Basic laws of electromagnetism irodov pdf


 

Download Basic Laws Of Electromagnetism PDF Book by I. E. Irodov - With a specific end goal to accentuate the most critical laws of electroattraction, and. Basic Laws of Electromagnetism Irodov - Ebook download as PDF File .pdf), Text File .txt) or read book online. MIR publishers, electricity, magnetiism. Title: Basic Laws of Electromagnetism - I. E. Irodov. Page number ISSUU Downloader is a free to use tool for downloading any book or publication on.

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Basic Laws Of Electromagnetism Irodov Pdf

Title: Basic Laws of Electromagnetism - I. E. Irodov, Author: Blog da Engenharia We substitute into (6')~) the charge pdF for q (where dV is the. Download Basic Laws of Electromagnetism Irodov. Basic laws of eledromagnefismI/I. E. HPOJJKOB OCHOBHLIE BAHOIIBI 3. HEKTPO1\·iAl`HE'1`IrI3A·IA Hanarenbcwuo, {Q} English t.

He was mostly known as a physics professor at the Moscow Institute of Physics and Engineering MEPHi and the author of a series of handbooks on general physics, which became lecture courses in physics in several countries. When he was eight his family moved to Moscow , where he lived until his death. During World War II he fought with various infantry units at the 1st and 4th Ukrainian Fronts , also acting as a drafter and cartographer. He went through Russia, Ukraine and Poland, ending the war in Czechoslovakia. In February , he entered the Physics Faculty of MEPHi, graduating with honors in November with a diploma of designer and operator of physics equipment. Its eighth and expanded edition, issued in , became the last substantial work by Irodov. He also published collections of problems in general physics in and He published the first part mechanics in , the second part electromagnetism in , and the full set in — In his handbooks Irodov aimed for brevity, crafting concise and clear definitions and removing unessential details and heavy calculus, and for relating theory with practical examples and problems.

For a very small l, we shall arrive at the concept of the surface charge density on the sphere. Find the projection of vector E onto thp. Let us first find vector E: Find the potential lp at the edge of a thin df! In order to simplify in. After substituting these expressions into integral 1. Fmd the dIstrIbution of the volume charge p r within the 'sphere.

Let us first find the field intensity. Find the force of interaction between two point dipoles with moments PI and P2, if the vectors PI anti P2 are directed along the straight line connecting the dipoles and the distance between the dipoles is l. A Conductor in an Electrostatic Field 2. Field in a Substance Micro- and Macroscopic Fields. The real electric field in any substance which is called the microscopic field varies abruptly both in space and in time.

It is different at different points of atoms and in the interstices. In order to find the intensity E of a real field at a certain point at a given instant, we should sum up the intensities of the fields of all individual charged particles of the substance, viz. The solution of this problem is obviously not feasible. In any case, the result would be so complicated. In many cases it is sufficient to have a simpler and rougher description which we shall be using henceforth. Under the electric field E in a substance which is called the rruu: This averaging is performed over what is called a physically infinitesimal volume, viz.

If any suhstance is introduced into an electric field, the positive and negative charges nuclei and electrons are displaced, which in turn leads to a parti. In certain regions of the suhstance, uncompensated charges of different signs appear.

This phenomenon is called the electrostatic. Induced charges create an additional electric field which in combination with the initial external field forms the resultant field. However, in many cases the situation is complicated by the fact that we do not kaow beforehand how all these charges are distributed in space, and the problem turns out to be not as simple as it eooJd seem at first sight.

It will be shown later that the distri-. We shall have to consider these questions in greater detail. Let us place a metallic conductor into an external electrostatic field or impart a certain charge to it. In both cases, the electric field will act on all the ch2.

This displacement current will continue until this practically takes a small fraction of a second a certain charge distribution sets in, at which the electric fi. This can be easily explained with the help of the Gauss theorem. And this means that there are no excess charges inside the conductor. Excess charges appear only on the conductor surface with a certain density J which is generally different for different points of the surface.

The absence of a field inside a conductor indicates in accordance with 1. The fact that the surface of a conductor is equipotential implies that in the immediate vicinity of this surface the field E at each point is directed along the normal to the surface. If the opposite were true, the tangential component of E would make the charges move over the surface of the.

Find the potential of an undlarged condudiug sphere provided that a point charge q is located at a distance r from its centre Fig. PotentialljJ is the same fur all points of the sphere. Thus we can calculate its value at the centre 0 of the sphere, r because only for this point it can be calculated in the most simple way:.

As a result of electric induction, the ch? As we move away from this The field itself in this case resembles more and more the field of a point charge q, viz.

The Field Near aConductor Surface. We shall show that the electric fwld intensity in the immediate vicinity of the surface of a conductor is connected with the local charge density at the conductor surface through a simple relation. This relation can be established with the help of the Gauss theorem. Suppose that the region of the conductor surface we are interested in borders on a vacuum. The field lines are normal to the conductor surface.

Hence for a closed surface we. Relation 2. This is not so. The intensity E is determined by all the charges of the system under consideration as well as the value of a itself. Forces Acting on the Surface of a Conductor Let us consider the case when a charged region of the surface of a conductor borders on a vacuum. S ;is the charge of this element and Eo is the field created by all the other charges of the system "in the region where the charge 0 f!

S is located. Let us find this relation, Le. S at the points that are very close to thIS element. In this region, it behaves as an infinite uniformly charged plane.

Then, in accordance with 1. S is the superposition of the fields Eo and Eo. On both sides of the area element f! Dividing both sides of this equation by f! S, we obtain the expression for the force acting on unit surface of a conductor: We can write this expression in a different form since the quantities a and E appearing in it are interconnected.

Indeed, in accordance with 2. Equation 2. Find the expression for the electric force acting in a vacuum on a conductor as a whole, assuming that the field intensities E are known at all points in the vicinity of the conductor surface. Multiplying 2. The resultant force acting on the entire conductor can be found by integrating this equation over the entire conductor lIUliace:. It was sho. This means that the excess charge is distributed on a conductor with a cavity in the same way as on a uniform conductor, viz.

Thus, in the absence of electric charges within the cavity the electric field is equal to zero in it. External charges, including the charges on the outer surface of the conductor, do not create any electric field in the cavity inside the conductor. This forms the basis of electrostatic shielding, Le. In practice, a solid conducting shell can be replaced by a sufficiently dense metallic grating. That there is no electric field inside an empty cavity can be proved in a different way.

Since the field E is equal to zero inside the conductor, the flux of E through the turface S is also equal to zero. Hence, in accordance with the Gauss theorem, the total charge inside S is equal to zero as well.

This does not exclude the situation depicted in Fig 2. Since the conducting medium is electrically neutral everywhere it does not influence the electric field in any way. Therefore, if we remove the medium, leavi. We arrive at the following important conclusion: They WIll also remam unchanged upon the displacement of charges outside the shell.

Naturally, the above arguments are applicable only in the framework of electrostatics. However, this assumption is prohibited by another theorem, viz. Indeed, let the contour r cross the cavity along one of the lines of E and be closed in the conductor material. It is clear that the line integral of vector E along this contour differs from zero, which is in contradiction with the theorem on circulation.

Let us now consider the case when the cavity is not empty but contains a certain electric charge g or several charges. Suppose also that the entire external space is filled by a conducting medium. In equilibrium, the field in this medium is equal to zero, which means that the medium is electrically neutral and contains no excess charges. According to the Gauss theorem, this means that the algebraic sum of the charges within this closed surface is equal to zero as well.

Thus, the algebraic sum of the charges induced on the cavity surface is equal in magnitude and opposite in sign to the algebraic sum of the charges inside the cavity. Find the potential cp at the point P lying outside the shell at a distance r from the centre 0 of the outer surface. The field at the point P is determined only by charges induced on. An infinite conducting plane is a special case of a closed conducting shell.

The -space on one side of this plane is. We shall repeatedly use this property of a closed conducting shell. Indeed, th'Iere is a one-toone correspondence 2. As a result. ThiS statement is called the uniqueness theorerr;. Ite obvIOUS: If there are more than one solution there wIll be. Thus we arrive at a phYSically absurd conclusion. Hence it immediately follows that the uniqueness of the field E determines the uniqueness of the charge distribution over the conductor surface.

The solution of Eqs. The analytic solutions i of these equations were obtained only for a few particular cases. If a solution of the problem satisfies the Lalplace or Poisson equation and the boundary conditions, we can stat6e that it is correct and unique regardless of the methods by which itt was obtained if only by guess.

Prove that in an empty cavity of a cconductor the field is absent. In ourr case such a simple problem is the problem about two changes: The field of this system is well known its equipotential surfaces and field lines are shown in Fig.

According to the uniqueness theorem, the field in the upper half-space will remain unchanged. The point charge g can be considered to be the limiting case of a small spherical conductor whose radius tends to zero and potential to infinity.

Thus, the boundary conditions for the potential in the upper half-space remain the same, and hence the field in this region is also the same Fig. It should he noted that we can arrive at this conclusion proceeding from the properties of a closed conducting shell [see Sec. Thus, in the case under consideration the field differs from zero only in the upper half-space. The fictitious charge g' creates in the upper half-space the same field as that of tlie charges induced on the plane.

This is precisely what is meant when we say that the fictitious charge produces the same "effect" as all the induced charges.

We must only bear in mind that the "effect" of the fictitious charge extends only to the half-space where the real charge q is located,. In another half-space the field is absent.

Summing up, we can say that the image method is essentially based on driving the potential to the boundary conditions, I. Let liS consider one more example. A point charge q is placed between two mutually perpendicular half-planes Fig.

Find the location of fictitious point. It can he seen from formula 2. Illg half-planes. These three hctl tlOllS cl. Jarges create just the same held within the "right angle" as the Fig. Capacitors Capacitance of an Isolated Conductor. The quantity 2. It is numerically. The capacitance depends on the SIze and shape of the conductor. Find the capacit: The basic charactens lC 0. The unit 'ofcapacitance is the capacitance of a conductor whose potential changes by 1 V when a charge of 1 C is supplied to it.

This unit of capacitance is called the farad F. The farad is a very large quantity. It corresponds to the capacitance of an isolated sphere 9 X km in radius, which is times the radius of the Earth the capacitance of the Earth is 0.

If a conductor is not isolated, its capacitance will considerably increase as other bodies approach it. This is due to the fact Jhat the field of the given conductor causes a redistribution of charges on the surrounding bodies, Le. Then negative induced charges will be nearer to the conductor than the positive charges. For this reason, the potential of the conductor, which is the algebraic sum of the potentials of its own charge and of the charges induced on other bodies will decrease when other uncharged bodies approach it.

This means that its capacitance increases. This circumstance made it possible to create the system of conductors, which has a considerably higher capacitance than that of an isolated conductor. Moreover, the capacitance of this system does not depend on surrounding bodies. Such a system is called a capacitor. The simplest capacitor consists' of two eonductors plates separated by a small distance.

In order to exclude the effect of external bodies on the capacitance of a eapacitor, its plates are arranged with respeet to one another in such a way that the field created by the charges accumulated on them is concentrated almost completely inside the capaeHor. This means that the lines of E emerging on one plate must terminate on the other,. This capacitor Capacitance of a Para e -p a d b a ap of width h.

Substituting this expression into 2. Let the radii of thE CapaCItance of a p. II inner and outer capaclt? IS g,. The voltage of the capacitor is charges on the outer surface. It is interesting that when the gap between the plates is small, Le. Capacitance of a Cylindrical Capacitor.

Like in the previous case, the obtained expression is reduced to 2. The influence of the medium on the capacitance of a capacitor will be discussed in Sec. On the determination or potential. A point charge q is at a distance r from the centre 0 of an uncharged spherical conducting layer, whose inner and outer radii are equal to a and b respectively.

According to the principle of superposition, the sought potential at the point 0 can be represented in the form q-J - -1 -4rtE o. It should be noted that the potential. What will be the dependence of potential cP on the distance r from the centre of the system? We write the expressions. The force acting on a surface charge. Find the magnitude of the resultant electnc force actmg on like charges.

According to 2. A point charge q is at a distance 1 from an infinite conducting plane. Find the work of the electric force acting on the charge q done upon its slow removal to a very large distance from the plane. Integrating this expression over the half-sphere i.

Image method. Find the density of surface charges induced on the plane as a function of the distance r from the base of the perpendicular dropped from the charge q onto the plane.

Using the image method, we find that the field at the point P Fig. By definition, the work of this force done upon an elementary displacement dx F;.

Integrating this equation over x between land 00, we find. ThIS IS because the re. However, In the reference system fixed to the conducting plane, the electric field of induced charges is not a potential field: Find 1 the surface charge density at a point of the plane, which is symmetric with respect to the ring and 2 the electric field potential at the centre of the ring.

It can be easily seen that in accordance with the image method, a fictitious charge -q must be located on a similar ring but. Indeed, only in this case the potential of the midplane between these rings is equal to zero, Le. Let us now use the formulas we already know. The expression for E on the R. In our case, this expression must he doubled. Three unlike point charges are arranged as shown in Fig.

Find 1 the total charge induced on the conducting half-planes and 2 the force acting on the charge -q. Hence the action of the charges induced on the conducting hali-planes Is equivalent to the action of the fictitious charge -q placed in the lower left comer of the dashed sauare. Thus we have already answered the first question: Find the capacitance of the WIres per umt length provided that Then, by defmition, the reqUIred capaCitance is Cu.

It follows from Fig. The intensity of the electric field created by one of the wires at a distance x from its axis can be easily found wi th ,the help of the Gauss theorem: Then b-a. Four identical metallic plates are arranged in air at the same distance h from each other.

The outer plates are connected by a conductor. The area of each plate is equal to S. Find the capacitance of this system between points 1 and 2, Fig. Let us charge the plates 1 and 2 by charges qo and -qo. Under the action of the dissipation field appearing between these plates edge effect , a charge will move in the connecting wire, after which the plate A will be charged negatively while the plate B will acquire a positive charge.

An electric field appears in the gaps between the plates, accompanied by the corresponding distribution of potential cp Fig. It should be noted that as follows from the symmetry of the system, the potentials at the Imiddle of the system as well as on its outer plates are equal to zero. On the other hand, it is clear that a. Eliminating E! X and E 2x from these equatIOns, we obtam a1. It would be difficult, however, to solve thiS problem WIth the help p. And since E ex a, we can state that according to 2 the charge qo on plate 1 is divided into two parts: Figure 2.

This also refers to the field intensity:. Distribution 01 an induced charge. A point charge q is placed between two large parallel conducting plates 1 and 2 separated by a distance 1. Find the tutal charges q1 and q2 induced on each plate, if the plates are connected by a wire and the charge q is located at a distance 11 from the left plate 1 Fig.

Let us use the superposition principle. We mentally place somewhere on a plane P the same charge q. If we now distribute uniformly on the surface P a certain cnarge with surface density a, the electric field can be easily calculated Fig. The plates are connected by the wire, and hence the potential.

Electric Field in Dielectrics 3. Polarization of Dielectrics Dielectrics. Dielectrics or insulators are substances that practically do not conduct electric current.. When even a neutral dielectric is introduced into an external electric field, appreciable changes are observed in the field and in the dielectric itself.

In order to understand the nature of these phenomena we must take into consideration that dielectrics consist either of neutral molecules or of charged ions located at the sites. The molecules can be either polar or nonpolar.

In a polar molecule, the centre of "mass" of the negative charge is displaced relative to the centre of "mass" of the positive charge.

As a result, the molecule acquires an intrinsic dipole moment p. Under the action of an external electric field, dielectric is polarized. This phenomenon consists in the following. L nder the action of an external field the dipole moments acquire predominant orientation in' the direction of the external field.

Finally, in dielectric crystals? It should be noted that under normal conditions the displacements of charges are very small even in comparison with the dimensions of the molecules. This property is inherent in dielectrics which are called electrets they resemble permanent magnets. Bulk and Surface Bound Charges.

As a result of polarization, uncompensated charges appear on the dielectric surf! Suppose that we have a plate made of a neutral inhomogeneous dielectric Fig. Switching on of the external field leads to a displacement of the positive charges along the field and of the negative charges against the field, and the two distributions will be shifted relative to one another Fig.

As a result, uncompensated charges will appear on the dielectric surface as well as in the bulk in Fig. It should be noted that the re-. Uncompensated charges appearing as a result of polarization of a dielectric are called polarization, or bound, charges. The latter term emphasizes that the displacements of these charges are limited. We shall denote bound charges by a prime q , p , J. Thus, in the general case the polarization of a dielectric! We shall call the charges that do not constitute dielectric molecules the extraneous charges.

The Field in a Dielectric. The field E in a dielectric is the term applied to the superposition of the field E of extraneous charges and the field E' of bound charges: Clearly, the field E in the dielectric defined in this way is also a macroscopic field. It is natural to describe polarization of a dielectric with the help of the dipole moment of a unit volume. If an. In order to characterize the polarization at a given POIllt, we must mentally isolate an infinitesimal volume!

This vector is numerically equal to the dipole moment of a unit volume of the substance. There are two more useful representations of vector P. Let a volume! We multiply and divide the right-hand side of 3.

Another expression for P corresponds to the model of a dielectric as a mixture of positive and negative "fluids". Let us isolate a very small volume! Dividing both sides of this formula by! Relation Between P and E. Experiments show that for a large number of dielectrics and a broad class of phenomena, 'Polarization P linearly depends on the field E in a dielectric. This quantity is independent of E and characterizes the properties of the dielectric itself.

These are some ionic crystals see footnote on page oR and ferroeZectrics. Then it is clear that the positive charge p: Besides, the negative charge p: But be know that the transport of a negative charge in a certain direction is equivalent to the transport of the positive charge in the opposite direction.

Taking this into account, we can write the expression for the total bound charge passing through the. IL dS cos a. We shall show that the field of P has the following remarkable and important property. It turns out that the flux of P through an. Next, according to 3. S, we find the total charge that left the volume enclosed by the surface S upon polarization. This charge is equal to. This equation expresses the Gauss theorem for vector P. Proof of the theorem.

Let an arbitrary closed surface S envelope a part of a dielectric Fig. Let us find the charge which passes through an element dS of the closed surface S in the outward direction Fig. Clearly, the charge leavlllg the volume must be equal to the excess bound charge remaining within the surface S, taken with the opposite sign.

Thus, we arrive at 3. Differential form of Eq. Equation 3. This equation can be obtained from 3. For this purpose, it is sufficient to replace E by P and p by p'.. When Is p' Equal to Zero in a Dielectric? We shall show that the volume density of excess bound charges in a dielectric is equal to zero if two conditions are simultaneously satisfied: Indeed, it follows from the main property 3.

The remaining integral is just the algebraic sum of all the charges-extraneous and bound-inside the closed surface S under consideration, Le. Then, after cancelling out dV, Eq. Thus, if we place a homogeneous isotropic dielectric of any shape into an arbitrary electric field, we can be sure that its polarization will give rise only to the surface bound charge, while the bulk excess bound charge will be zero at all points of such a dielectric, Boundary Conditions for Vector P.

Let us consider the behaviour of vector P at the interface between two homogeneous isotropic dielectrics.. We have just shown that in. For this purpose, e u ' 3. Let 11 be the common normal to t e interface at jl. We shall always draw vector n froDm. The SIgn of the pro-. Formula 3.

In accordance with 3. Here, too, the sign of En determines the sign of a'. A Remark about th ' Field of Vector P. Relations 3. The hound charge determines the flux of vector P through a closed surface S rather than the field of P.

Moreover, this flux is determined not hy the whole hound charge hut by its part enclosed hy the surface S. Since the sources of an electric field E are all the electric charges-extraneous and hound, we can write the Gauss theorem for the field E in the following form:.

The appearance of the hound charge q' complicates the analysis, and formula 3. Indeed, this formula expresses the properties of unknown field E in terms of the bound charge q' which in turn is determined hy unknown field E.

This difficulty, however, can he overcome hy expressing the charge q' in terms of the flux of P hy formula 3. Thus, we have defined an auxIlIary vector. Howe 't. The 2quantlty D IS measured in coulombs per square metre elm. Relation Between Vectors D and.

Substituting this expression into 3. Let us illustrate what was said above by several examples. Example t. The dielectric constant e as well as x is the basic electric characteristic of a dielectric. The value of 8 depends on the nature of the dielectric and varies between the values slightly differing from unity for gases and several thousands for some ceramics. For anisotropic dielectrics, these vectors are generally noncollinear.

The field D can be graphically represented by the lines of vector D, whose direction and density are determined in the same way as for vector E. The lines of E may emerge and terminate on extraneous as well as bound charges. We say that any charges may be the sources and sinks of vector E.

The sIJurces and sinks of field D, however, are only extraneous charges, since only on these charges the lines of D emerge and terminate. The lines of D pass without discontinuities through the regions of the field containing bound charges.

A Remark about the Field of Vector D. The field of vector D generally depends on extraneous as well as bound charges just as the field of vector E. However, in certain cases the field of vector D is determined only by extraneous charges. It is just the cases for which vector D is especially useful. At the same time, this may lead to the erroneous conclusion that vector D always depends only on extraneous charges and to au incorrect interpretation of the laws 3.

These laws express only a certain property of field D but do not determine this field proper. Find"'ihe projectiun 1,', of Iield intensity E as a functiun of the distance r frolll the centre of this sphere.

Hence we can lind lj, and then, usillg formula 3. Figure 3. IsotropIc dielectrrc Fig. Since in o. The removal of the dielectric will change the held E, and hence the lield D. However the flux of vector D through the surface SWill remalll the same in ;pite of the change in the lield D. Let us consider a system C?

What can we say about the fIeld Dt. If in the lower region of the contour we take the n n.. Fig, 3. Suppose that, for greater generality, an extraneous su: The required conditions cari be easily obtained with the help of two theorems: Boundary Condition for Veclor E.

Let the field near the lDterface be E 1 in dielectric 1 and E 2 in dielectric 2. Then, in accordance With the tllf'orem on circulation of vector E we IJave '.

Boundary Condition for Vector D. The cross sectIOn of the cylinder must be such that vector D is the same within each of its endfaces. Then, in accordance'with the Gauss theorem for vector D, we have D 2n.! It follows from this relation that the normal component of vector D generally has a discontinuity when passing Boundary Condition 'on the Conductor-Dielectric Interface.

If medium 1 is a conductor and medium 2 is a dielectric see Fig. This means that lines of D and E will form a larger angle with the normal to the interface in the dielectric with a larger value of 8 in Fig.

Let us represent graphically the fields E and D at the i II ["ri'ace hetween two homogeneous dielectrics 1 and 2, assuming that. Considering that the tangential component of vector E remains unchanged anrl using Fig. Taking into aceount the above conditions, we obtain the law of refraction of lines E, and hence of lines D: In this case the normal components of vector l do not have a discontinuity and tUI'll out to he tlte same on different sides of the interfaGo.

Thus, in the absence of extraneous charges at the interface hetween two homogeneous isotropic dielectrics, the components E, anfl J n vary continuollsly dnring a transition through this interface, while the components En and D, have discontinuities.

Hdl'action o[ E and D Lines. The boundary conditions which we obtained for the components of vectors E and D at the interface between two dielectrics indicate as will be shown later that these vectors have a break at this interface, Le.

This means, according to ;3. Bound Charge at the Conductor Surface. If a homogeneous dielectric adjoins a charged region of the surface of a COIlductor, bound charges of a certain r1en: Let us now apply the Gauss theorem to vector E in the same way as it was done while deriving formula 2. I t can be seen that the surface density G' of the bound charge in the dielectric is unambiguously connected with the surface density G of the extraneous charge on the conductor, the signs of these charges being opposite.

Field in a Homogeneous Dielectric. It was noted in Sec. It is only clear that the distribution of these charges depends on tho nature and shape of the substance as well as on the configuration of the external field Eo. Consequently, in the general case, while solving the problem about the resultant field E in a dielectric, we encounter serious difficulties: An exception is the case when the entire space where there is a field Eo is filled by a homogeneous isotropic dielectric.

Let us consider this case in greater detail. Suppose that we have a charged conductor or several conductors in a vacuum. Normally, extraneous charges are located on conductors. As we already know, in equilibrium the field E inside the conductor is zero, which corresponds to a certain unique distribution of the surface charge G. Let the fIeld created in the space surrounding the conductor be Eo.

Let us now fill the entire space of the field with a homogeneous dielectric. As a result of polarization, only surface bound charges G will appear in this dielectric at the interface with the conductor. According to 3. This means that the distribution of surface charges extraneous charges a and bound charges a' at the conductordielectric interface will be similar to the previous distribution of extraneous charges G , and the configuration of the resultant field E in the dielectric will remain the same as in the absence of the dielectric.

Only the magnitude of the field at each point will be different. It turns out that formulas 3. In the eases indicated above, the intensity E of the field of bound charges is connected by a simple relation with the polarization P of the dielectric, namely, E'. Thus, if a homogeneous dielectric fills the entire space oeeupied by a field, the intensity E of the field. Hence it follows that potential cp at all points will also decrease by a factor of e: The same applies to the potential difference: In the simplest case, when a homogeneous dielectric fills the entire space between the plates of a capacitor, the potential difference U between its plates will he by a factor of e less than that in the absence of dielectric naturally, at the same magnitude of the charge q on the plates.

It should be noted that this formula is valid when the entire space between the plates is filled and edge effects are ignored. Polarization of a dielectric and the bound charge.

Basic Laws of Electromagnetism - I. E. Irodov by Blog da Engenharia de Produção - Issuu

Find the volume density p' of a bound charge as a fuuction of r within the layer. We shall use Eq. The thickness of the platt! IS 2a. Plot schematic curves for the pro. Let t I' cross-sec 1 Then. Let us take the differential of this expression: Considermg a q as follows:. This result is valid for Il0th sides of the plate.

In order to find the volume density of the bound charge, we use Eq. A homogeneous dielectric has the shape of a spherical layer whose inner and outer radii are a and b. The curve for E r is shown in Fig. Besides, we must take into account the normalization condition: It should be noted that the curve corresponding to the function cp r is continuous. E we shall use the Gauss theaSolution. Let us find E with the help of the Gauss theorem for vector D: Capacitance of a cond t.

Suppose that we have a dielectric sphere which retains polarization. Dy definition th. If the sphere is. Then at an arbitrary point A inside the sphere we have. Find nEoE. It remains for us to take into account that in accordance with 3. The surface rmg. Substituting these expressions into 1 , we obtain. The dielectric's permittivity is e. Find 1 the surface density of the bound charge as a function of the distance r from the point charge '1 and analyse the obtained result; 2 the total hound charge on the surface of the dielectric.

Let liS Use the continuity of the normal component vector D at the dielectric-vacuum interface Fig. F' mtimte d the magmtudes 0 vectors ty e. II from the continuity of the normal Solution. Boundary conditions. In the vicinity of point A Fig. Indeed the displacement dl l of charge 1 in system K can be rep;esented as the displacement dl 2 of.

Approach to Interaction. The energy approach to lllteractlOn between electric charges is, as will be shown rather fruitful in respect of its applications. First of all, let us find out how we can arrive at the concept of the energy of interaction in a system of charges. Let us first consider a system of two point charges 1 and 2. We shall [lOd the algebraic sum of the elementary works of the forces F I and F 2 of interaction between the charges.

Suppose" that ill a certain system of reference K the charges were displaced by dl l and dl 2 during the time. In other words, the work 6A l. Z does not depend on the choice of the initial system of reference. Consequently, the work of the given force in the displacement dli can be represented as the decrease in the potential energy of interaction between the pair of charges under consideration:.

Thus, the energy of mteractlOn for a system of point charges is. Each term of this sum depends on the distance betwoon corresponding charges, and hence the energy W of the given system of charges is a function of its configuration.

Similar arguments are obviously valid for a system of any number of charges. Consequently, we can state that to each configuration of an arbitrary system of charges, there corresponds a certain value of energy W, and the work of all the forces of interaction upon a change in this configuration is equal to the decrease in the energy W 6A. Energy of Interaction. Let us find the expression for the energy W.

We represent each term W ik in a symmetric form: The energy of interaction fn.. Another approach to the solution of this problem is based on formula. The p,0t. Each sum in the parentheses is the energy WI of interaction between the ith charge with all the remaining charges. Hence the latter expression can be written in the form. Total Energy of Interaction. If the charges are arranged continuously, then, representing the system of charges.

This expression can be generalized to a system of an ubitrary number of charges, since the above line of reasoning. A similar expression can be written, for example, for a surface distribution of charges. For this pmpose, we must replace in 4. Expression 4.

Actually, this is not so since the two expressions differ essentially. The origin of this difference is in different meanings of the potential jJ appearing in these expressions.

Let us explain this difference with the help of the following example. Let us find the energy W of the given system by using both formulas. According to 4. Clearly, the result will be completely different: We shall return to this question in Sec.

Now, we shall use formula itA for obtaining several important results. Energy of an Isolated Conduetor. Let a conductor. The remallllllg! Energy of a Capacitor. The energies WI and W 2 are called the intrinsic energies of charges ql and q2' while W 12 is the energy of interaction between charge ql and charge q2' Thus, we see that the energy W calculated by formula 4.

U is the potential difference across its plates. We shall show that formula,s 4. For t! Integruting this expression over q' between 0 and q, we obtain A. Moreover, the expression o.

Thus, w. Obviously, all this applies to 4. Energy of Electric Field. On Loc? Formula 4. It turns out, however, t? This formula is valid for the uniform field which fills the volume V.

The integrand in this equation has the meaning of the? This leads us to a very Important and fruitful physic'al idea about localization ot energy in the field. This assumption was confirmed in experiments with fields varying in time. It is the domain where we encounter phenomena that can be explained with the help. Expenments show that electromagnetic waves carry energy. This circumstance confirms the idea that the field itself is a carrier of energy.

The last two formulas show that the electric energy is distributed in space with the volume density footE'. PosItIvely charged conductor. We I' y mes of E Fig.

I I' pro uct m of the tube andeh a 1. Let us consider two exam l ' I l. A point char I'. The e Fmd the electric energy contaiKed in thlJuadl. IS Ie ectrIC layer. We mentally isolate in the d' 1. I' energy localized in this. I t is assumed that the charge q of the capacitor remains unchanged and that the dielectric fills the entire space between the capacitor plates.

The capacitance of the capacitor in the abseace of the dielectric is C. The work against the electric forces in this system. Bearing in mind that the magnitude of vector D will not change as a result of the removal of the plate, 1.

An analysis of formula ,'1. At first glance this may seem strange: As a matter of fact. Therefore, under the energy of the field in the dielectric we must understand the -sum of the electric energy proper and an additional work which is accomplished during polarization of the dielectric.

In order to prove this, let lIS substitute into 4. P -2 2'. The first term on the ri ht-ha d ' '. Suppose that we have a system of two clwrged bodies in a vacuum.

Therefore, according ,to 4. The first two integrals in 4. The following impprtant circumstances should be mentioned in connection with formula 4. The intrinsic energy of each charged body is an essentially positive quantity. The total energy 1. This can be readily seen from the fact that the in-tegrand contains essentially positive quantities. The intrinsic energy of hod ies remains constant upon all possible displacements that do not. In such cases, the changes in Ware completely determined only by the changes in the interaction energy W 2' In particular, this is just the mode of behaviour of the energy of a system consisting of two point charges upon a change in the distance between them.

Unlike vector E, the energy of the electric field is not an additive quantity, Le. In particular, if E increases n times everywhere, the energy of the field increases n2 times. Here we assume that these displacements do not cause the transformation of electric energy into other kinds of energy. Integral 1. Electrostatic Field in a Vacuum This integral is taken along a certain line path and is therefore called the line integral. The quantity cp r defined in this way is.

The arrows on the contour indicate the direction of circumvention. This means that actually there are no closed lines of E in an electrostatic field: Till now we considered the description of electric field with the help of vector E. I electrostatic field shown in Fig.

This immediately becomes clear if we apply the theorem on circulation of vector E to the closed contour shown in the figure by the dashed line. Potential 27 Indeed.

It remains for us to consider the this case E 1 dl and E two horizontal segments of equal lengths. The figure shows that the contributions to the circulation from these regions are opposite in sign. Is the configuration of an I. It will be shown that the second method has a number of significant advantages.

With such a special choice of the contour. The fact that line integral 1. If we change wo by a certain value AT. This function will be the potential cp. The unit of potential is the volt V. We can make it even simpler. We can conditionally ascribe to an arbitrary point 0 of the field any value cpo of the potential. A comparison of 1. Let us use the fact that formula 1.

For this purpose. Let us apply this method for finding the potential of the. It is determined. Electrostatic Field in a Vacuum called the field potential. Potential of the Field of a Point Charge. The value of this constant does not play any role. Then the potentials of all other points of the field will be unambiguously determined by formula 1. In accordance with the principle of superposition.

The quantity appearing in the parentheses under the differential is exactly y r. Potential of the Field of a System of Charges. Here we also omitted an arbitrary constant. Let a system consist of fixed point charges q1. By using formula 1. Potential 29 field of a fixed point charge: This is in complete agreement with the fact that any real system of charges is bounded in space. Since the additive constant contained in the formula does not play any physical role.

Relation Between Potential and Vector E It is known that electric field is completely described by vector function E r.

Let us consider this question in greater detail. A comparison of this ex-. Let the displacement dl be parallel to the X-axis. If the charges are located only on the surface S. A similar expression corresponds to the case when the charges have a linear distribution. Knowing this function. The relation between cp and E can be established with the help of Eq. Electrostatic Field in a Vacuum If the charges forming the system are distributed continuously.

And what do we get by introducing potential? First of all. Taking this into consideration. This is exactly the formula that can be used for reconstructing the field E if we know the function p r. We write the righthand side of 1. Then with the help of formula 1. Relation between Potential and Vector t 81 pression with formula 1.

It can be seen that in this case the field E is uniform.

Find the field intensity E if the field potential has the form: Then Eq. Having determined E x. In a similar way. I Figure 1. Field intensity will be higher in the regions where equipotential surfaces are denser "the potential relief is steeper". Where is the magnitude of the potential gradient higher? At which point will the force acting on the charge be greater? It immediately shows the direction of vector E. We shall show that vector E at each point of the surface is directed along the normal to the equipotential surface and towards the decrease in the potential.

It is expedient to draw equipotential surfaces in such a way that the potential difference between two neighbouring surfaces be the same. This means that vector E is normal to the given surface. Such a representation can be easily visualized. Then the density of equipotential surfaces will visually indicate the magnitudes of field intensities at different points. Electrostatic Field in a Vacuum placement dl is equal to the directional derivative of the potential this is emphasized by the symbol of partial derivative.

Equipotential Surfaces. Since vector E is normal to an equipotential surface everywhere. Such a pattern can be used to obtain qualitative answers to a number of questions. The dashed lines correspond to equipotential surfaces. Let us introduce the concept of equipotential surface.

This means that the required work is equal to the decrease in the potential energy of the charge q' upon its displacement from point 1 to 2. Then what is the use of introducing potential? There are several sound reasons for doing that. If we know the potential cp r. The concept of potential is indeed very useful. This problem can be easily solved with the help of potential. Calculation of the work of the field forces with the help of formula 1. This means that we cannot calculate the work by evaluating the integral q E dl in this case.

Find the work of the field forces done in the displacement of a point charge q' from the centre of the ring to infinity.

This is a considerable advantage of potential. W2 1 where cp1and cp2 are the potentials at points 1 and 2. Let us note here that this does not apply to a comparatively small number of problems with high symmetry. Since the distribution of the charge q over the ring is unknown.

A charge q is distributed over a thin ring of radius a. It was noted earlier that electrostatic field is completely characterized by vector function E r. It turns out in many cases that in order to find electric field intensity E. The dipole field is axisymmetric.

When the dipole field is considered. Electrostatic Field in a Vacuum There are some other advantages in using potential which will be discussed later.

Let us first find the potential of the dipole field and then its intensity. This quantity corresponds to a vector directed along the dipole axis. Taking this into account. According to 1. Electric Dipole The Field of a Dipole. Electric Dipole 35 from the negative to the positive charge: The Force Acting on a Dipole..

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Let us place a dipole into a nonuniform electric field. Then the resultant force F acting on the dipole is Fig. It will be shown below that the behaviour of the dipole in an external field also depends on p. It can be seen from formula 1. In order to find the dipole field.

The derivative appearing in this expression is called the directional derivative of the vector. We suggest that the reader prove independently that it is really so. If we are interested in the projection of force F onto a certain direction X. Since the length of this segment is small. Electrostatic Field in a Vacuum of vector 1. This means that generally the force acts on a dipole only in a nonuniform field.

Vector F coincides in direction only with the elementary increment of vector E. We take the positive direction of the X-axis. The Moment of Forces Acting on a Dipole. Let a dipole with moment p be oriented along the symmetry axis of a certain nonuniform field E.

Let us consider behaviour of a dipole in an external electric field in its centre-of-mass system and find out whether the dipole will rotate or not. Such a position of the dipole is stable. For a sufficiently small dipole length. Electrostatic Field in a Vacuum follows: Both these motions are simultaneous. The Energy of a Dipole in an External Field. A dipole is the system of two charges.

If it is displaced from this. To within a quantity of the second order of smallness. It can be easily seen that 1 cr — dS cos O. In the immediate vicinity of the point 0. Problems 39 position. Find the electric field intensity E on the axis of this disc at the point from which the disc is seen at an angle Q. It is clear from symmetry considerations that on the disc axis vector E must coincide with the direction of this axis Fig. A thin nonconducting ring of radius R is charged with a linear density X.

Integrating 1 over cp between 0 and 2n. Find the magnitude and the direction of the field intensity at the point separated from the filament by a distance y and lying on the normal to the filament. In order to find the projection Ey. A semi-infinite straight uniformly charged filament has a charge k per unit length. Let us start with Ex.

In our case. Find the electric field intensity E at the centre of the ring. The problem is reduced to finding Ex and Ey.

The symmetry of this distribution implies that vector E at the point 0 is directed to the right. The given charge distribution is shown in Fig. Find the charge within a sphere of radius' R with the centre at the origin. Since the field E is axisymmetric as the field of a uniformly charged filament. In accordance with the Gauss theorem. The intensity of an electric field depends only on the coordinates x and y as follows: The Gauss theorem.

Electrostatic Field in a Vacuum the centre of the sphere. This conclusion is valid regardless of the ratio between the radii. Find the charge of the sphere for which the electric field intensity E outside the sphere is independent of r. Then at an arbitrary point A Fig. Using the Gauss theorem.

Let the sought charge of the sphere be q. Find the electric field intensity E in the region of intersection of two spheres uniformly charged by unlike charges with the volume densities p and —p.

Find the value of E. After integration. We can consider the field in the region of intersection of the spheres as the superposition of the fields of two iiniforlmy charged spheres. Let us consider two spheres of the same radius. For a very small 1. Problems 43 of the spheres and of the distance between their centres. The thickness of the charged layer at the points determined by angle 15 Fig.

In particular. Let us first find vector E: Using the solution of the previous problem. Suppose that the centres of the spheres are separated by the distance 1 Fig. The potential of a certain electric field has the z2. Find the distribution of the volume charge p r within the sphere. The potential of the field inside a charged sphere depends only on the distance r from its centre to the point under consideration in the following way: Let us first find the field intensity.

In order to simplify integration. After substituting these expressions into integral 1. In order to find the intensity E of a real field at a certain point at a given instant.

The solution of this problem is obviously not feasible. It is different at different points of atoms and in the interstices. A Conductor in an Electrostatic Field 2. The real electric field in any substance which is called the microscopic field varies abruptly both in space and in time. Field in a Substance Micro.

In any case. Find the force of interaction between two point dipoles with moments pi and p2. In certain regions of the substance. In many cases it is sufficient to have a simpler and rougher description which we shall be using henceforth. Induced charges create an additional electric field which in combination with the initial external field forms the resultant field. This phenomenon is called the electrostatic induction.

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It will be shown later that the distri-. This averaging is performed over what is called a physically infinitesimal volume. If any substance is introduced into an electric field. Under the electric field E in a substance which is called the macroscopic field we shall understand the microscopic field averaged over space in this case time averaging is superfluous.

A Conductor in an Electrostatic Field that it would be impossible to use it. The averaging over such volumes smoothens all irregular and rapidly varying fluctuations of the microscopic field over the distances of the order of atomic ones.

Knowing the external field and the distribution of induced charges. We shall have to consider these questions in greater detail. In both cases. And this means that there are no excess charges inside the conductor. The absence of a field inside a conductor indicates.

Let us place a metallic conductor into an external electrostatic field or impart a certain charge to it. Fields Inside and Outside a Conductor 47 bution of induced charges is mainly determined by the properties of the substance. The fact that the surface of a conductor is equipotential implies that in the immediate vicinity of this surface the field E at each point is directed along the normal to the surface.

Excess charges appear only on the conductor surface with a certain density o. It should be noted that the excess surface charge is located in a very thin surface layer whose thickness amounts to one or two interatomic distances. This can be easily explained with the help of the Gauss theorem. If the opposite were true.

This displacement current will continue until this practically takes a small fraction of a second a certain charge distribution sets in. As we move away from this system. Thus we can calculate its value at the centre 0 of the sphere. Potential cp is the same for all points of the sphere. As a result of electric induction. The field lines are normal to the conductor surface. We shall show that the electric field intensity in the immediate vicinity of the surface of a conductor is connected with the local charge density at the conductor surface through a simple relation.

Find the potential of an uncharged conducting sphere provided that a point charge q is located at a distance r from its centre Fig. Suppose that the region of the conductor surface we are interested in borders on a vacuum. The Field Near a Conductor Surface. Figure 2. The field of these charges will in turn cause a redistribution of charges on the surface of the left sphere. The solid lines in the figure are the lines of E.

This relation can be established with the help of the Gauss theorem. The field itself in this case resembles more and more the field of a point charge q. But since all induced charges are at the same distance a from the point 0 and the total induced charge is equal to zero. A Conductor in an Electrostatic Field conductor. Hence for a closed surface we. The intensity E is determined by all the charges of the system under consideration as well as the value of a itself.

Cancelling both sides of this expression by AS. Forces Acting on the Surface of a Conductor Let us consider the case when a charged region of the surface of a conductor borders on a vacuum. This is not so. Relation 2. Forces Acting on the Surface of a Conductor 49 shall take a small cylinder and arrange it as is shown in Fig. AS is the cross-sectional area of the cylinder and a is the local surface charge density of the conductor.

Then the flux of E through this surface will be equal only to the flux through the "outer" endface of the cylinder the fluxes through the lateral surface and the inner endface are equal to zero. Let us find this relation. It should be noted at the very outset that E0 is not equal to the field intensity E in the vicinity of the given surface element of the conductor. The resultant field both inside and outside the conductor near the area element AS is the superposition of the fields E0 and E0.

A Conductor in an Electrostatic Field. Let Ea be the intensity of the field created by the charge on the area element AS at the points that are very close to this element. On both sides of the area element AS the field E0 is practically the same. In this region. The resultant force acting on the entire conductor can be found by integrating this equation over the entire conductor surface: Equation 2.

Find the expression for the electric force acting in a vacuum on a conductor as a whole. This means that the excess charge is distributed on a conductor with a cavity in the same way as on a uniform conductor.

That there is no electric field inside an empty cavity can be proved in a different way. The quantity Fu is called the surface density of force. External charges. Properties of a Closed Conducting Shell It was shown that in equilibrium there are no excess charges inside a conductor. This forms the basis of electrostatic shielding. Multiplying 2. In practice. Suppose also that the entire external space is filled by a conducting medium.

Since the field E is equal to zero inside the conductor. Let us now consider the case when the cavity is not empty but contains a certain electric charge q or several charges. It is clear that the line integral of vector E along this contour differs from zero. This does not exclude the situation depicted in Fig 2. In equilibrium. In equilibrium the charges induced on the surface of the cavity are arranged so as to compensate completely.

A Conductor in an Electrostatic Field of the conductor. They will also remain unchanged upon the displacement of charges outside the shell. The same refers to the field inside the cavity if it contains charges and to the distribution of charges induced on the cavity walls.

Find the potential cp at the point P lying outside the shell at a distance r from the centre 0 of the outer surface. Properties of a Closed Conducting Shell 53 space outside the cavity.

The field at the point P is determined only by charges induced on the outer spherical surface since. The space on one side of this plane is electrically independent of the space on its other side. Since the conducting medium is electrically neutral everywhere. This must be interpreted as follows: We arrive at the following important conclusion: We shall repeatedly use this property of a closed conducting shell. A point charge q is within an electrically neutral shell whose outer surface has spherical shape Fig.

Image Method Frequently. Using the uniqueness theorem. It should be recalled that if we know the potential cp r.

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Thus we arrive at a physically absurd conclusion. It can be proved theoretically that this problem has a unique solution. The Poisson and Laplace equations. This statement is called the uniqueness theorem. We must find the potential cp r at any point of the field between the conductors.

From the physical point of view. Let us derive the differential equation for the function cp potential. The analytic solutions of these equations were obtained only for a few particular cases. The solution of Eqs. The solution of the Laplace equation satisfying this condition can immediately be found. Hence it immediately follows that the uniqueness of the field E determines the uniqueness of the charge distribution over the conductor surface.

As for the uniqueness theorem. Prove that in an empty cavity of a conductor the field is absent. Let us consider this method by using a simple example of a point charge q near an infinite conducting plane Fig.

The potential cp must satisfy the Laplace equation 2. This is an artificial method that makes it possible to calculate in a simple way the electric field in some unfortunately few cases.

Image Method. The idea of this method lies in that we must find another problem which can be easily solved and whose solution or a part of it can be used in our problem. In our case such a simple problem is the problem about two charges: If a solution of the problem satisfies the Laplace or Poisson equation and the boundary conditions.

In accordance with the uniqueness theorem. General Problem of Electrostatics 55 static case the charge is distributed over the surface of a conductor in a unique way as well. It should be noted that we can arrive at this conclusion proceeding from the properties of a closed conducting shell [see Sec. We must only bear in mind that the "effect" of the fictitious charge extends only to the half-space where the real charge q is located.

Find the location of fictitious point. This is precisely what' is meant when we say that the fictitious charge produces the same "effect" as all the induced.

In another half-space the field is absent. Summing up. Let us consider one more example. According to the uniqueness theorem. The point charge q can be considered to be the limiting case of a small spherical conductor whose radius tends to zero and potential to infinity. A Conductor in an Electrostatic Field The field of this system is well known its equipotential surfaces and field lines are shown in Fig.

A point charge g is placed between two mutually perpendicular half-planes Fig. In order to calculate this field. The fictitious charge q' creates in the upper half-space the same field as that of the charges induced on the plane.

The image method proves to be very effective if this can be done with the help of sufficiently simple configurations. Having found this configuration of point charges another problem. The quantity C 2. Capacitors 57 charges whose action on the charge q is equivalent to the action of all charges induced on these half planes. Let us consider a solitary conductor.

It is numerically equal to the charge that must be supplied to the conductor in order to increase its potential by unity. Capacitors Capacitance of an Isolated Conductor. These three fictitious charges create just the same field within the "right angle" as the Fig. Find the capacitance of an isolated conductor which has the shape of a sphere of radius R.

It can be seen from formula 2. Only with such a configuration of the system of four charges we can realize the required "trimming". One or two fictitious charges are insufficient in this case. The capacitance depends on the size and shape of the conductor. Experiments show that the charge q of this conductor is directly proportional to its potential cp we assumed that at infinity potential is equal to zero: This is due to the fact that the field of the given conductor causes a redistribution of charges on the surrounding bodies.

This circumstance made it possible to create the system of conductors. This means that the lines of E emerging on one plate must terminate on the other. It corresponds to the capacitance of an isolated sphere 9 x km in radius. For this reason. A Conductor in an Electrostatic Field tial cp. This means that its capacitance increases. If a conductor is not isolated. The simplest capacitor consists of two conductors plates separated by a small distance.

This unit of capacitance is called the farad F. Such a system is called a capacitor. In order to exclude the effect of external bodies on the capacitance of a capacitor. In accordance with 1. The farad is a very large quantity. Then negative induced charges will be nearer to the conductor than the positive charges. In actual practice. The capacitance of a capacitor depends on its geometry size and shape of its plates.

Capacitance of a Parallel-plate Capacitor. Let the radii of the inner and outer capacitor plates be a and b respectively. If the charge of the capacitor is q. Let us derive the expressions for the capacitances of some capacitors assuming that there is a vacuum between their plates.

Substituting this expression into 2. Unlike the capacitance of an isolated conductor. The capacitance of a real plane capacitor is determined by this formula the more accurately the smaller the gap h in comparison with the linear dimensions of the plates. The basic characteristic of a capacitor is its capacitance.

Capacitance of a Spherical Capacitor. Er 4nso r-. This capacitor consists of two parallel plates separated by a gap of width h. Capacitors 59 i. The influence of the medium on the capacitance of a capacitor will be discussed in Sec. By using the same line of reasoning as in the case of a spherical capacitor. As a result of electrostatic induction.

Like in the previous case. According to the principle of superposition. On the determination of potential. A point charge q is at a distance r from the centre 0 of an uncharged spherical conducting layer. Capacitance of a Cylindrical Capacitor. The force acting on a surface charge. An uncharged metallic sphere of radius R is placed into an external uniform field. We write the expressions for potentials outside the system pH and in the region between the spheres cpi: What charge q2must be placed onto the outer sphere of radius R2 to make the potential of the inner sphere equal to zero?

What will be the dependence of potential cp on the distance r from the centre of the system? Plot schematically the graph of this dependence.

Problems charges on the outer surface. A system consists of two concentric spheres. Its value can be easily found from the boundary condition: According to 2. The cp r. Find the magnitude of the resultant electric force acting on like charges. Using the image method. A Conductor in an Electrostatic Field as the sum integral of the projections of elementary forces 1 onto the Z-axis: Find the density of surface charges induced on the plane as a function of the distance r from the base of the perpendicular dropped from the charge q onto the plane.

Integrating this expression over the half-sphere i. Image method. An attempt to solve this problem in a different way through potential leads to an erroneous result which differs from what was obtained by us by a factor of two. A thin conducting ring of radius R. A point charge q is at a distance 1 from an infinite conducting plane.

It can be easily seen that in accordance with the image method. Problems where the minus sign indicates that the induced charge is opposite to sign to the point charge q. Find 1 the surface charge density at a point of the plane. Find the work of the electric force acting on the charge q done upon its slow removal to a very large distance from the plane. The half-planes forming the angle AOB go to infinity. Thus we have already answered the first question: Three unlike point charges are arranged as shown in Fig.

Hence the action of the charges induced on the conducting half-planes is equivalent to the action of the fictitious charge q placed in the lower left corner of the dashed square. A Conductor in an ElectrostaticField on the other side of the conducting plane Fig. The magnitude of each of the charges is I q I and the distances between them are shown in the figure.

Let us now use the formulas we already know. By reducing the system to four point charges. Find 1 the total charge induced on the conducting half-planes and 2 the force acting on the charge q. It follows from Fig. Capacitance of parallel wires. Problems the required force see Fig. Let us mentally charge the two wires by charges of the same magnitude and opposite signs so that the charge per unit length is equal to X. Two long straight wires with the same cross section are arranged in air parallel to one another.

This also refers to the field intensity: An electric field appears in the gaps between the plates. The area of each plate is equal to S. Let us use the superposition principle. If we now distribute uniformly on the surface P a certain charge with surface density o. Find the capacitance of this system between points 1 and 2.

We mentally place somewhere on a plane P the same charge q. Find the total charges q1and q2induced on each plate. Under the action of the dissipation field appearing between these plates edge effect. A point charge q is placed between two large parallel conducting plates 1 and 2 separated by a distance 1. The outer plates are connected by a conductor. A Conductor in an Electrostatic Field It follows from 1.

The plates are connected by the wire. Four identical metallic plates are arranged in air at the same distance h from each other. It should be noted that as follows from the symmetry of the system.

Let us charge the plates 1 and 2 by charges go and — go. Distribution of an induced charge. Polarization of Dielectrics Dielectrics. The formulas for charges q1 and q2 in terms of q have a similar form. E2x the minus sign indicates that the normal n2 is directed oppositely to the unit vector of the X-axis.

Polarization of Dielectrics difference between them is equal to zero. This means that in contrast. Dielectrics or insulators are substances that practically do not conduct electric current. It would be difficult.

Eliminating Eix and E2. Under the action of an external field. It should be noted that under normal conditions the displacements of charges are very small even in comparison with the dimensions of the molecules. Electric Field in Dielectrics When even a neutral dielectric is introduced into an external electric field. In order to understand the nature of these phenomena. For further discussion it is only important that regardless of the polarization mechanism. This property is inherent in dielectrics which are called electrets they resemble permanent magnets.

Under the action of an external electric field. If a dielectric consists of polar molecules. In a polar molecule. The molecules can be either polar or nonpolar. This phenomenon consists in the following. If a dielectric is made up by nonpolar molecules.

This is because the dielectric is acted upon by a force. Nonpolar molecules do not have intrinsic dipole moments. As a result of polarization.

Suppose that we have a plate made of a neutral inhomogeneous dielectric Fig. To understand better the mechanism E. In the absence of an external field. Bulk and Surface Bound Charges. Switching on of the external field leads to a displacement of the positive charges along the field and of the negative charges against the field. Polarization of Dielectrics 69 ably lower than the intensities of internal electric fields in the molecules.

This figure shows that in the absence of external field. It should be noted that the re-. We denote by and V the magnitudes of the volume densities of the positive and negative charges in the material these charges are associated with nuclei and electrons. We shall denote bound charges by a prime cr. They can move only within electrically neutral molecules.

In order to characterize the polarization at a given point. It can be easily seen that in the case of a plate made ' x and of a homogeneous dielectric. Polarization Definition. The field E in a dielectric is the term applied to the superposition of the field E0 of extraneous charges and the field E' of bound charges: We shall call the charges that do not constitute dielectric molecules the extraneous charges. Electric Field in Dielectrics version of the field direction changes the sign of all these charges.

If an external field or a dielectric or both are nonuniform. It is natural to describe polarization of a dielectric with the help of the dipole moment of a unit volume. The Field in a Dielectric. Uncompensated charges appearing as a result of polarization of a dielectric are called polarization. The latter term emphasizes that the displacements of these charges are limited. Dividing both sides of this formula by AV. Another expression for P corresponds to the model of a dielectric as a mixture of positive and negative "fluids".

Let us isolate a very small volume AV inside the dielectric. This vector is numerically equal to the dipole moment of a unit volume of the substance. We multiply and divide the right-hand side of 3. Experiments show that for a large number of dielectrics and a broad class of phenomena. This quantity is independent of E and characterizes the properties of the dielectric itself. Relation Between P and E.

Let a volume AV contain AN dipoles. There are two more useful representations of vector P. For an isotropic dielectric and for not very large E. Upon polarization. Let us find the charge which passes through an element dS of the closed surface S in the outward direction Fig.

Electric Field in Dielectrics consider only isotropic dielectrics for which relation 3. These are some ionic crystals see footnote on page 68 and ferroelectrics. It turns out that the flux of P through an P a Fig. We shall show that the field of P has the following remarkable and important property. The relation between P and E for ferroelectrics is nonlinear and depends on the history of the dielectric.

Let an arbitrary closed surface S envelope a part of a dielectric Fig. Proof of the theorem. When an external electric field is switched on. Then it is clear that the positive charge Equation 3. But be know that the transport of a negative charge in a certain direction is equivalent to the transport of the positive charge in the opposite direction. Differential form of Eq.

Properties of the Field of P 73 of the positive and negative bound charges as a result of polarization. This equation can be obtained from 3. Boundary Conditions for Vector P. Let us consider the behaviour of vector P at the interface between two homogeneous isotropic dielectrics. We have just shown that in. Electric Field in Dielectrics i.

The remaining integral is just the algebraic sum of all the charges—extraneous and bound—inside the closed surface S under consideration. When Is p' Equal to Zero in a Dielectric? We shall show that the volume density of excess bound charges in a dielectric is equal to zero if two conditions are simultaneously satisfied: Let us find the relation between polarization P and the surface density a' of the bound charge at the interface between the dielectrics.

Disregarding the flux of P through the lateral surface of the cylinder. Considering that the projection of vector P onto the normal n' is equal to the projection of this vector onto the opposite common normal n. The sign of the pro-. We choose the closed surface in the form of a flat cylinder whose endfaces are on different sides of the interface Fig. Let n be the common normal to the interface at a given point.

Properties of the Field of P 75 such a dielectric there is no excess bound bulk charge and only a surface bound charge appears as a result of polarization. Pen — P ln This means that at the interface between dielectrics the normal component of vector P has a discontinuity.

We shall always draw vector n from dielectric 1 to dielectric 2. We shall assume that the height of the cylfp. Formula 3. In accordance with 3. Here, too, the sign of En determines the sign of a'.

A Remark about the Field of Vector P.