Download 3. Electrostatics

Electrostatics
EEL 3472
Electrostatics
Electrostatics
An electrostatic field is produced by a static (or time-invariant) charge
distribution. A field produced by a thunderstorm cloud can be viewed as an
electrostatic field.
Coulomb’s law deals with the force a point charge exerts on another point
charge. The force F between two charges Q1 and Q2 is
1. Along the line joining Q1 and Q2
2. Directly proportional to the product Q1 Q2 of the charges
3. Inversely proportional to the square of the distance r between the
charges
F 
2
kQ1Q2
r2
EEL 3472
Electrostatics
In the international system of units (SI), Q1 and Q2are in coulombs (1C is
approximately equivalent to 6x101 8 electrons since one electron charge
e =  1.6019 x10 1 9 C), the distance r is in meters, and the force F is in newtons
so that:
k 
1
4o
 o  8.854 x10
or
k
F 
3
1
4o
1 2
10 9
F/m

36
the permittivity
of free space
 9x109 m/F
Q1Q2
4o r 2
EEL 3472
Electrostatics
Electric field intensity (or electric field strength) is defined as the force per unit
charge that a very small stationary test charge experiences when it is placed in
the electric field.
F
N V

E  lim
,
Qt 0 Q
C m
t
In practice, the test charge should be small enough not to disturb the field
distribution of the source.
For a single point charge Q located at the origin
E 
F
Q

er
Qt
4or 2
where er is a unit vector pointing in the radial direction (away from the charge).
 o is valid for vacuum and air. For other materials  o should be replaced by
   r  o (typically  r  1  100 ).
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EEL 3472
Electrostatics
Suppose that there are several charges Qi located at positions ri with respect to
the origin. The observation point is at position r . The vector from the ith charge
to the observation point is ( r  ri ), and the distance is r  ri .
Source point
The total field due to N
charges is
r  ri
E 
N

i 1
Observation (field)

point
e rri
5
r  ri

r  ri
N

i 1
Qi
4 r  ri
Qi
4 r  ri
3
2
erri
(r  r i )
EEL 3472
Electrostatics
Example: Find the electric field at P due to Q and –Q.
The vector from Q to P lies in the x-y plane. Therefore, Ez (the z component
of E ) vanishes at P.
y components due to Q and –Q are equal and opposite and therefore add to
zero.
Thus, E has only an x component.
E
Q
x
 E  e x  E cos  

6
Q

4 (b2  d 2 )
Qb
3
4(b2  d 2 )2
1
2

2


b
 2
1 
2 2
 (b  d ) 
ExQ  ExQ
Ex  ExQ  ExQ  2ExQ  
2Qb
3
4(b2  d 2 )2
EEL 3472
Electrostatics



Various charge distributions.
Charge density
Consider a charged wire. Charge on an elemental wire segment l is q so that
the charge per unit length   q / l . The total electric field is the sum of the
field contributions from the individual segments.
E(r)  
i

q
4 r  ri
1
4

i

If we let l  0 we get
E(r) 
7

1
4 

3
(r  ri )
(q / l )l
r  ri
3
(r  ri )
 (r)dl
(r  r)
3
r  r
  q / l
Position of the field point
Position of the source point
EEL 3472
Electrostatics
If the charge is distributed on a surface (surface charge density   q / A ), we
can write
E(r) 
1
4 

 ( r)dA
3
r  r
(r  r) 
1
4 

 (r)dA
2
r  r
e rr
Surface integral
Unit vector directed
from the source point
to the field point

If charge is distributed through a volume (  q / v) ,
E(r) 
1
4 

 ( r)dv 
3
r  r
(r  r) 
1
4 

 (r)dv 
2
r  r
e rr
Integral over the
volume containing

the charge
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EEL 3472
Electrostatics
Example: A circular disk of radius b with uniform surface charge density  , lies
in the x-y plane with its center at the origin. Find the electric field at P
(z=h).
z

All the contributions in the x and
y directions are cancelled, and
the field at P contains only z
component.
Ez 
1
4
r  r

b

0
h
2
u
2
9

h u
2
cos 
2
2udu
dA
h 
1


2  h2  u2

b

2
h
b

1 
2
 0

h
udu

2 0 h2  u2 

h 
1
1  
h
1 
 

2  h2  b2 
h  2 
h2  b2

3
2
1
2





EEL 3472
Electrostatics
The Source Equation
As we have seen, the electric field can be expressed as an integral of the charges
that produce it. The charge can be expressed in terms of derivatives of the
electric field:
 
  E  
source equation
divergence operator
Let
E(x, y, z)  E x (x, y, z)e x  Ey (x, y, z)ey  E z (x, y, z)e z
Then
10
E 
E x Ey E z


x
y
z
scalar
EEL 3472
Electrostatics
The Source Equation continued
Example: Let E be the field of a single point charge Q at the origin. Show that
  E is zero everywhere except at the origin.
E
Q
4 r
2 er 
Q
4  r
3
r
er 
r
r
r  xe x  y ey  z e z
where
E 
The x component of E is


Ex 
Q
x
3
4 (x 2  y 2  z 2 )2
E x
Q y 2  z 2  2x 2

5
x
4  (x 2  y 2  z 2 ) 2
Ey
Q x 2  z 2  2y 2

5
y
4 (x 2  y 2  z 2 )2

2
2
2
E z
Q x  y  2z

5
z
4 (x 2  y 2  z 2 )2
11
Q xe x  y ey  z e z
3
4 (x 2  y 2  z 2 )2
The divergence of a
vector field at a given
point is a measure of
how much the field
diverges or emanates
from that point.
E x Ey E z


0
x
y
z
except at the origin
(x=0,y=0,z=0) where the
derivatives are undefined.
EEL 3472
Electrostatics
The Source Equation continued
In cylindrical coordinates
E 
1 
1 E EZ
(EP ) 

 
 
z
In spherical coordinates
E 
1  2
1

1 E
(
r
E
)

(sin

E
)

r

r sin 
r sin 
r 2 r
Example: Let E be the electrostatic field of a point charge Q at the origin.
Express this field in spherical coordinates and find its divergence.
E 
Er 
Q
4r 2
E  0
Q
er
2
4r
E  0
E 
1   Q 

0
r 2 r  4 
everywhere except at the origin, where   E is undefined (because the field is
infinite).

12
EEL 3472
Electrostatics
Gauss’ Law
Essentially, it states that the net electric flux through any closed surface is equal
to the total charge enclosed by that surface. Gauss’ law provides an easy means
of finding E for symmetrical charge distributions such as a point charge, an
infinite line charge, and infinite cylindrical surface charge, and a spherical
distribution of charge.

Let us choose an arbitrary closed surface S. If a vector field V is present, we can
construct the surface integral of V over that surface:
 V  ds
S
The total outward flux of V through S
The divergence of V (a scalar function of position) everywhere inside the surface
S is   V.We can integrate this scalar over the volume enclosed by the surface
S.
   Vdv
V
13
Over the volume enclosed by S
EEL 3472
Electrostatics
Gauss’ Law continued
According to the divergence theorem
 V  ds     Vdv
S
V
 
We now apply the divergence theorem to the source equation    E  
integrated over the volume v:

V
 E  ds 
S
Gaussian
surface
 
   E dv 
Qenc

V
D E
dv
  E ds  

S

V
  E cos dA

dv  Qenc
dA  dS
Gauss’ Law
the angle between E
and the outward
normal to the surface
S
En - the normal component of E
14
EEL 3472
Electrostatics
Gauss’ Law continued
Example: Find the electric field at a distance r from a point charge q using
Gauss’ law.
Let the surface S (Gaussian surface) be a sphere of radius r centered on the
o
charge. Then   0 and cos  1 . Also E  Er (only radial component is present).
  E cos dA   
S
S
Er dA  Qenc  q
Since Er = constant everywhere on the surface, we can write
Er  dA 
S
q

Er 
q
4r 2
4r 2
15
EEL 3472
Electrostatics
Gauss’ Law continued
Example: A uniform sphere of charge with charge density o and radius b is
centered at the origin. Find the electric field at a distance r from the
origin for r>b and r<b.

As in previous example the electric field is radial and spherically symmetric.
(1) r > b
4b 3
4r E r 
o
3
2
2

r




0
0
2
2
sindd
b o
Er 
3r 2
3


b
r





0
2
sindrd d
0 r 0
decreases as 1 / r 2
(2) r < b
Er 
ro
3
increases linearly with r
Gauss’ law can be used for finding E when E is constant on the Gaussian surface.
16
EEL 3472
Electrostatics
Gauss’ Law continued
Example: Determine the electric field intensity of an infinite planar charge with a
uniform surface charge density  .
coincides with
the xy-plane
The E field due to a charged sheet of an infinite extent is normal to the sheet.
We choose as the Gaussian surface a rectangular box with top and bottom faces
of an arbitrary area A, equidistant from the planar charge. The sides of the box
are perpendicular to the charged sheet.
 E  dS 
17
Qenc

EEL 3472
Electrostatics
Gauss’ Law continued

On the top face


E  dS  E z e z  dse z  E z ds



E  dS   E z e z   dse z  E z ds
On the bottom face
There is no contribution from the side faces.
 E  dS  2E  ds  2E
z
S
The total enclosed charge is
A
2E z A 


18
z
A
S
Qencl  A
Ez 

2
E z e z ,z  0
E 
 E z e z ,z  0
EEL 3472
Electrostatics
Ohm’s Law
Ohm’s law used in circuit theory, V=IR, is called the “macroscopic” form of
Ohm’s law. Here we consider “microcopic” (or point form). (Applicable to
conduction current only). Let us define the current density vector J . Its
direction is the direction in which, on the average, charge is moving. J is current
per unit area perpendicular to the direction of J . The total current through a
surface is
I   J  dS
(if J  J o , S  a2,   0 ,I  Ioa2 )
S
According to Ohm’s law, current density has the same direction as electric field,
and its magnitude is proportional to that of the electric field:

J  E E
where
E
is the electric conductivity. The unit for  E is siemens per meter
( S / m  (  m)1 ).
Copper
Seawater
Glass
19
5.80 x107 S/m
4 S/m
10 1 2 S/m
EEL 3472
Electrostatics
Ohm’s Law continued
The reciprocal of  E is called resistivity, in ohm-meters (  m).
Suppose voltage V is applied to a conductor of length l and uniform cross section
S. Within the conducting material J   E E ( E  constant)
E 
J
V
l
I
V
  EE   E
S
l
 l 
 l 
I   I  RI
V  
S
 ES 
“macroscopic” form of
Ohm’s law
Resistance of the conductor
20
EEL 3472
Electrostatics
Electrostatic Energy and Potential
A point charge q placed in an electric field E experiences a force:
F  qE
If a particle with charge q moves through an electric field from point P1 to point
P2, the work done on the particle by the electric field is
W 
P2
P2
P1
P1
 F  dl  q  E  dl
This work represents the difference in electric potential energy of charge q
between point P1 and point P2
W = W1 - W2
(W is negative if the work is done by an external agent)
The electric potential energy per unit charge is defined as electric potential V
(J/C or volts)
P2
Electrostatic potential
V1  V2   E  dl
difference or voltage
P1
between P1 and P2
W1
W2
(independent of the path
q
q
between P1 and P2 )
21
EEL 3472
Electrostatics
Electrostatic Energy and Potential continued
V1 and V2 are the potentials (or absolute potentials) at P1 and P2, respectively,
defined as the potential difference between each point and chosen point at which
the potential is zero (similar to measuring altitude with respect to sea level). In
most cases the zero–potential point is taken infinity.
Example: A point charge q is located at the origin. Find the potential difference
between P1 (a,0,0) and P2 (0,b,0)
E due to q is radial. Let us choose a path of
integration composed of two segments.
Segment I (circular line with radius a) and
Segment II (vertical straight line).
V1  V2 
P2
 E  dl
P1
22
EEL 3472
Electrostatics
Electrostatic Energy and Potential continued
Example continued:
Segment I contributes nothing to the integral because E is perpendicular to dl
everywhere along it. As a result, the potential is constant everywhere on
Segment I; it is said to be equipotential.
On Segment II, e r  e y ,
E 
dl  dye y , and r  y
q
q
e

ey
r
4r 2
4y 2
qdy
qdy
E  dl 
e

e

y
y
4y 2 
4y 2
b
V1 V 2
qdy
q 1 1

2
a 4y 4  a  b 
 dy
1
  2   
y
 y
If we move P2 to infinity and set V2=0,
V1 
q
4a
point
where a is the distance between the observation point and the source
In general,
  q
4 r  r
Vr 
qi
i
i
23
dl
i
for line charge
ds for surface charge
dv for volume charge
EEL 3472
Electrostatics
Gradient of a Scalar Field
By definition, the gradient of a scalar field V is a vector that represents both the
magnitude and direction of the maximum space rate of increase of V.
The gradient operator () is a differential operator. It will act on the potential (a
scalar) to produce the electric field (a vector).
Let us consider two points P1(x1, y1, z1) and P2(x2  x, y2  y, z2  z) which are
quite close together.
If we move from P1 to P2 the potential will change from V1 to
scalar
V 
V
V
V
x 
y 
z
x
y
z
V1  V where
due to the
change in z
The displacement vector (express the change in position as we move from P1 and
P2) is
vector
l  x e x  y e y  z e z
24
EEL 3472
Electrostatics
Gradient of a Scalar Field continued
The relation between V and l can be expressed using the gradient of V:
V 
V
V
V
ex 
ey 
ez
x
y
z
vector
V  V  l
On the other hand, if we let P1 and P2 to be so close together that E is nearly
constant,
V  V2  V1  (V1  V2 )  E  l
(potential difference equation)

V  V 
2
 1



E

dl
P

1

P2
E  V
The negative sign shows that the direction of E is opposite to the direction in
which V increases.
25
EEL 3472
Electrostatics
Gradient of a Scalar Field continued
Example: Determine the electric field of a point charge q located at the origin,
by first finding its electric scalar potential.
V 
q
4 x 2  y 2  z 2

q
4r
V
qx

x
4 x 2  y 2  z 2


V
qy

y
4 x 2  y 2  z 2


V
qz

z
4 x 2  y 2  z 2


V 
3
2
3
2
3
2
V
V
V
ex 
ey 
ez
x
y
z
E  V
E 
26
q xe x  y ey  z e z
q
q

r

er
3
3
2
4 x 2  y 2  z 2 2
4 r
4 r


EEL 3472
Gradient of a Scalar Field continued
Example continued: Determine the electric field of a point charge q located at
the origin, by first finding its electric scalar potential.
dV  V  dl  V dl cos 
vector
the angle between dl and V
dV
 V cos 
dl
dV  dV 

 V

dl
dl

MAX
  0o
  90
27
o
dV
0
dl
(V=const)
V points in the direction of the
maximum rate of change in V.
V at any point is perpendicular to
the constant V surface which
passes through that point
(equipotential surface).
EEL 3472
Electrostatics
Gradient of a Scalar Field continued
The electric field is directed from the conductor at higher potential toward one at
lower potential. Since E  V , E is always perpendicular to the equipotentials.
If we assume that for an ideal metal  E   , no electric field can exist inside the
metal (otherwise J   E E   , which is impossible).
Considering two points
P1 and P2 located inside a metal or on its surface – and
P
recalling that V1  V2   E  dl we find that V1=V2. This means the entire metal is
2
P1
at the same potential, and its surface is an equipotential.
V
1 V
V
e 
e 
ez

 
z
V
1 V
1 V
V 
er 
e 
e
r
r 
r sin 
For cylindrical coordinates V 
For spherical coordinates
28
EEL 3472
Electrostatics
Capacitors
Any two conductors carrying equal but opposite charges form a capacitor. The
capacitance of a capacitor depends on its geometry and on the permittivity of
the medium.
C 
The capacitance C of the capacitor
is defined as the ratio of the
magnitude of Q of the charge on
one of the plates to the potential
difference V between them.
Area A
Q
V
 E  dS 
Qenclosed
 E  dS 
E A  A / 
S
d
d is very small compared with
the lateral dimensions of the
capacitor (fringing of E at the
edges of the plates is
neglected)

S
E 

A
C 
d

D   
A parallel-plate capacitor.
Cylindrical
Top face (inside the metal)
Side surface
dS
Bottom face
A
E 0
Gaussian
Surface
E
Close-up view of the upper plate of the parallel-plate capacitor.
29
EEL 3472
Electrostatics
Laplace’s and Poisson’s Equations
The usual approach to electrostatic problems is to begin with calculating the
potential. When the potential has to be found, it is easy to find the field at any
point by taking the gradient of the potential E  V

the divergence operator
E 
vector



the source equation
   V  


  V   


the gradient operator
the Laplacian operator
 2V  
scalar
If no charge is present

30


Poisson’s equation
  0 ,
 2V  0
Laplace’s equation
EEL 3472
Electrostatics
Laplace’s and Poisson’s Equations continued
The form of the Laplacian operator can be found by combining the gradient and
divergence operators.
 V

V
V
2V    V     
ex 
ey 
e z 
y
z
 x

2V 2V 2V



x 2 y 2 z 2
1 
V 
 
2
- in rectangular coordinates
 V  1 2V 2V
 
  2

2
2 - in cylindrical coordinates






z


1   2 V 
1
 
V 
1
2V -in spherical
V  2
r

 sin

  r 2 sin2   2
r r  r  r 2 sin  
coordinates
2
Electrostatic problems where only charge and potential at some boundaries are
known and it is desired to find E and V throughout the region are called
boundary-value problems. They can be solved using experimental, analytical, or
numerical methods.
31
EEL 3472
Electrostatics
Method of Images
A given charge configuration above an infinite grounded perfectly conducting
plane may be replaced by the charge configuration itself, its image, and an
equipotential surface in place of the conducting plane. The method can be used
to determine the fields only in the region where the image charges are not
located.
Original problem
Construction for solving
original problem by the
method of images. Field
lines above the z=0
plane are the same in
both cases.
32
By symmetry, the potential in
this plane is zero, which
satisfies the boundary
conditions of the original
problem.
EEL 3472