Download A dipole in an electric field

13. Fundamentals of electrostatics
The origins of electromagnetism go back to ancient times:
IV cent. BC – Thales of Miletus (rubbed amber attracts bits of straw)
Essential development of the science of electromagnetism:
XVIII cent. – B. Franklin introduces two kinds of electric
charges (positive on the rubbed glass and negative
on the plastic rod)
- C. Coulomb discovers the law of interaction
between charged particles
q q
(Coulomb’s law)
F~ 1 2
r2
XIX cent. - M. Faraday (important empirical discoverythe law of induction)
- H. Oersted, A. Ampere (fundamental relations
between electricity and magnetism)
- J.C. Maxwell (basic laws of electromagnetism)
- H. Hertz (discovers electromagnetic waves)
Recent theory of electromagnetism assumes the hypothesis of
charge conservation (the total charge of an isolated system, i.e. the
algebraic sum of positive and negative charges is constant).
γ
e+ + e- (pair production )
From HRW 3
Charges of the same sign
repel each other (a) and of
opposite signs attract each
other (b).
1
Positron annihilation
e+ + e−
2γ
PET (Positron Emission Tomography) The system detects pairs of gamma rays emitted indirectly by
a positron-emitting radionuclide (tracer), which is introduced into the body on a biologically active
molecule. Three-dimensional images of tracer concentration within the body are then constructed by
computer analysis.
Schema of a PET acquisition process
2
13.1. Electric field
Instead of considering interaction at a distance between particles 1 and 2, one
introduces a concept of an electric field: charged particle 1 sets up an electric field in
the surrounding space and this field acts with a force on charged particle 2 placed in
this space: charge q  generates electric field E  E exerts a force F on q
1
The electric field

 F
E
q0

E
2
is defined through a force acting on a positive test charge q0
(charge q0 is small enough and does not alter the electric field)
(13.1)
Calculate the electric field generated by two point charges: a positive q1 and a
negative q2 at the point where the test charge q0 is placed.
The net force is a vector sum of forces exerted by
unit vectors
q1 and q2 on q0:



q1q0 
q2q0A 
F  F1  F2  k 2 r01  k 2 r02
r1
r2

 
F   E1  E2 q0




F  E q0
For N point charges (sources) the electric field at (x,y,z) is

qi 
E x , y ,z    k 2 r0i
ri
i 1
N
k
1
40
in vacuum in SI units
(13.2)
3
Electric field of a dipole
A system of two equal charges of opposite signs and separated
by a distance d is called a dipole.
Calculate the electric field on the dipole symmetry axis
perpendicular to the line connecting both charges. From the
superposition principle


+
-

E  E1  E2
For a point charge
q
E1  E2  k 2
a r2
The magnitude of E is a length of the rombus diagonal
E  2E1 cos   2k
q
a r2
2
a
a r
2
2

2kqa
a
2
r2

3
2
Electric field generated by an
electric dipole, visualized
by electric field lines. At any
point the field vector is tangent
to the field line.
For the case r >> a one obtains
Ek
2qa kp
1 p


r3
r 3 40 r 3
+q/2
+q/2
where p = 2aq is a dipole moment
In measurements of E we cannot determine a and q separately
but only their product p.
Real dipoles are not formed by systems of point charges but
exhibit the properties of ideal dipoles. An example is the water
molecule with a permanent dipole moment.
-q
The water molecule
as a dipole. The valence
electrons remain
4 closer
to the oxygen atom
13.2. Forces acting in an electric field
The force acting on a point
charge q in an electric field
is equal


(13.3)
F  qE
The charge (q<0) in a uniform electric
field generated by two plates with
charges of opposite signs
A dipole in an electric field
A dipole of moment p is placed in a uniform electric field E. The net torque, trying to
rotate a dipole about point O along the field lines, is equal to the product of a force
and a distance between the forces
  Fd  F 2a sin  2q Ea sin  pE sin
(13.4)
Eq.(13.4) can be written in a more general vector form
  
(13.5)
  p E
During rotation of a dipole the work dW is done by the field
A
which can be related to a change in the potential energy dEp
dW  dEp  dW '
dW’ - a work done against field forces
The potential energy of a dipole at any angle θ is




2
2
θ – angle between a dipole
moment and an electric field
 

Ep   dW     d   pE sin  d  pE  cos     pE cos    p  E
2
(13.6)
In eq.(13.6) the reference energy Ep=0 was chosen for θ=π/2 (Epmin = -pE).
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13.3. Gauss’ law
The flux of a vector field will be defined first. As an example we take the velocity
vector v. In this case the flux is a volume rate flowing through the loop
0
  vS
  vS cos 

Introducing the area vector S (perpendicular to the surface with a magnitude equal to
an area of the loop) one can express the flux as   v S
(13.7)
The flux of an electric field will be defined in an analogous way. For a non-uniform
field the elementary flux is defined first


d  E d S
The flux through the closed surface (called a Gaussian surface)


is    E d S
(13.8)
(the loop on the integral sign means the integration over the closed surface)
Eq.(13.7) indicates that the flux through a Gaussian surface is proportional to the
6 net
number of field lines passing through that surface.
Gauss’ law, cont.
The meaning of a net number of field lines is illustrated in the figure below


 E d S 
0
the field lines
are outward
0
the numbers of
inward and outword
lines are equal
0
qenc
0
Gauss’ law
(13.9)
The flux of electric field through any
closed surface is equal to the net charge
q enclosed by this surface
the field lines
are inward
The Gauss’ law holds for all fields proportional to 1/r2, e.g. for the gravitational field.
If we know qenc one can calculate E and vice versa.
Eq.(13.9) indicates that the exact distribution of a charge inside the Gaussian surface
is of no concern but in practice we apply the Gauss’ law for the cases with a certain
degree of symmetry, where calculation of the integral in eq.(13.9) is simple.
7
Applications of Gauss’ law
The field of a point charge
Due to a symmetry the Gaussian surface is taken as a
sphere of radius r centered on a point charge q. At any
point the electric field is perpendicular
to the surface, thus

the angle θ between E and dS is zero.
The Gauss’ law is expressed as


q
(13.10)
E d S 

0
The left side of eq.(13.10) is equal
 E dS cos    E dS  E  dS  ES
S
S
(13.11)
S
Thus, from eq.(13.10) we have
ES 
q
0
E  4 r 2 
q
0

E
q
40 r 2
1
(13.12)
The force acting on point charge q0 in the field (13.12) is
F  q0 E 
1 qq0
40 r 2
(13.13)
Eq.(13.13) is a Coulomb’s law. Thus, one can derive the Coulomb’s law from the
Gauss’ law. Gauss’ law is one of the fundamental laws of electromagnetism. 8
Applications of Gauss’ law, cont.
The field of a conducting sphere
As the conducting sphere is charged, the excess charge Q
quickly distributes itself moving to the surface and the
internal electric field becomes zero. This agrees with the
Gauss’ law. For 0  r  R


 E d S  0
 E0
For r  R the situation is analogous to the case of a point
charge, what gives E  1 Q
40 r 2
Zeroing of an electric field inside a conductor is called screening.
Electric field of a uniformly charged nonconducting plane
The nonconducting thin sheet is charged on one side with a uniform
surface charge density σ. Due to the symmetry of electric field we
choose the Gaussian surface as a closed cylindrical surface passing
through the sheet. Electric field is perpendicular
to the end caps of a cylinder and tangent to the
lateral curved surface. From the Gauss’ law we


have  E d S  EA  EA  0  2EA   A  E 9 
0
2 0
Problem 1
A small nonconduting ball of mass m and charge q hangs from an insulating thread
that makes an angle a with a vertical, uniformly charged nonconducting sheet of large
dimensions.
Taking into account the gravitational force and the force acting on the ball in an
electric field, calculate the surface charge density  of the sheet.
Electric field produced by the sheet is equal E  
2 0
and the electric force acting on charge q is
equal qE. The vector sum of all forces acting on charge q, i.e.
gravitational, electric and reaction forces, is zero.
From the figure it can then be inferred that
qE
 tan a
mg
mg
E 
tan a
q
Substituting for E one can calculate the charge density 

mg

tan a
2 0
q
2mg 0

tan a
q
10
13.4. Electric potential
The electrostatic field is conservative and a potential energy can be associated with it.
This simplifies the calculations of work done by an electric field as the principle of
conservation of mechanical energy can be applied. Following the discussion in
Section 5 one can write the expression for the change in potential energy dEp in the
field of electric forces as




q0
(13.14)
dE  dW   F d l  q E d l
p
0
The elementary change in electric potential, independent
of the test charge q0, is defined as


dEp
(13.15)
 dV   E d l
q0
The total change in electric potential between arbitrary points
1 and 2 is obtained by integrating (13.15)
2 
Q
Mooving q0 along the path
ACB we do the same work
as in mooving along the
field line between A and B.
(13.16)

V2  V1    E d l
1
Choosing arbitrarily the starting reference point in infinity where V1 = 0, one can
write
xyz 

V x , y ,z     E d l
(13.17)

The potential depends only on the coordinates of a given point and electric field11 E.
Electric potential of a point charge
According to eq.(13.3) the potential difference in
a radial


field of a point charge is (as the path is radial dl  dr )
B
1 Q^ 
r d r
2
4

r
0
A
VB  VA   
dr
Q 1 B
Q 1 1
  
VB  VA  


40 rA r 2 40 r rA 40  rB rA 
Q
rB
r
With a reference potential equal to zero (for rB
one obtains VA  V  Q  1
40 r
∞)
(13.18)
The positive point charge Q
produces a radial electric field

r̂ - a unit vector
E
Q r̂
40 r 2
From (13.5) it follows that for r = const V= const, what determines the co called
equipotential surface. For a point charge the equipotential surfaces are spheres.
Cross sections of equipotential
surfaces (dashed lines) shown
A
together with respective field lines
for the fields generated by a point
charge (left figure) and by
an electric dipole (right figure).
Equipotential surfaces are always
perpendicular to electric field
lines.
From HRW 3
12
13.5. Capacitance
Two conductors isolated electrically from each other form a capacitor; when
charged, the charges on the conductors (plates) are
equal in magnitude but opposite signs.
The potential difference between the plates, called
a voltage U, and the charge q are proportional to each
other
q  CV1  V2   CU
(13.19)
The charges on the capacitor
plates create an electric field in
The proportionality constant C is called a capacitance
the surrounding space.
and its value depends on the capacitor geometry and
the type of dielectric between the capacitor plates.
A parallel plate capacitor
This capacitor consists of two parallel plates
of area A each, separated by a small distance d.
The electric field between the plates is uniform
(neglecting the nonuniform edge field) and can
be calculated from Gauss’ law


 E d S 
q
0
A
(13.20)
q is a charge enclosed by a Gaussian surface on the positive plate
13
Parallel plate capacitor, cont.


Vectors E and
q
EA 
0
dS
 E
in eq.(13.20) are parallel, then this equation reduces to
(13.21)


q
0A
0
The potential difference between the plates can be calculated using eq.(13.16)
(13.22)
V  V    E d l    E dl cos    E dl
2 
2

2
2
1
1
1
1
Substituting for E from eq.(13.21) into eq.(13.22) one
obtains

d
(13.23)
V  V  U   dl 


2
2
1
1
0
0
From the definition of C (13.19) and using (13.23) one
obtains for the capacitance of a parallel plate capacitor
q q
q A  A
(13.24)
C 


0
U
d
0
qd
The positive potential is that of the
upper plate hence the integration
path from the plate 1 to the plate 2
in eq.(13.22) is directed against the
electric field line.
0
d
Inserting between the capacitor plates a dielectric, it becomes polarized in the electric
field and this leads to a decreasing of U (at constant q)
U' 
U
r
εr – the relative dielectric permitivity (dielectric constant) of a material
In effect the capacitance increases εr times
C 
q r q
A

 rC  0 r
U U
d
(13.25)
14
Connection of capacitors
Sample problem
Calculate the equivalent capacitance for the connections of capacitors as in the figures
below; C1 = 10μF, C2 = 30μF, C = 20μF
Parallel connection
Each capacitor has the same
potential difference V, which
produces charges q1 and q2 on
the capacitors
q1 = C1V, q2 = C2V.
The total charge
q=q1 + q2=(C1 + C2)V=CeqV
Ceq=40 μF
Serial connection
From the mechanism of charging
it follows that each capacitor has
the same charge q and the sum of
potential differences across each
capacitor equals the applied
potencial difference V.
V1 = q/C1, V2 = q/C2.
V = V1 + V2=q(1/C1+ 1/C2)=q/Ceq
Ceq = C1C2/(C1 + C2)
Ceq=7.5 μF
This connection is neither
serial nor parallel.
The analysis indicates that
the potentials at points a
and b are equal, so the
A
capacitor between these
points can be disconnected.
In this case
Ceq= C/2 + C/2 = 20 μF
15
13.6. Energy stored in an electric field
Charging of a capacitor is connected with some work which has to be done. One can
say that this work is stored in a form of electric potential energy in the field between
the plates. This energy can be recovered
by discharging the capacitor.
To transfer a charge dq’ between the
plates across which the voltage U’ exists
requires the work
dW  U' dq' 
q'
dq'
C
To transfer the total charge q requires
the work W which is equal to the potential
q
energy
q'
q 2 CU 2
W  Ep  
C
0
dq' 
2C

2
Charing of a capacitor by transferring a charge
dq’ from the negative to the positive plate.
(13.26)
The density of potential energy, i.e. the energy per unit volume of a capacitor is
Ep CU 2  0 A U 2
1
u


  0E 2
A d 2 Ad
d 2 Ad 2
(13.27)
Eq.(13.27) holds not only for the capacitor but can be used in each case where an
electric field exists in a space.
16