Transcript electrostatic 3

Electric Field Concepts
Rules for constructing filed lines
• A convenient way to visualize
the electric field due to any
charge distribution is to draw a
field line diagram. At any point
the field line has the same
direction as the electric field
vector
• Field lines begin at positive
charge and end at negative
charge
• The number of field lines shown
diverging from or converging
into a point is proportional to the
magnitude of the charge.
• Field lines are spherically
symmetric near a point charge
• If the system has a net charge,
the field lines are spherically
symmetric at great distances
• Field lines never cross each
other.
ELECTRIC POTENTIAL (The Volt)
To develop the concept of electric potential and
show its relationship to electric field intensity.
In moving the object from point a to b, the
work can be expressed by:
b
W   F  dL
a
dL is differential length vector along some
portion of the path between a and b
ELECTRIC POTENTIAL (cont’d)
The work done by the field in moving the charge
from a to b is
b
WE  field  Q  E  dL
a
If an external force moves the charge against the
field, the work done is negative:
b
W  Q  E  d L
a
ELECTRIC POTENTIAL (cont’d)
We can defined the electric potential difference, Vab as the work done by
an external source to move a charge from point a to point b as:
Vba
Where,
b
W

   E  dL
Q
a
Vab  Vb  Va
‘a’ is the initial point while ‘b’ is the
final point
If Vab is negative, there is a loss in potential energy in moving Q from
‘a’ to ‘b’; this implies that the work is being done by the field. However.
If Vab is positive, there is a gain in potential energy in the movement,
an external agent performs the work
Vab is independent of the path taken
Vab is measured in joules per coulomb, commonly referred to as volts (V)
ELECTRIC POTENTIAL (cont’d)
Consider the potential difference between two
points in space resulting from the field of a
point charge located at origin, where the
electric field intensity is radially directed, then
move from point a to b to have:
b
b
a
a 4 0 r
Vba    E  dL   
Q
2
a r  dra r
ELECTRIC POTENTIAL (cont’d)
Thus,
Vba 
Q
r b
4 0 r r  a
Q 1 1

    Vb  Va
4 0  b a 
The absolute potential at some finite radius
from a point charge fixed at the origin:
V
Q
4 0r
ELECTRIC POTENTIAL (cont’d)
If the collection of charges becomes a continuous
distribution, we could find:
V 
Where,
 L dL
V 
4 0 r
 S dS
V 
4 0 r
V dV
V 
4 0 r
dQ
4 0 r
Line charge
Surface charge
Volume charge
ELECTRIC POTENTIAL (cont’d)
The principle of superposition, where applied to
electric field also applies to potential difference.
Q1
Q2
V 

4 0 r  r1 4 0 r  r2
QN
... 
4 0 r  rN
Or generally,
1
N
Qk
V 

4 0 k 1 r  rk
ELECTRIC POTENTIAL (cont’d)
Three different paths to
calculate work moving
from the origin to point
P against an electric
field.
Based on figure, if a closed path is chosen, the
integral will return zero potential:
 E  dL  0
EXAMPLE 10
Two point charges -4 μC and 5 μC are located at (2,1-,3) and (0,4,-2)
respectively. Find the potential at (1,0,1).
SOLUTION - EXAMPLE 10
Let
So,
Q1  4C and Q2  5C
Q1
Q2
V 

4 0 r  r1 4 0 r  r2
Where, r  r1  1,0,1  2,1,3   1,1,2  6
r  r2  1,0,1  0,4,2  1,4,2 
Therefore, V 1,0,1 
26
Q1
Q2

4 0 r  r1 4 0 r  r2
 4  10  6
5  10  6


4 0 6
4 0 26
 
V 1,0,1  5.872 kV


ELECTRIC POTENTIAL (cont’d)
The electrostatic potential
contours
from
a
point
charge form equipotential
surfaces surrounding the
point charge. The surfaces
are always orthogonal to the
field lines. The electric field
can be determined by
finding the maximum rate
and direction of spatial
change of the potential
field.
ELECTRIC POTENTIAL (cont’d)
Therefore,
E  V
The negative sign indicates that the field is
pointing in the direction of decreasing potential.
By applying to the potential field:
 Q
Q
E  V  
ar 
a
2 r
r 4 0r
4 0r
IMPORTANT !!
Three ways to calculate E:

If sufficient symmetry, employ Gauss’s Law.

Use the Coulomb’s Law approach.

Use the gradient equation.
EXAMPLE 11
Consider a disk of charge ρS, find the potential at point
(0,0,h) on the z-axis and then find E at that point.
SOLUTION TO EXAMPLE 11
Find that,
dQ   S dS
and
r  h2   2
  S dd
dQ
With
V 
then,
 S a 2 dd
V
 
4 0  0  0 r
4 0 r
SOLUTION TO EXAMPLE 11 (Cont’d)
How to calculate the integral?
Let
u  h2   2
Integral
1 2
u
du

and
then,
leads to
du  2 d
S
V
h2   2
2 0

S
2 0
h
2
 a
 0
 a2  h

To find E, need to know how V is changing with position. In this case
E varies along the z-axis, so simply replace h with z in the answer for
V, then proceed with the gradient equation.
E  V  
V
az
z


S  1 2 z
S 
z

 1a z 

1  2
a z
2
2
2
2 0  2 z  a
2 0 
z a 

Conductors and Insulators
• A conductor is a substance that allows current to
flow through it :- they transfer charge across them.
• In metals, the current is composed of moving
electrons.
• Electrolytic solutions also conduct current but by the
movement of flow of ions.
• Insulators have few mobile electrons or ions and
the flow of current is inhibited- They keep tight tabs
on their electrons.
• As fields are increased, dielectric breakdown of
insulators occurs and the current is discharged as a
surge.
• The dielectric strength is the maximum field an
insulator can support.
Resistance and Ohm’s Law
• Resistance is a measure of resistance to flow of
electricity. It is defined by Ohm’s Law as follows:
V
R
I
(ohm’s Law)
Therefore, resistance is in the units of volts per ampere.
One volt per ampere is called an ohm (Ω).
The reciprocal of resistance is conductance
Capacitance
The amount of charge that accumulates as a
function of potential difference is called the
capacitance.
Q
C
V
The unit is the farad (F) or coulomb per volt.
Capacitance (Cont’d)
Two methods for determining capacitance:

Q Method
Assume a charge +Q on plate ‘a’ and a
charge –Q on plate ‘b’.
•
Solve for E using the appropriate method
(Coulomb’s Law, Gauss’s Law, boundary
conditions)
•
Solve for the potential difference Vab
between the plates (The assumed Q will divide
out)
•
Capacitance (Cont’d)

V Method
•
Assume Vab between the plates.
•
Find E , then D using Laplace’s equation.
Find ρS, and then Q at each plate using
conductor dielectric boundary condition
(DN = ρS )
•
•
C = Q/Vab (the assumed Vab will divide out)
Example 12
Use Q method to find the capacitance for
the parallel plate capacitor as shown.
Solution to Example 12
Place charge +Q on the inner surface of the top
plate, and –Q charge on the upper surface of the
bottom plate, where the charge density,
S  Q S
from
Q   S dS
Use conductor dielectric boundary, to obtain:
Q
D

 az 
S
from
DN  S
Solution to Example 12
We could find the electric field intensity, E
Q
E
az
 0 r S
The potential difference across the plates is:
a
Vab    E  dL
b
d
Q
Qd
 
a z  dza z 
 0 r S
0  0 r S
Solution to Example 12
Finally, to get the capacitance:
Q
C

Vab
C 
Q
 0 r S
d
Qd
 0 r S
Bioelectrical Impedance Analysis
• Bioelectrical impedance analysis (BIA) is a
commonly used method for estimating body
composition.
• Since the advent of the first commercially available
devices in the mid-1980s the method has become
popular owing to its ease of use, portability of the
equipment and its relatively low cost compared to
some of the other methods of body composition
analysis.
• It is familiar in the consumer market as a simple
instrument for estimating body fat.
• BIA actually determines the electrical impedance,
or opposition to the flow of an electric current, of
body tissues, which can be used to calculate an
estimate of total body water (TBW).
• TBW can be used to estimate fat-free body mass
and, by difference with body weight, body fat.
Bioelectrical Impedance Theory
When constant electric current is applied between two
electrodes through a biological medium and the
corresponding voltage is measured between the two source
poles, the resultant impedance or bioimpedance is
determined by Ohm’s law.
The recorded voltage is the sum of the potential difference
contributions due to the electrical conductivity properties of
the tissue medium. The exchange of electrons from source
to sink occurs from electrons of the metal electrode (such
as platinum or silver-silver chloride) to ions of the tissue
medium. The electrode is the site of charge carrier
exchange between electrons and ions and thus serves as a
transducer of electrical energy.
Impedance measurements most commonly use a twoelectrode
(bipolar)
of
four-electrode
(tetrapolar)
arrangement.
Bioelectrical Impedance Theory
The Maxwell equation most relevant to bioimpedance is
Eq. (1)
  H  D t  J
D  E  P
Eq. (2)
where
H - magnetic field strength [A/m],
D - electric flux density [coulomb/m2],
J - current density [A/m2],
E - electric field strength [V/m],
  - permittivity of vacuum [farad (F)/m], and
P - electric polarization dipole moment pr. volume [coulomb/m2].
• If the magnetic component is ignored, Equation 1 is reduced to:
D t   J
Eq. (3)
• Equations 1-3 are extremely robust and also valid under
nonhomogeneous, nonlinear, and anisotropic conditions. They relate the
time and space derivatives at a point to the current density at that point.
Bioelectrical Impedance Theory
• Impedance and
permittivity in their
simplest forms are based
on a basic capacitor
model.
• Basic equation of
bioimpedance is then
(time vectors)
Y  G  jC
Typical body segment resistance values
SUMMARY (1)
The force exerted on a charge Q1 on charge Q2 in a
medium of permittivity ε is given by Coulomb’s Law:
F12 
Where
Q1Q2
4 R12
R12  R12a12
2
a12
is a vector from charge Q1 to Q2
Electric field intensity E1 is related to force F12 by:
F12
E1 
Q2
SUMMARY (2)
The Coulomb’s Law can be rewritten as:
E
Q
4 0 R
2
aR
For a continuous charge distribution:
dQ
E
a
2 R
4 0 R
For a point charge at origin:
E
Q
4 0 r
2
ar
SUMMARY (3)
•For an infinite length line charge ρL on the z axis
L
E
a
2
•For an infinite extent sheet of charge ρS
S
E
aN
2
Electric flux density, D related to field intensity by:
D   r  0E
Where εr is the relative permittivity in a linear, isotropic and
homogeneous material.
SUMMARY (4)
Electric flux passing through a surface is given by:
   D  dS
Gauss’s Law states that the net electric flux through any
closed surface is equal to the total charge enclosed by
that surface:
 D  dS  Qenc
Point form of Gauss’s Law is
  D  V
SUMMARY (5)
The electric potential difference Vab between a pair
of points a and b in an electric field is given by:
b
Vab   E  dL  Vb  Va
a
Where Va and Vb are the electrostatics potentials at a
and b respectively.
For a distribution of charge in the vicinity of the origin,
where a zero reference voltage is taken at infinite
radius:
dQ
V 
4r
SUMMARY (6)
E is related to V by the gradient equation:
E  V
Which for Cartesian coordinates is:
V
V
V
V 
ax 
ay 
az
x
y
z
• The conditions for the fields at the boundary between
a pair of dielectrics is given by:
ET 1  ET 2
and
a 21  D1  D2    S
SUMMARY (7)
Where ET1 and ET2 are the electric field components
tangential to the boundary, a21 is a unit vector from
medium 2 to 1 and ρS is the surface charge at the
boundary. If no surface charge is present, the
components of D normal to the boundary are equal:
D N1  D N 2
At the boundary between a conductor and a
dielectric, the conditions are:
ET  0
and
DN  S
SUMMARY (8)
Poisson’s equation is:
V
 V 

2
Where the Laplacian of V in Cartesian coordinates is
given by:
2
2
2

V

V

V
2
V 2  2  2
x
y
z
In a charge free medium, Poisson’s equation reduces to
Laplace’s equation
 V 0
2
SUMMARY (9)
• Capacitance is a measure of charge storage capability
and is given by:
Q
C
V
For coaxial cable:
L
Vab 
ln b a 
2
So,
2L
C
ln b a 
For two concentric spheres:
Q 1 1
Vab 
  
4  a b 
So,
4
C
1 a 1 b