This course of lectures is a comprehensive lecture notes on the course “Electricity and Magnetism”, taught in English at
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ALFARABI KAZAKH NATIONAL UNIVERSITY
A. E. Davletov L. T. Yerimbetova
ELECTRICITY AND MAGNETISM Course of lectures
Almaty «Kazakh University» 2015
UDC 537 (075.8) LBC 22.33 D 43 Recommended for publishing by Scientific Council of the physical and technical faculty and the EditorialPublishing Board of AlFarabi Kazakh National University
Reviewers: Doctor of physical and mathematical sciences, Professor Prikhodko O.Yu. Doctor of physical and mathematical sciences, Professor Somsikov V.M.
Davletov А.Е. electricity and magnetism: course of lectures / A.E. Davletov, L.T. YerimD 43 betova. – Almaty: Kazakh University, 2015. – 196 с. ISBN 9786010408043 This course of lectures is a comprehensive lecture notes on the course “Electricity and Magnetism”, taught in English at the Department of Physics and Technology of AlFarabi Kazakh National University for more than a decade. It is intended for second year students of special groups of universities, majoring in the specialties of “Physics”, “Nuclear Physics” and “Technical Physics”. The authors are grateful to the Ministry of Education and Science of the Republic of Kazakhstan for the support of one of them (A.E. Davletov) rendered by granting the title of “The Best University Professor – 2013”.
UDC 537 (075.8) LBC 22.33
ISBN 9786010408043
© Davletov A.E. , Yerimbetova L.T., 2015 © AlFarabi KazNU, 2015
Foreword The content of the present tutorial is a summary of the lecture course on “Electricity and Magnetism” which has been given during the last few years for undergraduates of the Department of Physics and Technology of AlFarabi Kazakh National University majoring in specialties of “Physics”, “Nuclear Physics” and “Technical Physics”. The whole course consists of 15 compulsory lectures, lasting for 2 academic hours each, and is accompanied by demonstration experiments and videos. Each lecture is supported by a seminar designed for developing the problem solving skills and is assumed to be held for 1 academic hour. It should be noted that without ability in solving problems, both typical and nonstandard, it is impossible to acquire competences necessary for application of theoretical knowledge to practical situations. The main system of physical units, chosen for the present course, is SI, i.e. the International System, which is mandatory for use in all general physics courses of higher educational institutions of the Republic of Kazakhstan. It is very convenient when it comes to real physical experiments since the course itself is a prerequisite for the corresponding laboratory practicum. From the theoretical physics point of view the SI system has some disadvantages compared to the CGS, however, the transition from one to the other should be extremely easy for a professional physicist or physics engineer. It is hoped that the present tutorial will be useful for students of various undergraduate specialties of universities, for which the general physics is the major course. These comprehensive lecture notes will be helpful in preparation for exams and various tests, such as the external evaluation of educational achievements (EEEA). At the end of this tutorial a list of recommended references is provided to contain both textbooks and problem books. Particular attention should be paid to the literature in English language because the knowledge of English terminology is crucial in bringing up a new generation of high quality professionals in the relevant fields of science and technology.
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Lecture 1.
4
These lecture notes are mostly intended for students of special groups that are taught general physics in English. In reality each student is assumed to read every lecture beforehand in order to get familiar with the terminology and the corresponding vocabulary. This significantly improves their understanding of the lectures when they really take place and stimulates close involvement of the audience into the process. The structure of the course is truly original and totally complies with the syllabus of relevant specialties, although the material has been selected from a variety of open sources, listed at the end of the tutorial, to ensure the overall clarity of presentation. The authors are very grateful to Yu.V. Arkhipov, the Professor of the Plasma and Computer Physics Chair of AlFarabi Kazakh National University, for numerous and useful discussions of the lectures, as well as to V.V. Kashkarov and A. U. Aldiyarov, the tutors of the Thermophysics and Technical Physics Chair of AlFarabi Kazakh National University, for preparation of rather informative experimental demonstrations supporting this course. The authors also acknowledge the support of one of them (A.E. Davletov) rendered by granting the title of “The Best University Professor  2013”.
Lecture notes on
“Electricity and magnetism”
Lecture 1.
Introduction Fundamental interactions In physics, fundamental interactions (sometimes called interactive forces) are the ways that the simplest particles in the universe interact with one another. An interaction is fundamental when it cannot be described in terms of other interactions. The four known fundamental interactions, all of which are noncontact forces, are electromagnetism, strong interaction, weak interaction (also known as “strong” and “weak nuclear force”) and gravitation. With the possible exception of gravitation, these interactions can usually be described as being mediated by the exchange of special kind of particles, called bosons. In the conceptual model of fundamental interactions, matter consists of fermions, which carry properties called charges and spin ±1⁄2 (intrinsic angular momentum ±ħ/2, where ħ is the reduced Planck constant). They attract or repel each other by exchanging bosons. The interaction of any pair of fermions in perturbation theory can then be modeled thus: two fermions go in → interaction by boson exchange → two changed fermions go out. The exchange of bosons always carries energy and momentum between the fermions, thereby changing their speed and direction. The exchange may also transport a charge between the fermions, changing the charges of the fermions in the process (e.g. turn them from one type of fermion to another). Because an interaction results in fermions attracting and repelling each other, an older term for “interaction” is force. According to the present understanding, there are four fundamental interactions or forces: gravitation, electromagnetism, the weak interaction, and the strong interaction. Their magnitude and behavior vary greatly, as described in the table below. Modern physics attempts to explain every observed physical phenomenon by these fundamental interactions. Moreover, reducing the number of different interaction types is seen as desirable. Two cases in point are the unification of electric and magnetic
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Lecture 1.
force into electromagnetism as well as the electromagnetic interaction and the weak interaction into the electroweak interaction. The table below demonstrates our current knowledge on the characteristics of the known fundamental interactions. Interaction
Current theory
Strong
Mediators
Relative LongRange Strength Distance (m) Behavior 1038 1 1015
Quantum Gluons chromodynamics Weak Electroweak theory W and Z 1025 bosons Gravitation General Relativity Gravitons 1 (hypothetical) Electromagnetic Quantum Photons 1036 Electrodynamics
d æç exp(mW ,Z r )ö÷÷ ç ÷÷ dr ççè r ÷ø
1 r2 1 r2
1018
¥ ¥
The modern quantum mechanical view of the fundamental forces other than gravity is that particles of matter (fermions) do not directly interact with each other, but rather carry a charge, and exchange virtual particles (gauge bosons), which are the interaction carriers or force mediators. For example, photons mediate the interaction of electric charges, and gluons mediate the interaction of color charges.
Charge carriers In physics, a charge carrier denotes a free (mobile, unbound) particle carrying an electric charge, especially the particles that carry electric currents in electrical conductors. Examples are electrons and ions. In metals, the charge carriers are electrons. One or two of the outer valence electrons from each atom is able to move about freely within the crystal lattice of the metal. This cloud of free electrons is referred to as a Fermi gas. In ionic solutions, such as salt water, the charge carriers are the dissolved cations and anions. Similarly, cations and anions Quark model of a proton Quark model of a neutron of the dissociated liquid serve as charge carriers in liquids and melted ionic solids (see eg. the HallHeroult process for an example of electrolysis of a melt). In plasma, such as an electric arc, the electrons and cations of ionized gas and vaporized material of electrodes act as charge carriers. (The electrode vaporization occurs in vacuum too, but then the arc is not technically occurring in vacuum, but in lowpressure electrode vapors.) The proton is a subatomic particle with an electric charge of +1 6 elementary charge. One or more protons are present in the nucleus of each
atom, along with neutrons. The proton is also stable by itself and has a Fundamental second identity as the hydrogen ion, H+. The proton is composed of three interactions fundamental particles: two up quarks and one down quark. It is about 1.61.7 Fm in diameter. The neutron is a subatomic particle with no net electric charge and a mass slightly larger than that of a proton. With the exception of hydrogen, nuclei of atoms consist of protons and neutrons, which are therefore collectively referred to as nucleons. The number of protons in a nucleus is the atomic number and defines the type of element the atom forms. The number of neutrons is the neutron number and determines the isotope of an element. For example, the abundant carbon12 isotope has 6 protons and 6 neutrons, while the very rare radioactive carbon14 isotope has 6 protons and 8 neutrons. While bound neutrons in stable nuclei are stable, free neutrons are unstable; they undergo beta decay with a mean lifetime of just under 15 minutes (885.7±0.8 s). Even though it is not a chemical element, the free neutron is sometimes included in tables of nuclides. It is then considered to have an atomic number of zero and a mass number of one, and is sometimes referred to as neutronium. Protons and neutrons are both nucleons, which may be bound by the nuclear force into atomic nuclei. The nucleus of the most common isotope of the hydrogen Charge distribution in proton and neutron atom is a lone proton. The nuclei of the heavy hydrogen isotopes deuterium and tritium contain one proton bound to one and two neutrons, respectively. All other types of atoms are composed of two or more protons and various numbers of neutrons. The number of protons in the nucleus determines the chemical properties of the atom and thus which chemical element is represented; it is the number of both neutrons and protons in nucleus which determine the particular isotope of an element.
Elementary charge and its invariance The elementary charge, usually denoted as e, is the electric charge carried by a single proton, or equivalently, the absolute value of the electric charge carried by a single electron. This elementary charge is a fundamental physical constant. To avoid confusion over its sign, e is sometimes called the “elementary positive charge”. This charge has a measured value of approximately 1.602176487(40)×10−19 coulombs. The magnitude of the elementary charge was first measured in Robert A. Millikan’s noted oil drop experiment in 1909. Charge quantization is the concept that every stable and independent object (meaning an object that can exist independently for a prolonged period of time) has a charge which is an integer multiple of the elementary charge e. Thus, e.g., a charge can be exactly 0e, or exactly 1e, −1e,
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Lecture 1.
2e, etc., but not, say, 1⁄2e, or −3.8e, etc. (This statement must not be interpreted to include quarks or quasiparticles, since neither quarks nor quasiparticles possess the ability to exist on their own for prolonged periods of time. Quarks have charges that are integer multiples of 1⁄3e.). This is the reason for the terminology “elementary charge”: it is meant to imply that it is an indivisible unit of charge. Charge invariance refers to the fixed electrostatic potential of a particle, regardless of speed. For example, an electron has a specific rest charge. Accelerate that electron, and the charge remains the same (as opposed to the relativistic mass and energy increasing). The key word here is relativistic. Some particle characteristics are relativistically invariant (charge, spin, and magnetic moment). Others are relativistic (mass, energy, and de Broglie wavelength).
Experiment of Robert A. Millikan Robert Millikan’s apparatus incorporated a parallel pair of horizontal metal plates. By applying a potential difference across the plates, a uniform electric field was created in the space between them. A ring of insulating material was used to hold the plates apart. Four holes were cut into the ring, three for illumination by a bright light, and another to allow viewing through a microscope. Robert A fine mist of oil droplets was sprayed into a chamber above the A. Millikan plates. The oil was of a type usually used in vacuum apparatus and (18681953) was chosen because it had an extremely low vapor pressure. Ordinary oil would evaporate under the heat of the light source causing the mass of the oil drop to change over the course of the experiment. Some oil drops became electrically charged through friction with the nozzle as they were sprayed. Alternatively, charging could be brought about by including an ionizing radiation source (such as an Xray tube). The droplets entered the space between the plates and, because they were charged, could be made to rise and fall by changing the voltage across the plates. Initially the oil drops are allowed to fall between the plates with the electric field Robert Millikan’s apparatus turned off. They very quickly reach a terminal velocity because of friction with the air in the chamber. The field is then turned on and, if it is large enough, some of the drops (the charged ones) will start to rise. (This is because the upwards electric force FE is greater for them than the downwards gravitational force W , in the same way bits of paper can be picked up by a charged rubber rod). A likely looking drop is selected and kept in the middle of the field of view by alternately switching off the voltage until all 8
the other drops have fallen. The experiment is then continued with this one drop. The drop is allowed to fall and its terminal velocity v1 in the absence of an electric field is calculated. The drag force acting on the drop can then be worked out using Stokes’ law:
Fundamental interactions
Fd = 6pr hv1 where v1 is the terminal velocity (i.e. velocity in the absence of an electric field) of the falling drop, h is the viscosity of the air, and r is the radius of the drop. The weight W is the volume V multiplied by the density r and the acceleration due to gravity g . However, what is needed is the apparent weight. The apparent weight in air is the true weight minus the upthrust (which equals the weight of air displaced by the oil drop). For a perfectly spherical droplet the apparent weight can be written as:
W =
4 3 pr g(r  r air ) . 3
At terminal velocity the oil drop is not accelerating. Therefore the total force acting on it must be zero and the two forces Fd and must cancel one another out (that is, Fd = W ). This implies Fd = W
r2 =
9hv1 . 2g(r  r air )
Once r is calculated, W can easily be worked out. Now the field is turned back on, and the electric force on the drop is FE = qE ,
where q is the charge on the oil drop and E is the electric field between the plates. For parallel plates E=
V, d
where V is the potential difference and d is the distance between the plates. One conceivable way to work out q would be to adjust V until the oil drop remained steady. Then we could equate FE with W . But in practice this is extremely difficult to do precisely. Also, determining V proves difficult because the mass of the oil drop is difficult to determine without reverting back to the use of Stokes’ Law. A more practical approach is to turn V up slightly so that the oil drop rises with a new terminal velocity v2 . Then v qE W = 6phrv2 = W 2 v1
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Lecture 1.
Charge and current densities The linear, surface, or volume charge density is the amount of electric charge in a line l , surface s , or volume r , respectively. It is measured in coulombs per meter (C/m), square meter (C/m²), or cubic meter (C/m³), respectively. Since there are positive as well as negative charges, the charge density can take on negative values. Like any density it can depend on position. The integral of the charge density l(r) , s(r) , r (r) over a line L , surface V , or volume V , is equal to the total charge Q of that region, defined to be:
ò l(r)dl ,
Q=
L
Q=
ò s(r)dS , S
Q=
ò r (r)dV . V
This relation defines the charge density mathematically. Note that the symbols used to denote the various dimensions of charge density vary between fields of studies. Electric current is a coarse, average quantity that tells what is happening in an entire wire. The distribution of flow of charge is described by the current density: J(r, t ) = qn(r, t )v d (r, t ) = r (r, t )v d (r, t ) ,
where J(r, t ) is the current density vector at location r at time t (SI unit: amperes per square meter), n(r, t ) is the particle density in count per volume at location r at time t (SI unit: m3), n(r, t ) is the charge of the individual particles with density n(r, t ) (SI unit: coulombs), r (r, t ) = qn(r, t ) is the charge density (SI unit: coulombs per cubic meter), and v d (r, t ) is the particles’ average drift velocity at position t at time t (SI unit: meters per second) The current I through a surface area S perpendicular to the flow can be calculated using a surface integral: I =
ò J × dS , S
where the current is in fact the integral of the dot product of the current density vector and the differential surface element dS , in other words, the net flux of the current density vector field flowing through the surface S . 10
Continuity equation or charge conservation law Because charge is conserved, the net flow out of a chosen volume must equal the net change in charge held inside the volume:
ò
J × dS = 
S
d dt
ò V
Fundamental interactions
æd r ö÷ ÷dV , çè dt ø÷÷
rdV = ò ççç V
where r is the charge density per unit volume, and V is a surface element of the surface V enclosing the volume V . The surface integral on the left expresses the current outflow from the volume, and the negatively signed volume integral on the right expresses the decrease in the total charge inside the volume. From the divergence theorem (called Gauss theorem),
ò J × dS = ò (Ñ × J)dV = ò div JdV , S
V
div J =
V
dJ x dJ y dJ z . + + dx dy dz
Hence æd r ö
ò (Ñ × J)dV = ò (div J)dV = ò çççè dt ÷÷÷÷ødV V
V
V
Because this relation is valid for any volume, no matter how small, no matter where located: Ñ × J = div J = 
dr dt
which is called the continuity equation or charge conservation law. Questions 1. What kind of interaction is called fundamental? 2. How many do fundamental interactions exist? 3. Are proton and neutron elementary particles? 4. What is the main idea behind Millikan’s experiment with oil drops? 5. Is the net electric charge of the Universe conserved?
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Lecture 1.
Calculation of electric fields Coulomb’s law
Lecture 2.
Calculation of electric fields Coulomb’s law Coulomb’s law, or Coulomb’s inverse square law, is a law of physics describing the electrostatic interaction between electrically charged particles. It was studied and first published in 1783 by French physicist Charles Augustin de Coulomb and was essential to the development of the theory of electromagnetism. Nevertheless, the dependence of the electric force with distance had been proposed previously by Joseph Priestley and the dependence with both distance and charge had been discovered, but not published, by Henry Cavendish, prior to Coulomb’s works. It states that: “The magnitude of the electrostatics force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them.” The scalar form of Coulomb’s law is an expression for the magnitude and sign of the electrostatic force between two idealized point charges, small in size compared to their separation. This force F acting simultaneously on point charges q1 and q 2 , is given by
F =k
CharlesAugustin de Coulomb (17361806)
q1q2 r2
where k is the separation distance and k is a proportionality constant. A positive force implies it is repulsive, while a negative force implies it is attractive. The proportionality constant k , called the Coulomb constant (sometimes called the Coulomb force constant), is related to defined properties of space and can be calculated based on knowledge of empirical measurements of the speed of light. k=
1 4pe0
= 8.987 ´ 109 N × m 2 / C 2 13
Lecture 2.
and
e0 = 8.854 ´ 1012 F × m 1
In the more useful vectorform statement, the force in the equation is a vector force acting on either point charge, so directed as to push it away from the other point charge; the righthand side of the equation, in this case, must have an additional product term of a unit vector pointing in one of two opposite directions, e.g., from q1 to q 2 if the force is acting on q 2 ; the charges may have either sign and the sign of their product determines the ultimate direction of that force. Thus, the vector force pushing the charges away from each other (pulling towards each other if negative) is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The square of the distance part arises from the fact that the force field due to an isolated point charge is uniform in all directions and gets “diluted” with distance as much as the area of a sphere centered on the point charge expands with its radius. The law of superposition allows this law to be extended to include any number of point charges, to Torsion balance of Coulomb derive the force on any one point charge by a vector addition of these individual forces acting alone on that point charge. The resulting vector happens to be parallel to the electric field vector at that point, with that point charge (or “test charge”) removed.
Vector form In order to obtain both the magnitude and direction of the force on a charge, q1 at position r , experiencing a field due to the presence of 1 another charge, q 2 at position r2 , the full vector form of Coulomb’s law is required.
F=
where r is the separation of the two charges. This is simply the scalar definition of Coulomb’s law with the direction given by the unit vector, r21 , parallel with the line from charge q2 to charge q1 . If both charges have the same sign (like charges) then the product q1q 2 is positive and the direction of the force on q1 is given by r21 ; the charges repel each other. If the charges have opposite signs then the product q1 is negative and the direction of the force on q1 is given by r21 ; the charges attract each other. 21
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1 q1q2 (r1  r2 ) 1 q1q2 = r 3 4pe0  r1  r2  4pe0 r 2 21
Graphical representation Below is a graphical representation of Coulomb’s law, when q1q2 > 0. The vector F1 is the force experienced by. The vector F2 is the force experienced by q 2 . Their magnitudes will always be equal. The vector r21 is the displacement vector between two given charges ( q 2 and q 2 ). The electric field is defined as the force per unit charge that would be experienced by a stationary point charge at a given location in the field:
E=
Calculation of electric fields Coulomb’s law
F, q
where q is the electric force experienced by the particle, q is its charge, E is the electric field wherein the particle is located. Taken literally, this equation only defines the electric field at the places where there are stationary charges present to experience it. Furthermore, the force exerted by another charge q will alter the source distribution, which means the electric field in the presence of q differs from itself in the absence of q . However, the electric field of a given source distribution remains defined in the absence of any charges with which to interact. This is achieved by measuring the force exerted on successively smaller test charges placed in the vicinity of the source distribution. By this process, the electric field created by a given source distribution is defined as the limit as the test charge approaches zero of the force per unit charge exerted thereupon.
E = limq ® 0
Electric fields
F. q
This allows the electric field to be dependent on the source distribution alone. As is clear from the definition, the direction of the electric field is the same as the direction of the force it would exert on a positivelycharged particle, and opposite the direction of the force on a negativelycharged particle. Since like charges repel and opposites attract (as quantified below), the electric field tends to point away from positive charges and towards negative charges. Based on Coulomb’s law for interacting point charges, the contribution to the Efield at a point in space due to a single, discrete charge located at another point in space is given by the following:
E=
q , r 4pe0 r 2 1
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Lecture 2.
where q is the charge of the particle creating the electric force, r is the distance from the particle with charge q to the Efield evaluation point, r is the unit vector pointing from the particle with charge q to the Efield evaluation point.
Superposition principle According to above stated, electric field is dependent on position. The electric field due to any single charge falls off as the square of the distance from that charge. Electric fields follow the superposition principle. If more than one charge is present, the total electric field at any point is equal to the vector sum of the respective electric fields that each object would create in the absence of the others.
E total = å Ei = E1 + E2 + E3 ... i
If this principle is extended to an infinite number of infinitesimally small elements of charge dq located at , the following formula r results for the electric field at position :
r
E=
1 4pe0
r r'
ò dq  r  r '  .
For a linear charge distribution (a good approximation for charge in a wire) where l(r ') gives the charge per unit length at position r ' , and dl ' is an infinitesimal element of length,
dq = l(r ')dl ' . For a surface charge distribution (a good approximation for charge on a plate in a parallel plate capacitor) where s(r ') gives the charge per unit area at position r ' , and dS ' is an infinitesimal element of area,
dq = s(r ')dS ' For a volume charge distribution (such as charge within a bulk metal) where r (r ') gives the charge per unit volume at position r ' , and dV ' is an infinitesimal element of volume,
dq = r (r ')dV ' Gauss’s law
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In physics, Gauss’ law, also known as Gauss’ flux theorem, is a law relating the distribution of electric charge to the resulting electric field. Gauss’ law states that: the electric flux through any closed surface is proportional to the enclosed electric charge.
The law was formulated by Carl Friedrich Gauss in 1835, but was not published until 1867. It is one of four of Maxwell’s equations which form the basis of classical electrodynamics, the other three being Gauss’s law for magnetism, Faraday’s law of induction, and Ampère’s law with Maxwell’s correction. Gauss’s law can be used to derive Coulomb’s law, and vice versa. Gauss’s law can be derived from Coulomb’s law, which states that the electric field due to a stationary point charge is (see above):
E=
Calculation of electric fields Coulomb’s law
q . r 4pe0 r 2 1
Using the expression from Coulomb’s law, we get the total field at r by using an integral to sum the field at s due to the infinitesimal charge at each other point s in space, to give
E(r) =
r
1 4pe0
ò
r (s)(r  s)  r  s 3
ds
where is the charge density. If we take the divergence of both sides of this equation with respect to r , and use the known theorem
æ s ö Ñ × ççç 3 ÷÷÷ = 4pd(s) , çè s  ÷ø
where
Johann Carl Friedrich Gauss (1777–1855)
d(s) is the Dirac delta function, the result is Ñ × E(r) =
1
e0
ò r (s)d(r  s)ds .
Using the “sifting property” of the Dirac delta function, we arrive at
Ñ × E(r) = div E(r) =
r (r) , e0
which is the differential form of Gauss’s law, as desired. Note that since Coulomb’s law only applies to stationary charges, there is no reason to expect Gauss’s law to hold for moving charges based on this derivation alone. In fact, Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. The integral form of Gauss’s law is:
Q
ò E × dA = e S
0
for any closed surface S containing charge theorem, this equation is equivalent to
Q
ò Ñ × EdV = e V
0
Q.
By the divergence
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Lecture 2.
for any volume V containing charge Q . By the relation between charge and charge density, this equation is equivalent to:
1
ò Ñ × EdV = e ò rdV 0 V
V
for any volume V . In order for this equation to be simultaneously true for every possible volume V , it is necessary (and sufficient) for the integrands to be equal everywhere. Therefore, this equation is equivalent to:
Ñ × E(r) = div E(r) =
r (r) . e0
Mathematical Appendix The Dirac delta function, or d function, is (informally) a generalized function depending on a real parameter such that it is zero for all values of the parameter except when the parameter is zero, and its integral over the parameter from −∞ to ∞ is equal to one. It was introduced by theoretical physicist Paul Dirac. The Dirac delta can be loosely thought of as a function on the real line which is zero everywhere except at the origin, where it is infinite,
ìï+¥, ïï0, î
d(x ) = ïí
x =0 x ¹0
and which is also constrained to satisfy the identity ¥
ò d(x )dx = 1 .
¥
This is merely a heuristic definition. The Dirac delta is not a true function, as no function has the above properties. Moreover there exist descriptions of the delta function which differ from the above conceptualization. For example, sinc(x / a ) / a (where sinc is the sinc function sinc(x ) = sin(px ) / (px ) ) becomes the delta function in the limit as a ® 0 , yet this function does not approach zero for values of x outside the origin, rather it oscillates between 1 / x and 1 / x more and more rapidly as a approaches zero. The Dirac delta function has the property ¥
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ò
¥
f (x )d(x  a )dx = f (a ) .
Electric field lines An electric field can be visualized on paper by drawing lines of force, which give an indication of both the size and the strength of the field. Lines of force are also called field lines. Field lines start on positive charges and end on negative charges, and the direction of the field line at a point tells you what direction the force experienced by a charge will be if the charge is placed at that point. If the charge is positive, it will experience a force in the same direction as the field; if it is negative the force will be opposite to the field. The fields from isolated, individual charges look like this: The electric field is stronger where the field lines are close together than where they are farther apart. These 2 images show the field lines for both a set of positive point charges, as well as a positive and a negative charge. In vector calculus a conservative vector field is a vector field which is the gradient of a function, known in this context as a scalar potential. Conservative vector fields have the property that the line integral from one point to another is independent of the choice of path connecting the two points: it is path independent. Conversely, path independence is equivalent to the vector field being conservative. An conservative vector field is called a Laplacian vector field because it is the gradient of a solution of Laplace’s equation. Definition. A vector field E is said to be conservative if there exists a scalar field j suchthat: E =  grad j = Ñj . Here Ñ denotes the gradient of j . When the above equation holds, j is called a scalar potential for E . The fundamental theorem of vector calculus states that any vector field can be expressed as the sum of a conservative vector field and a solenoidal field. Mathematically, the gradient of a scalar function j in Cartesian coordinates is written as
grad j = Ñj = i
Calculation of electric fields Coulomb’s law
Conservative fields
¶j ¶j ¶j +j +k . ¶x ¶y ¶z
Path independence. A key property of a conservative vector field is that its integral along a path depends only on the endpoints of that path, not the particular route taken. Suppose that V is a region of threedimensional space, and that L is a rectifiable path in V with start point A and end point B . If E =  grad j = Ñj is a conservative vector field then
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Lecture 2.
ò E × dr = j(A)  j(B ) . L
An equivalent formulation of this is to say that
ò E × dr = 0 for every closed loop in E . The converse is also true: if the circulation of E around every closed loop in an open set V is zero, then E is a conservative vector field. Irrotational fields. A vector field E is said to be irrotational if its curl is zero. That is, if
Ñ´ E = curl E = rot E = 0 . For this reason, such vector fields are sometimes referred to as curlfree vector fields. Mathematically, the curl of a vector function E in Cartesian coordinates is written as æ i j k ö÷ çç ÷÷ çç ¶ ÷÷ ¶ ¶ ÷= Ñ´ E = curl E = rot E = çç çç ¶x ¶y ¶z ÷÷÷ ÷ çç E çè x Ey Ez ÷÷ø æ æ ¶E ö æ ö ¶Ey ÷ö ç ¶E ÷÷  j çç ¶E z  ¶E x ÷÷÷ + k çç y  ¶E x ÷÷÷. =i çç z ç ççè ¶x ¶z ÷÷ø ¶z ÷ø ¶y ÷÷ø çè ¶y çè ¶x
It is an identity of vector calculus that for any scalar field
j:
Ñ´Ñj = 0 . Therefore every conservative vector field is also an irrotational vector field.
Kelvin–Stokes theorem The classical Kelvin–Stokes theorem is written as:
Σ = ò curl E × d Ó = ò rot E ò E × dr =ò Ñ´ E × d Ó ¶S
S
S
Σ = ò rot E × d Ó Σ ò E × dr =ò Ñ´ E × d Ó = ò curl E × d Ó ¶S
20
S
S
S
and relates the surface integral of the curl of a vector field over a surface
¶S in Euclidean threespace to the line integral of the vector field over
S
its boundary. The curve of the line integral, ¶S , must have positive orientation, meaning that dr points counterclockwise when the surface normal, d Σ Ó, points toward the viewer, following the righthand rule.
Calculation of electric fields Coulomb’s law
Conservativeness of the electrostatic field Conservativeness of the electrostatic field and the electric field potential and the electric field potential The electric field of a pointlike charge is conservative because it can be tten as The electric field of a pointlike charge is conservative because it can be written as
E=
1 q q xi + yj + zk r = .. 4pe0 r 2 4pe0 (x 2 + y 2 + z 2 )3 / 2
Applying the curl to this expression yields
Applying the curl to this expression yields Ñ´ E = curl E = rot E = 0 . Ñ´ E = curl E = rot E = 0 .
It is also elementary to show that the electrostatic potential of a pointlike charge is found as
It is also elementary to show that the q electrostatic potential of a pointlike j = rge is found as 4pe0r
d they are
and they are related to each other by
q j= E = Ñj =  grad j. 4pe r 0
As long as the superposition principle holds, the electrostatic field of any canother be represented as a sum of pointlike charges, which relatedsystem to each by means that the electrostatic field of any charge system is conservative.
E = Ñj =  grad j .
Calculation of the electric field using the Coulomb law
As longDueastothe superposition principle holds, theelectric electrostatic the superposition principle the potential of the field at a field of any r = ( x , y , z ) point is calculated as: tem can be represented as a sum of pointlike charges, which means that the the system of charge pointlikesystem charges is conservative. ctrostatic field of any qi , 1 j r = ( ) Calculation of the electric fieldå using the Coulomb law 4pe i ri Due to the superposition principle the0 potential of the electric field at a point = (x , y, z ) is calculated as: ri = (charges x  X i )2 + (y Yi )2 + (z  Z i )2 and R = (X ,Y , Z ) system of with pointlike i i i being the charges coordinates; the system of surface charge
1 1 s(sq) j(r)j=(r) = ò å i ,dS , 4pe04pe  r  rsi  S 0 i 2
2
2
21
h ri = (x  Xi ) + (y Yi ) + (z  Z i ) and R = (X i ,Yi , Z i ) being the charges ordinates; the system of surface charge
Lecture 2.
with integration over the surface S of the charge distribution; the system of volume charge
(r) =
1
4
(s)
ò  r  s  dV ,
0 S
with integration over the volume V of the charge distribution.
Calculation of the electric field using Gauss’s law Electric Field of a Uniformly Charged Plane Consider an innite plane which carries the uniform charge per unit area . Suppose that the plane coincides with the y  z plane (i.e., the plane which satises x = 0 ). By symmetry, we expect the electric eld on either side of the plane to be a function of x only, to be directed normal to the plane, and to point away from/towards the plane depending on whether is positive/negative. Let us draw a cylindrical gaussian surface, whose axis is normal to the plane, and which is cut in half by the plane (see Fig.) Let the cylinder run from x = a to x = +a , and let its crosssectional area be A . According to Gauss’ law,
A
2E (a )A =
0
x = +a is the electric eld strength at x = +a . Here, the lefthand side represents
where
the electric ux out of the surface. Note that the only contributions to this ux come from the at surfaces at the two ends of the cylinder. The righthand side represents the charge enclosed by the cylindrical surface, divided by . It 0 follows that
E=
22
2
. 0
Note that the electric eld is uniform (i.e., it does not depend on x ), normal to the charged plane, and oppositely directed on either side of the plane. The electric eld always points away from a positively charged plane, and vice versa. Consider the electric eld produced by two parallel planes which carry equal and opposite uniform charge densities ± . We can calculate this eld by superposing the electric elds produced by each plane taken in
isolation. It is easily seen, from the above discussion, that in the region between the planes the field is uniform, normal to the planes, directed from the positively to the negatively charged plane, and of magnitude
E=
s e0
Calculation of electric fields Coulomb’s law
,
(Outside this region, the electric field cancels to zero. The above result is only valid for two charged planes of infinite extent. However, the result is approximately valid for two charged planes of finite extent, provided that the spacing between the planes is small compared to their typical dimensions. Electric Field of Line Charge The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss’ law. Considering a Gaussian surface in the form of a cylinder at radius r , the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder.
E 2prL =
lL , thus E = l 2pe0r e0
.
This expression is a good approximation for the field close to a long line of charge.
The Laplace and Poisson equations
One of the cornerstones of electrostatics is the posing and solving of problems that are described by the Poisson equation. Finding j for r is an important practical problem, since this is the usual some given way to find the electric potential for a given charge distribution. The derivation of Poisson’s equation in electrostatics follows. SI units are used and Euclidean space is assumed. Starting with Gauss’ law for electricity (also part of Maxwell’s equations) in a differential control volume, we have:
Ñ × E(r) = div E(r) =
r (r) . e0
23
Since the curl of the electric field is zero, it is defined by a scalar electric potential field, j .
Lecture 2.
E = Ñj =  grad j Eliminating equation:
E
by substitution, we have a form of the Poisson
Ñ × Ñj = Ñ2j = Dj = (1749 – 1827)
Here
D=
2 Here ¶2 D¶=
¶ + + ¶x 2 ¶y 2 ¶z 2 2
r e0
¶2 ¶2 ¶2 + + is the Laplace 2 ¶ ¶y 2 operator. ¶z 2 is xthe Laplace
operator. Solving Poisson’s equation potential requires knowing the Solving Poisson's equation for for thethepotential requires knowing the c charge density distribution. If the charge density is zero, then Laplace’s density distribution. If the charge density is zero, then Laplace's equation results equation results Siméon Denis Poisson (1781–1840) Example
Dj j= = 00 .. D
Example Consider a spherically symmetric volume charge distributed as
Consider a spherically symmetric volume charge distributed as
r (r ) =
for
r(r ) =
r 0r
a r0r for 0£r £a, a0 £ r £ a ,
with r 0constants. and a being constants. with r0 and a being Using thepolar spherical polar coordinates the Poisson equation is Using the spherical coordinates the Poisson equation is rewritten as rewritten as
PierreSimon, marquis de Laplace (1749 – 1827)
r(r ) . 1 d æç 2 d j ÷ö ÷÷ = . çr 2 e0 r dr èç dr ÷ø
Integrating this differential equation under condition that Integrating this differential equation under thethecondition thatthethe potenti potential j should be finite at the origin r = 0 , one gets the solution should be finite at the origin r = 0 , one gets the solution
r 0 æç 3 r 33 ö÷ . j(r ) = r æça  r ö÷ j(r ) = 3e0 a çççççèa 3  4 ÷÷÷ø . 0 ÷ 3e0a çè
4ø
Earnshaw's theorem Earnshaw’s theorem Earnshaw's theorem states that a collection of point charges canno maintained in a Earnshaw’s stable stationary equilibrium configuration solely bybethe electro theorem states that a collection of point charges cannot interaction of the charges. This stationary was first provenconfiguration by Britishsolely mathematician Sa maintained in a stable equilibrium by the electrostatic interaction of the charges. This was first provenbut by British Earnshaw in 1842. It is usually referenced to magnetic fields, originally appli 24 electrostatic fields. It applies to the classical inversesquare law forces (electric gravitational) and also to the magnetic forces of permanent magnets paramagnetic materials or any combination, (but not diamagnetic materials). Informally, the case of a point charge in an arbitrary static electric field
mathematician Samuel Earnshaw in 1842. It is usually referenced to magnetic fields, but originally applied to electrostatic fields. It applies to the classical inversesquare law forces (electric and gravitational) and also to the magnetic forces of permanent magnets and paramagnetic materials or any combination, (but not diamagnetic materials). Informally, the case of a point charge in an arbitrary static electric field is a simple consequence of Gauss’s law. For a particle to be in a stable equilibrium, small perturbations (“pushes”) on the particle in any direction should not break the equilibrium; the particle should “fall back” to its previous position. This means that the force field lines around the particle’s equilibrium position should all point inwards, towards that position. If all of the surrounding field lines point towards the equilibrium point, then the divergence of the field at that point must be negative (i.e. that point acts as a sink). However, Gauss’s Law says that the divergence of any possible electric force field is zero in free space. In mathematical notation, an electrical force qE deriving from a potential j will always be divergenceless (satisfy Laplace’s equation):
Calculation of electric fields Coulomb’s law
Ñ × E = Ñ × (Ñj ) = Ñ2j = 0 . Therefore, there are no local minima or maxima of the field potential in free space, only saddle points. A stable equilibrium of the particle cannot exist and there must be an instability in at least one direction. To be completely rigorous the existence of a stable point does not require that all neighboring force vectors point exactly toward the stable point; the force vectors could spiral in towards the stable point, for example. One method for dealing with this invokes the fact that, in addition to the divergence, the curl of any electric force field in free space is also zero (note that zero curl is more or less equivalent to conservation of energy). This theorem also states that there is no possible static configuration of ferromagnets which can stably levitate an object against gravity, even when the magnetic forces are stronger than the gravitational forces. Earnshaw’s theorem has even been proven for the general case of extended bodies, and this is so even if they are flexible and conducting, provided they are not diamagnetic, as diamagnetism constitutes a (small) repulsive force, but no attraction. There are, however, several exceptions to the rule’s assumptions which allow magnetic levitation. Questions 1. Is Coulomb’s law valid for finite size electric charges? 2. What is the direction of Coulomb’s force between two point charges? 3. How is the electric field strength defined? 4. In which situations is Gauss’ law is preferred to use rather than Coulomb’s law? 5. Which of the electric field characteristics is a vector and which is a scalar?
25
Lecture 1.
26
Electric fields in dielectrics electric dipole
Lecture 3.
Electric fields in dielectrics electric dipole An electric dipole is a separation of positive and negative charges. The simplest example of this is a pair of electric charges of equal magnitude but opposite sign, separated by some (usually small) distance. Dipoles can be characterized by their dipole moment, a vector quantity. For the simple electric dipole given above, the electric dipole moment points from the negative charge towards the positive charge, and has a magnitude equal to the strength of each charge times the separation between the charges. In physics, the electric dipole moment is a measure of the separation of positive and negative electrical charges in a system of charges, that is, a measure of the charge system’s overall polarity. In the simple case of two point charges, one with charge +q and one with charge q , the electric dipole moment p is:
p = qd where d is the displacement vector pointing from the negative charge to the positive charge. Thus, the electric dipole moment vector p points from the negative charge to the positive charge. There is no inconsistency here, because the electric dipole moment has to do with orientation of the dipole, that is, the positions of the charges, and does not indicate the direction of the field originating in these charges. 27
Lecture 3.
An idealization of this twocharge system is the electrical point dipole consisting of two (infinite) charges only infinitesimally separated, but with a finite p = qd . The potential of an electric dipole can be found by superposing the point charge potentials of the two charges:
j(r ) =
1 æç q q ö÷÷ çç ÷ 4pe0 çè 4pe0r+ 4pe0r ÷ø
The potential of a dipole is of most interest where r >> d . The standard approximations are:
r  r+ » d cos q , r+r » r 2 . Thus, the dipole potential is obtained as
j(r ) »
1 qd cos q pr cos q p×r . = = 4pe0 r 2 4pe0r 3 4pe0r 3
The electric field is simply found by the relation
E = Ñj =  grad j =
3p × r p . r5 4pe0r 4pe0r 3
Schematically, the electric field of a dipole is presented above by means of the force lines.
Dipole moment of a continuous charge distribution More generally, for a continuous distribution of charge confined to a volume V , the corresponding expression for the dipole moment is:
p=
ò r (r)rdr, V
28
where r locates the point of charge distribution and dr denotes an elementary volume in V . For an array of point charges, the charge density becomes a sum of Dirac delta functions:
n
Electric fields in dielectrics electric dipole
r (r) = å qi d(r  ri ), i =1
where each ri is a vector from some reference point to the charge Substitution into the above integration formula provides: n
qi .
n
p = å qi ò rd(r  ri )dr = å qi ri . i =1
i =1
V
This expression is equivalent to the previous expression in the case of charge neutrality and n = 2 . For two opposite charges, denoting the location of the positive charge of the pair as r+ and the location of the negative charge as r :
p = q1r1 + q2 r2 = qr+  qr = q(r+  r ) = qd, showing that the dipole moment vector is directed from the negative charge to the positive charge because the position vector of a point is directed outward from the origin to that point. The dipole moment is most easily understood when the system has an overall neutral charge; for example, a pair of opposite charges, or a neutral conductor in a uniform electric field. For a system of charges with no net charge the electric dipole moment is independent of the origin:
p' =
ò
r r ' dr ' =
V
= r+a = = dr
r' = r + a = dr ' = dr
ò r (r + a)dr = ò r rdr + a ò rdr = p, V
V
V
ò r (r + a)dr = ò r rdr + a ò rdr = p, V
V
V
since the net neutrality of the system assumes that
ò rdr = 0. V
Thus, because of overall charge neutrality, the dipole moment is independent of the observer’s position r .
Dipole moment density and polarization density The dipole moment of an array of charge determines the degree of polarity of the array, but for a neutral array it is simply a vector property of the array with no directions about where the array happens to be located. The dipole moment density of the array p(r) , that is
29
Lecture 3.
the dipole moment of the unit volume, contains both the location of the array and its dipole moment. When it comes time to calculate the electric field in some region containing the array, Maxwell’s equations are solved, and the information about the charge array is contained in the polarization density P(r) of Maxwell’s equations. Depending upon how finegrained an assessment of the electric field is required, more or less information about the charge array will have to be expressed by P(r) . As clearly explained below, sometimes it is sufficiently accurate to take P(r) = p(r) . Sometimes a more detailed description is needed (for example, supplementing the dipole moment density with an additional quadrupole density) and sometimes even more elaborate versions of P(r) are necessary.
Bound volume and surface charges As described next, a model for polarization moment density p(r) results in a polarization p(r) restricted to the same model. For a smoothly varying dipole moment distribution p(r) , the corresponding bound charge density is simply Ñ × p(r) = r b . However, in the case momentof ata a boundary betweenantwo regions, Ñ × p(r) exhibits a surface charge abrupt step in dipole moment at a boundary p(r) that exhibits component of bound charge. This charge acan be treated through a surface between two regions, surface charge component Ñ × psurface (r) exhibits integral,oforbound by using discontinuity conditions as illustrated in the charge. This surface charge at canthebeboundary, treated through a surface by using discontinuity conditions at the boundary, as variousintegral, examplesor below. illustrated in the various below.to polarization, consider a medium As a first example relatingexamples dipole moment As a first example relating dipole consider a made up of a continuous charge density rmoment (r) and to a polarization, continuous dipole moment medium made up of a continuous charge density r (r) and a continuous r is:potential at a position r is: distribution r) . The potential at a position dipolep(moment distribution p(r) . The
j(r) =
1 4pe0
r(r0 )
ò rr
0

dr0 +
1 4pe0
ò
p(r0 ) × (r  r0 )  r  r0 3
dr0 ,
r(r) is rthe (free) charge density, p(r) isand thepdipole where where is the unpaired (free) chargeand density, the (r)unpaired (r) ismoment dipole moment density. Using an identity: density. Using an identity:
(r  r )
1
Ñr
=(r  r0 ) 0 3 1 Ñ0 r r  r =   r rr 3 r0  0  r  r 0 0 0
the polarization integral can be transformed:
the polarization integral can be transformed:
1
4pe0
ò1
ò
pr(r r0 ) ) ×r(0r 0 3
dr0 =
ò
3
dr0 =
1
41 pe0
4 pe0ö ò
1
ò p(r )Ñ 1  r  r 0
p(r0 )Ñr
r0
0

dr0 =.
dr =  r  r0 Ñ  0 p(r )  r æ r0  1 1 ÷ 1 ÷÷dr0  Ñ p(r ) r0 0 dr0 . = Ñr çççpæ(r0 ) ö 0 1 1 1 4pe r0 0 r  r  ÷ 4pe =0 Ñçèr çççp(r0 ) r  r0÷÷÷drø0 dr . 0 0 ò 0 ç 4pe0  r  r0 ÷ø 4pe0 ò  r  r0  0 è 4pe0
30
p(r0 ) × (r  r0 )
0
ò
The first term can be transformed to an integral over the surface bounding the volume of integration, and contributes a surface charge density, discussed later. Putting this result back into the potential, and ignoring the surface charge for now:
The first term can be transformed to an integral over the surface bounding the volume of integration, and contributes a surface charge density, discussed later. Putting this result back into the potential, and ignoring the surface charge for now:
j(r) =
1 4pe0
ò
r (r0 )  Ñr p(r0 ) 0
 r  r0 
Electric fields in dielectrics electric dipole
dr0 ,
where the volume integration extends only up to the bounding surface, and does not include this surface. The potential is determined by the total charge, which, the above shows, consists of:
r total (r0 ) = r (r0 )  Ñr p(r0 ) 0
showing that:
Ñr p(r0 ) = r b . 0
In short, the dipole moment density p(r) plays the role of the polarization density P(r) for this medium. Notice, p(r) has a nonzero divergence equal to the bound charge density (as modeled in this approximation). Above, discussion was deferred for the leading divergence term in the expression for the potential due to the dipoles. This term results in a surface charge. The figure at the right provides an intuitive idea of why a surface charge arises. The figure shows a uniform array of identical dipoles between two surfaces. Internally, the heads and tails of dipoles are adjacent and cancel. At the bounding surfaces, however, no cancellation occurs. Instead, A uniform array of identical dipoles is on one surface the dipole heads create a positive equivalent to a surface charge. surface charge, while at the opposite surface the dipole tails create a negative surface charge. These two opposite surface charges create a net electric field in a direction opposite to the direction of the dipoles. This idea is given mathematical form using the potential expression above. The potential is:
j(r) =
1 4pe0
ò
æ 1 Ñr çççp(r0 ) 0 ç  r  r0 è
ö÷ 1 ÷÷dr0 ÷ø 4pe0
Ñr p(r0 )
ò rr 0
0

dr0 .
Using the divergence theorem, the divergence term transforms into the surface integral: 31
Lecture 3.
1 4pe0
ò
æ 1 Ñr çççp(r0 ) 0 ç  r  r0 è
ö÷ 1 ÷÷dr0 = ÷ø 4pe0
p(r0 )
ò rr
0

dA 0
dA0 an element of surface area of the volume. In the event that p(r) is a constant, only the surface term survives:
with
j(r) =
p(r0 )
1
ò rr
4pe0
0

dA 0
with dA0 being an elementary area of the surface bounding the charges. In words, the potential due to a constant p inside the surface is equivalent to that of a surface charge s = p × dA , which is positive for surface elements with a component in the direction of p and negative for surface elements pointed oppositely (Usually the direction of a surface element is taken to be that of the outward normal to the surface at the location of the element). Relation between
P
(or
p ) and E
in various materials
In a homogeneous linear and isotropic dielectric medium, the polarization is aligned with and proportional to the electric field E :
P = e0cE,
where c is the electric susceptibility of the medium. In an anisotropic material, the polarization and the field are not necessarily in the same direction. Then, the i’th component of the polarization is related to the j’th component of the electric field according to:
Pi = å e0cij E j , j
32
where cij is the electric susceptibility tensor of the medium. This relation shows, for example, that a material can polarize in the x direction by applying a field in the z direction, and so on. The case of an anisotropic dielectric medium is described by the field of crystal optics. As in most electromagnetism, this relation deals with macroscopic averages of the fields and dipole density, so that one has a continuum approximation of the dielectric materials that neglects atomicscale behaviors. The polarizability of individual particles in the medium can be related to the average susceptibility and polarization density by the ClausiusMossotti relation. If the polarization P is not linearly proportional to the electric field E , the medium is termed nonlinear and is described by the field of nonlinear optics. To a good approximation (for sufficiently weak fields, assuming no permanent dipole moments are present), P is usually
given by a Taylor series in susceptibilities:
E
whose coefficients are the nonlinear
(2) (3) Pi / e0 = å cij(1)E j + å cijk E j Ek + å cijkl E j Ek El + ... j
j ,k
Electric fields in dielectrics electric dipole
j ,k ,l
where cij is the linear susceptibility, cijk is the secondorder susceptibility (describing phenomena such as the Pockels effect, optical rectification (3) and secondharmonic generation), and cijkl is the thirdorder susceptibility (describing thirdorder effects such as the Kerr effect and electric fieldinduced optical rectification). In ferroelectric materials, there is no onetoone correspondence between P and E at all because of hysteresis. (1)
(2)
Electric displacement and Gauss’s law in simple dielectrics In a dielectric material the presence of an electric field E causes the bound charges in the material (atomic nuclei and their electrons) to slightly separate, inducing a local electric dipole moment. The electric displacement field D is defined as
D = e0E + P , where P is the (macroscopic) density of the permanent and induced electric dipole moments in the material, called the polarization density. P is a vector field whose divergence depends on the density of bound charge r b in the material:
r b = ÑP =  div P . Using this definition and the differential form of Gauss’s law, we find
div D = ÑD = r  r b = r f , where
rf
is the density of free charge in the material. is not determined exclusively by the free charge. Consider the relationship:
D
Ñ´ D = e0 (Ñ´ E) + Ñ´ P ,
which, by the fact that evaluates to:
Ñ´ D = Ñ´ P,
has a curl of zero in electrostatic situations,
Ñ´ D = Ñ´ P, In a linear, homogeneous, isotropic dielectric with instantaneous response to changes in the electric field, depends linearly on the electric field as P = e0cE
33
P = e0cE ,
Lecture 3.
c is called the electric susceptibility
where the constant of proportionality of the material. Thus
D = e0 (1 + c)E = e E , where e = e 0er is the permittivity, and er = 1 + c the relative permittivity of the material. In linear, homogeneous, isotropic media e is a constant. However, in linear anisotropic media it is a matrix, and in nonhomogeneous media it is a function of position inside the medium. It may also depend upon the electric field (nonlinear materials) and have a time dependent response. Using the equation ÑD = div D = r and integrating it over a volume one gets the integral form of Gauss’s law in dielectrics
ò div DdV V
=
ò rdV
= Q,
V
ò DdS = Q . S
Displacement field in a capacitor Consider an infinite parallel plate capacitor placed in space (or in a medium) with no free charges present except on the capacitor. In SI units, the charge density on the plates is equal to the value of the D field between the plates. This follows directly from Gauss’s law, by integrating over a small rectangular pillbox straddling one plate of the capacitor:
ò D × dA = Q
free
.
A
On the sides of the pillbox, dA is perpendicular to the field, so that part of the integral is zero, leaving, for the space inside the capacitor where the fields of the two plates add
D = Q free / A,
34
where A is surface area of the top face of the small rectangular pillbox and Q free / A is just the free surface charge density on the positive plate. Outside the capacitor, the fields of the two plates cancel each other and
D = 0 . If the space between the capacitor plates is filled with a linear homogeneous isotropic dielectric with permittivity e the electric field between the plates is constant: E = Q free / eA .
Electric fields in dielectrics electric dipole
Boundary conditions for electric field Let the interface separating two dielectrics with permittivities e1 and e2 be denoted by S with surface normal n directed from medium 2 into medium 1. At any point r Î S the dielectric field vector E j (r) , j = 1,2 , on either side of the interface S may be decomposed into tangential Etj (r) and normal Enj (r) components with respect to that surface S at that point as
E1(r) = Et 1(r) + En 1(r) , E2 (r) = Et 2 (r) + En 2 (r) , for all r Î S . Notice that n changes as r Î S varies over the interface surface S and that this field decomposition will then also vary as this direction changes. Application of the integral form of the conservativeness of the electric field to an infinitesimally small loop C in the plane of the incident and transmitted electric field vectors about the point r Î S , the upper side tangent to S in medium 1 and the lower side tangent to S in medium 2. Gives b
ò E × dl = ò E C
a
d
2
× dl + ò E1 × dl = 0 , c
where the contributions from the sides vanish as In addition
Dh ® 0
about
S.
E2 × dl = Et 2Dl , E1 × dl = Et 1Dl in the limit as Dl ® 0 on S . Hence, in this limit, The conservativeness of the electric field gives Et 2Dl  Et 1Dl = 0 , or
Et 1(r) = Et 2 (r), Thus, the tangential component of interface S .
r ÎS .
E is
continuous across the
35
Lecture 3.
Application of the integral form of Gauss’ law to an infinitesimally small pillbox of thickness Dh ® 0 with upper interface S parallel to S in medium 1 with outward normal n1 = n and the lower surface S 2 parallel to S in medium 2 with outward normal n 2 = n , gives
ò D × dS = ò D
1
S
S1
× n1dS + ò D2 × n2dS = s f DS , S1
where the contributions from the sides vanish as Dh ® 0 about S . r Î S denotes the surface free charge density Here s f = s f (r), residing on the interface S . In the limit as DS ® 0 , one obtains
n × (D1(r)  D2 (r)) = s f (r), or
Dn 1(r)  Dn 2 (r) = s f (r),
r ÎS r ÎS
.
The normal component of D changes discontinuously across the interface S by an amount given by the surface charge density s f at that point. At the interface between two dielectrics with s f = 0 , the boundary conditions are
e1En 1 = e2En 2 , Et 1 = Et 2 . Let E1 be at the angle n with respect to the surface normal n and E2 be at the angle q2 with respect to the surface normal n , where
æ E ö÷ t1 ÷ , çè En 1 ÷÷ø
q1 = arctan ççç
æ E ö÷ t2 ÷ çè En 2 ÷÷ø
q2 = arctan ççç Then
tan q2 = 36
Et 2 eE e = 2 t 1 = 2 tan q1 , En 2 e1En 1 e1
tan q2
e2
=
tan q1
e1
.
Electric fields in dielectrics electric dipole
Notice that
e1 > e2
Þ
tan q1 > tan q2 ,
e2 > e1
Þ
tan q2 > tan q1
Questions 1. What system of electric charges is called a dipole? 2. How rapidly does the electric field of a dipole vanish with distance? 3. What is a relation between the polarization vector and the electric field strength in a linear isotropic medium? 4. Can the permittivity of a linear isotropic medium be smaller than unity? 5. Why do the electric field lines refract at the interface between two dielectric media?
37
Lecture 1.
38
Dielectrics. Forces in electric field dielectrics
Lecture 4.
Dielectrics. Forces in electric field dielectrics A dielectric is an electrical insulator that can be polarized by an applied electric field. When a dielectric is placed in an electric field, electric charges do not flow through the material, as in a conductor, but only slightly shift from their average equilibrium positions causing dielectric polarization. Because of dielectric polarization, positive charges are displaced toward the field and negative charges shift in the opposite direction. This creates an internal electric field that partly compensates the external field inside the dielectric. If a dielectric is composed of weakly bonded molecules, those molecules not only become polarized, but also reorient so that their symmetry axis aligns to the field. Although the term “insulator” implies low electrical conduction, “dielectric” is typically used to describe materials with a high polarizability. The latter is expressed by a number called the dielectric permittivity, or dielectric constant. A common, yet notable example of a dielectric is the electrically insulating material between the metallic plates of a capacitor. The polarization of the dielectric by the applied electric field decreases the capacitor’s surface charge. The study of dielectric properties is concerned with the storage and dissipation of electric and magnetic energy in materials. It is important to explain various phenomena in electronics, optics, and solidstate physics. The term “dielectric” was coined by William Whewell (from “diaelectric”) in response to a request from Michael Faraday.
William Whewell (1794 – 1866)
Dielectric polarization In the classical approach to the dielectric model, a material is made up of atoms. Each atom consists of a cloud of negative charge (electrons) bound to and surrounding a positive point charge at its center. Because
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Lecture 4.
Atomic polarization
of the comparatively huge distance between them, none of the atoms in the dielectric material interact with one another (Note that this model is not attempting to describe the structure of matter, but the interaction between an electric field and matter). In the presence of an electric field the charge cloud is distorted, as shown in the top right of the figure. This can be reduced to a simple dipole using the superposition principle. A dipole is characterized by its dipole moment, a vector quantity shown in the figure as the blue arrow labeled p . It is the relation between the electric field and the dipole moment that gives rise to the behavior of the dielectric. (Note that the dipole moment is shown to be pointing in the same direction as the electric field. This is not always correct, and it is a major simplification, but it is suitable for many materials.) When the electric field is removed the atom returns to its original state. The time required to do so is the socalled relaxation time. This is the essence of the model in physics. The behavior of the dielectric now depends on the situation. The more complicated the situation the richer the model has to be in order to accurately describe the behavior.
Nonpolar dielectrics By definition, charges in an insulator are not free to move. This is not the same thing as saying they can not move. An electron in an insulator is like a guard dog tied to a tree — free to move around, but within limits. Placing the electrons of an insulator in the presence of an electric field is like placing a tied dog in the presence of a mailman.
Polarization by stretching
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The electrons will strain against the field as far as they can in much the same way that our hypothetical dog will strain against its leash as far as it can.
Electrons on the atomic scale are more cloudlike than doglike, however. The electron is really spread out over the whole volume of an atom and is not concentrated in any one location. When the atoms or molecules of a dielectric are placed in an external electric field, the nuclei are pushed with the field resulting in an increased positive charge on one side while the electron clouds are pulled against it resulting in an increased negative charge on the other side. This process is known as polarization and a dielectric material in such a state is said to be polarized. There are two principal methods by which a dielectric can be polarized: stretching and rotation. Let a oneelectron atom be placed in an external electromagnetic field E = E 0 cos wt , at w ® 0 the limit of the constant field is realized. In addition to this field the electron is subject to the elastic force F = kx . The equation of motion of the electron has the form:
Dielectrics. Forces in electric field dielectrics
mx = kx  eE cos wt. Let us find the stationary solution x = A cos wt which is to be substituted into the equation above. Then, the resulting dipole moment in the limit of zero frequency is obtained as
p = ex 0 = at
e 2E 0
m w02
,
w ® 0.
Using the expression for the ionization energy of a hydrogen atom for the estimation of w0
me 4 , e = w0 ~ 0.5 4pe0 2 (here to be used is the formula from the course “Atomic Physics “, which is studied after the course of “Electricity and Magnetism”), the final expression for the polarization of nonpolar atoms is found as
P = Np =
(8pe0 )2 6 m 3e 6
E0 .
It has to be noted that the polarization is independent of the medium temperature.
Polar dielectrics Polar molecules generally polarize more strongly than nonpolar molecules. Water (a polar molecule) has a dielectric strength 80 times that of nitrogen (a nonpolar molecule that is the major component of
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Lecture 4.
air). This happens for two reasons — one of which is usually trivial. First, all molecules stretch in an electric field whether they rotate or not. Nonpolar molecules and atoms stretch, while polar molecules stretch and rotate. This combination of actions only has a tiny effect on the overall degree to which a substance will polarize, however. What is more important is that polar molecules are already strongly stretched — naturally. The way the hydrogen atoms sit themselves on the arms of an oxygen atom’s electron clouds distorts the molecule into a dipole.
Polarization by rotation
All of this takes place on an interatomic or molecular scale. At such tiny separations, the strength of the electric field is relatively huge for what would otherwise be an unremarkable voltage. (13.6 V for an electron in a hydrogen atom, for example). Polar dielectrics are insulators in which atoms or molecules initially have a dipole moment p . When placing a dielectric in an external electric field the dipoles begin to orient along the field. They, thus, acquire the additional potential energy
W = p × E . Then, using the Boltzmann distribution and the normalization condition, we obtain an expression for the polarization of the polar dielectric in terms of the Langevin function
P = np{cth b  1 / b } , where the expression in curly brackets represents the Langevin function. At b = pE / (kBT )