dz  0, edge  0 hence
Dn1  Dn 2   s
The change in the normal component of D at a
boundary is equal to the surface charge density
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Poisson’s and Laplace Equations
A useful approach to the calculation of electric potentials
Relates potential to the charge density.
The electric field is related to the charge density by the divergence relationship
The electric field is related to the electric potential by a gradient relationship
Therefore the potential is related to the charge density by Poisson's equation
In a charge-free region of space, this becomes Laplace's equation
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Potential of a Uniform Sphere of Charge
outside
inside
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Poisson’s and Laplace Equations
From the point form of Gaus's Law
Del_dot_D
v
L aplace’s E quation
Definition D
if
D
E
Del_dot_ D
Del_Del
and the gradient relationship
E
DelV
Del_D
v
Del_  E
Del_DelV
 v

Del_dot_  DelV
v
0
v
Laplacian
T he divergence of the
gradient of a scalar function
is called the L aplacian.
Poisson’s Equation
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Poisson’s and Laplace Equations
 d  d V(x, y , z)   d  d V(x, y , z)   d  d V(x, y , z)  

 
 

 dx dx
 dy  dy
 d z d z

LapR  
1 d  d
1 d  d

 d  d



LapC      V  , , z  
   V  , , z    V  , , z 
 d  d

  2 d d
 dz dz
1
1
1 d  2d

d 
d

d d 
LapS     r  V r ,  ,    
  sin    V r ,  ,   

V r ,  , 
2 dr
dr
d
  r2 sin  d 
 r2 sin 2 dd
r 
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Examples of the Solution of Laplace’s Equation
D7.1
Given
V( x, y , z) 
 x 1
 y    2 
   
 z 3
4 y  z
2
x 1
Find: V @ and v at P
 12
o  8.854 10
V( x, y , z)  12
 d  d V( x, y , z)   d  d V( x, y , z)   d  d V( x, y , z)  






 dx dx
 dy  dy
 d z d z

LapR  
v  LapR o
LapR  12
 10
v  1.062  10
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Uniqueness Theorem
Given is a volume V with a closed surface S. The function V(x,y,z) is completely
determined on the surface S. There is only one function V(x,y,z) with given
values on S (the boundary values) that satisfies the Laplace equation.
Application: The theorem of uniqueness allows to make statements about the
potential in a region that is free of charges if the potential on the surface of
this region is known. The Laplace equation applies to a region of space that is
free of charges. Thus, if a region of space is enclosed by a surface of known
potential values, then there is only one possible potential function that
satisfies both the Laplace equation and the boundary conditions.
Example: A piece of metal has a fixed potential, for example, V = 0 V. Consider
an empty hole in this piece of metal. On the boundary S of this hole, the value
of V(x,y,z) is the potential value of the metal, i.e., V(S) = 0 V. V(x,y,z) = 0
satisfies the Laplace equation (check it!). Because of the theorem of
uniqueness, V(x,y,z) = 0 describes also the potential inside the hole
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Examples of the Solution of Laplace’s Equation
Assume V is a function only of x – solve Laplace’s equation
V
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V o x
d
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Examples of the Solution of Laplace’s Equation
Finding the capacitance of a parallel-plate capacitor
Steps
1 – Given V, use E = - DelV to find E
2 – Use D = E to find D
3 - Evaluate D at either capacitor plate, D = Ds = Dn an
4 – Recognize that s = Dn
5 – Find Q by a surface integration over the capacitor plate
C
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Q
 S
Vo
d
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Examples of the Solution of Laplace’s Equation
Cylindrical
ln 

V
C





Vo 
b 
ln 
 
a 
b
2    L
ln 



a
 
b
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Examples of the Solution of Laplace’s Equation
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Examples of the Solution of Laplace’s Equation
(spherical coordinates)
1
V
r
Vo 
1
a
C

1

1
b
b
4  
1
a

1
b
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Examples of the Solution of Laplace’s Equation
 


ln  tan   
  2 
V Vo 
   
ln  tan   
  2 
C
2   r1


  

2
 
ln  cot 
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Examples of the Solution of Poisson’s Equation
1
1
v( x)
0
charge density
1
1
10
5
0
 10
5
x
10
10
0
E( x)
E field intensity
1
 10
0.5
V( x)
10
0.5
0
 0.5 0.5
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x
potential
10
 10
5
0
x
5
10
10
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Product Solution Of Laplace's Equation
Referring to the figure below and the specific boundary conditions for the potential on the four
sides of the structure ...
... choose the dimensions a and b of the box and the potential boundary condition V0:
b  0.5
Length of box in the x direction (m).
a  0.5
Length of box in the y direction (m).
V0  2
Impressed potential on the wall at x = b (V).
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Product Solution Of Laplace's Equation
Define potential function
The solution for the potential everywhere inside the rectangular box structure is given as an
infinite series. It is not possible to numerically add all of the infinite number of terms in this
series. Instead, we will choose the maximum number of terms nmax to sum:
nmax  41
Maximum n for summation.
n  1 , 3 .. nmax
Only the odd n terms are summed since all even n terms are zero.
The potential V everywhere inside the structure was determined in Example 3.24 to be:
V( x , y) 
4 V0



n
1
 n b 
nsinh

a


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 n x   n y 
 sin

a
a

 

sinh
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Product Solution Of Laplace's Equation
Plot V versus x at y = a/2
We will generate three different plots of this potential. The first is V as a function of x through
the center of the box structure. The other two plots will show the potential within the interior
of the box in the xy plane. Choose the number of points to plot V in the x and y directions:
npts  50
Number of points to plot V in x and y.
xend  b
yend  a
x and y ending points (m).
Generate a list of xi and yj points at which to plot the potential:
i  0 .. npts  1
xi  i
j  0 .. npts  1
xend
yj  j 
npts  1
yend
npts  1
Now plot the potential as a function of x through the center of the box at y = a/2:
V at y = a/2
2.5
V (Volts)
2
For a rectangular box with
b  0.5 (m)
a  0.5 (m)
and
V0  2 (V)
1.5
1
0.5
0
0
0.1
0.2
0.3
x (meters)
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0.4
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Product Solution Of Laplace's Equation
Computed


V b ,
a
  2.0303
2
Exact
(V)
V0  2.0000
(V)
For nmax  41 , the percent error in the potential at this point is:


V b ,
Error 
a
  V0
2
100
V0
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Error  1.515
(%)
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Product Solution Of Laplace's Equation
Plot V throughout the inside of the box
Now we will plot the potential throughout the interior of the rectangular box structure. First
compute V at the matrix of points xi and yj:
Potentiali , j  V  xi , yj
Now generate a contour plot of Potentiali,j:
Plot of V(x,y)
yend  0.500
( m)
For a box with
b  0.5 (m)
a  0.5 (m)
and
V0  2 (V)
y = 0 (m)
Potential
x = 0 (m)
xend  0.500
( m)
We can observe in this plot that the potential is a complicated function of x and y. (The surface
plot below may help in visualizing the variation of V throughout the interior of this box.) The
potential is symmetric about the plane y = a/2 which we would expect since the box and the
boundary conditions are
both symmetric
about this same plane.
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AP-ECE
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Product Solution Of Laplace's Equation
V(x,y)
For a box with
b  0.5 (m)
a  0.5 (m)
and
V0  2 (V)
Potential
The jagged edge on the potential at the far wall is due to numerical error and is a nonphysical
result. The potential along that wall should exactly equal
V0  2 (V)
since that is the applied potential along that wall. This jaggedness in the numerical solution can
be reduced by increasing the number of terms in the infinite summation (nmax) for V and/or
increasing the number of points to plot in the contour and surface plots (npts).
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