wildlifeprotection.info Laws Basic Laws Of Electromagnetism Irodov Pdf

BASIC LAWS OF ELECTROMAGNETISM IRODOV PDF

Sunday, June 9, 2019


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.


Basic Laws Of Electromagnetism Irodov Pdf

Author:MARGIT MANGRICH
Language:English, Spanish, French
Country:Chile
Genre:Personal Growth
Pages:208
Published (Last):02.08.2016
ISBN:877-5-18144-945-1
ePub File Size:17.75 MB
PDF File Size:13.27 MB
Distribution:Free* [*Regsitration Required]
Downloads:38002
Uploaded by: RALEIGH

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.

The problems are closely related to the text and often complement it. Hence they should be analYfed together with the text. In order to emphasize the most important laws of electromagnetism, and especially to clarify the most difficult topics, the author has endeavored to exclude the less important topics.

In an attempt to describe the main ideas concisely, clearly and at the same time correctly, the text has been kept free from superfluous mathematical formulas, and the main stress has been laid on the physical aspects of the phenomena. With the same end in view, various model representations, simplifying factors, special cases, symmetry considerations, etc. SI units of measurements are used throughout the book. However, considering that the Gaussian system of units is still widely used, we have included in Appendices 3 and 4 the tables of conversion of the most important quantities and formulas from SI to Gaussian units.

The most important statements and terms are given in italics. More complicated material and problems involving cumbersome mathematical calculations are set in brevier type. This material can be omitted on first reading without any loss of continuity. The brevier type is also used for problems and examples. The book is intended as a textbook for students specializing in physics for undergraduate students specializing in physics in the framework of the course on general physics.

It can also be used by university teachers. This book can be downloaded from here. In the books attributed to Irodov above, I could not find any data on the book mentioned on the side cover of Problems in General Physics namely, A Laboratory Course in Optics. I tried worldcat, but there seems to be no entry for this particular book in English. So, I do not know whether an English edition of this book was ever published.

Also, I do not know whether there are any books of Irodov which were translated into Indian languages. So please let me know if this is the case. Figure 1. Electric Field 15 problem: It will be shown below that in. These properties. Geometrical Description of Electric Field. The field E defined above has two very important properties. The knowledge of these properties helped to deeper understand the very concept of the field and formulate its laws.

Let us reduce this equation to the form convenient for integration. If we know vector E at each point. In this case 1 k dl cos a.

The density of the lines. On the General Properties of Field E. Such a line is drawn so that a tangent to it at each point coincides with the direction of vector E.

This pattern gives the idea about the configuration of a given electric field. If we have an arbitrary surface S. It should be noted that the choice of the direction of n and hence of dS is arbitrary. This vector could be directed oppositely. Although we considered here the flux of E.

Let us go over to a systematic description of these properties. This quantity is just the flux dO of E through the area element dS. If a surface is closed it is customary to direct the normal n outside the region enveloped by this surface. For the sake of clarity. In a more compact form. The Gauss Theorem Flux of E. Henceforth we shall always assume that this is the case. Then the number of lines piercing the area element dS. This expression is essentially the Gauss theorem: The flux of E through an arbitrary closed surface S has a remarkable property: It should be noted that for a more complicated shape of a closed surface.

Let us first consider the field of a single point charge q. We enclose this charge by an arbitrary closed surface S Fig. The integration of this expression over the entire surface S is equivalent to the integration over the entire solid angle. In accordance with what was said above. In order to prove this. The sum is equal to zero.. To complete the proof of the theorem. In this case. Then the flux of E can be written in the form Then the integration of Eq. Then on the right-hand side of 1.

We must pay attention to the following important circum-. Let us now consider the case when the electric field is created by a system of point charges q1 q2. Electrostatic Field in a Vacuant In particular. In terms of field lines or lines of E. Applications of the Gauss Theorem 19 stance: On the other hand. Can its equilibrium be stable? In order to answer this question.

Suppose that we have in vacuum a system of fixed point charges in equilibrium. For the sake of definiteness.. For the equilibrium of this charge to be stable. On the impossibility of stable equilibrium of a charge in an electric field.

What a remarkable property of electric field! This means that if we displace the charges. Applications of the Gauss Theorem Since the field E depends on the configuration of all charges. Let us consider one of these charges. Only in this case any small displacement of the charge q from the equilibrium position will give rise to a restoring force.

But such a configuration of the field E around the charge q is in contradiction to the Gauss theorem: Let us consider some examples and then formulate several general conclusions about the cases when application of the Gauss theorem is the most expedient. In a more exact form. This field can be easily found as superposition of the fields created by each plane separately Fig. A charge a AS is enclosed within the cylinder. Electrostatic Field in a Vacuum Hence it follows that in any electrostatic field a charge cannot be in stable equilibrium.

Example 3. Suppose that the surface charge density is a. It is clear from the symmetry of the problem that vector E can only be normal to the charged plane. The fact that E is the same at any distance from the plane indicates that the corresponding electric field is uniform both on the right and on the left of the plane. According to the Gauss theorem. Such a configuration of the field indicates that a right cylinder should be chosen for the closed surface as shown in Fig.

The obtained result is valid only for an infinite plane surface. Here the upper arrows correspond. The field of a uniformly charged plane. The field of two parallel planes charged uniformly with densities o. This indicates that a closed surface here should be taken in the form of a coaxial right cylinder Fig. Example 4. Then the flux of E through the endfaces of the cylinder is equal to zero. In the space between the planes the intensities of the fields being added have the same direction.

Applications of the Gauss Theorem 21 to the field from the positively charged plane. It can be easily seen that outside this space the field is equal to zero. This result is approximately valid for the plates of finite dimensions as well. The field of an infinite circular cylinder uniformly charged over the surface so that the charge X. General Conclusions. The field of a spherical surface uniformly charged by the charge q. L13 where Er is the projection of vector E onto the radius vector r coinciding with the normal n to the surface at each of its points.

Example 5. The field of a uniformly charged sphere. Suppose that a charge q is uniformly distributed over a sphere of radius a. The sign of the charge q determines the sign of the projection ET in this case as well. The results obtained in the above. The curve representing the dependence of E on r is shown in Fig.

Hence it determines the direction of vector E itself: Example 6. It is clear that for such a configuration of the field we should take a concentric sphere as a closed surface. Outside this surface the field decreases with the distance r in accordance with the same law as for a point charge. It can be easily seen that for the field outside the sphere we obtain the same result as in the previous example [see 1. This field is obviously centrally symmetric: The Gauss theorem can be effectively applied to calculation of fields only when a field has a special symmetry in most cases plane.

Differential Form of the Gauss Theorem A remarkable property of electric field expressed by the Gauss theorem suggests that this theorem be represented in a different form which would broaden its possibilities as an instrument for analysis and calculation. The number of problems that can be easily solved with the help of the Gauss theorem is limited. The symmetry.

The simple solution of the problems considered above may create an illusive impression about the strength of the method based on the application of the Gauss theorem and about the possibility of solving many other problems by using this theorem.

Differential Form of the Gauss Theorem 23 examples could be found by direct integration 1. If these conditions are not satisfied. In contrast to 1. The expression obtained for the divergence will depend on the choice of the coordinate system in different systems of coordinates it turns out to be different.

It follows from definition 1. Then we substitute this expression into Eq. The eperator v itself does not have any meaning. In order to obtain the expression for the divergence of the field E. In Cartesian coordinates. In this case.. The form of many expressions and their applications can be considerably simplified if we introduce the vector differential operator v.

It becomes meaningfu I. Electrostatic Field in a Vacuum For this purpose. The Gauss theorem in the differential form is a local theorem: We shall be using the latter.

The field lines emerge from the field sources and terminate at the sinks. This property is inherent in the electrostatic field. Potential Theorem on Circulation of Vector E. This is one of the remarkable properties of electric field.

Circulation of Vector K Potential 25 only in combination with a scalar or vector function by which it is symbolically multiplied. If we take a unit positive charge for the test charge and carry it from point 1 of a given field E to point 2.

At the points of the field where the divergence of E is positive. It is known from mechanics that any stationary field of central forces is conservative. A field having property 1. The theorem on circulation of vector E makes it possible to draw a number of important conclusions without resorting to calculations. Since line integral 1. The field lines of an electrostatic field E cannot be closed.

We shall now show that from the independence of line integral 1. Therefore In order to prove this theorem. On the and 2. 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.

Basic laws of electromagnetism pdf download

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?

irodov-basic-laws-of-electromagnetism.pdf

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.. 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. 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. However, in certain cases the Gauss theorem proves to be a very effective analytical instrument since it gives answers to certain principle questions without solving the problem and allows us to determine the field E in a very simple way. Let us consider some examples and then formulate several general conclusions pbout the cases when application of the Gauss theorem is the most expedient.

On the impossibility 01 stable eqUilibrium 01 a charge in an electric field. Suppose that we have in vacuum a system of fixed point charges in equilibrium. Let us consider one of these charges, e. In order to answer this question, let us envelop the charge q by a small closed surface S Fig. For the equilibrium of this charge to be stable, it is necessary that the field E creatl'd by all the remaining charges of the system at all the points of the surface S he directed towards the charge q.

Only in this case any small displacement of the charge q from the eqUilibrium position will gi ve rise to a restoring force, and the equilibrium state will actually he stable. But such a configuration of the field E around the charge q is in contradiction to the Gauss theorem: On tile other hand, the fact that E is equal to zero indicates that at S Illl' p ints of thl' surface S vl'clor E is directed inside it and at some other lliJints it is llirect.

Example 2. The field of a unHormly charged plane. Suppose that the surface charge density is o. It is clear from the symmetry of the problem that vector E can only be normal to the charged plane. Such a configuration of the field indicates that a right cylinder should be chosen for the closed surface as shown in Fig. In the space between the planes the intensities of the fields bdng added have tlw same Hrection, hence the result 1.

J sLands for the magnitude of the surface charge density. It can be easily seen that outside this space the field is equal to zero. The flux through the lateral surface of this cylinder is equal to zero, and hence the total flux through the entire cylindrical surface is 2E l!.

S, where l!. S is the area of each endface. A charge J AS is enclosed within the cylinder. The fact thiit E IS the same at any distance from the plane indicates thatJ the corresponding electric field is uniform both on the right and on the left of. The obtained result is valid only for an infinite plane surface, since only in this case we can use the symmetry considerations discussed above.

However, this result is approximately valid for the region near the middle of a finite uniformly charged plane surface far from its ends. Example 3. The field of two parallel planes charged uniformly with densities J and - J by unlike charges. This fIeld can be easily found as superposition of the f,elds created by each plane separately Fig. Here the upper arrows correspond.

This result is approximately valid for the plates of finite dimensions as well, if only the separation between the plates is considerably smaller than their linear dimensions parallel-plate capacitor. In this case, noticeable deviations of the field from uniformity are observed. In this case, as follows from symmetry considerations, the field is of a radial nature, i.

This indicates that a closed surface here should be taken in the form of a coaxial right cylinder Fig. The sign of the charge q determines the sign of the projection E r in this case! Hence it determines the direction of vector E itself: In other words, inside a uniformly charged spherical surface the electric field is absent. Outside this surface the field decreases with the distance r in accordance with the same law as for a point charge. Example 6. The field of a uniformly charged sphere.

Suppose that a charge g is uniformly distributed over a sphere of radius a.

Obviously, the field of such a system is centrally symmetric, and hence for determining the field we must take a concentric sphere as a closed surface.

It can be easily seen that for the field outside the sphere we obtain the same result as in the previous example [see 1. However, inside the sphere the expression for the field will be different. S since in our case the ratio of charges is equal to the ratio of volumes and is proportional to the radii to the third power. The curve representing the dependence of E on r is shown in Fig.

It follows from 1. We shall be using the latter, more convenient notation. Then, for example, the Gauss theorem 1. The quantity which is the limit of the ratio of t. The dive! D8 out to be. For example, if we forin the scalar product of vector V and vector E, we obtain.

We now make the volume Y Je d. IVI e ot Its sides by. The Gauss theorem in the differential form is a local theorem: This is one of the remarkable properties of electric field. For example, the field E of a point charge is different at different points.

Generally, this refers to the spatial derivatives oE: At the points of the field where the divergence of E is positive, we have the sources of the fieIa positive charges , while at the points where it is negative, we have Binks negative charges.

The field lines emerge from the field sources and terminate at the sinks. Theorem on Circulation of Vector E. It is known from mechanics that any stationary field of central forces is conservative, Le. This property is inherent in the electrostatic field, viz.

If we take a unit positive charge for the test charge and carry it from point 1 of a given field E to point 2. This integral is taken alon '. We shall now show that fm integral 1. This statement is' called th h vector E. Since line integral 1. Therefore ut taken In the opposite n.

A field haVing property 1 I Hence, any electrostatic field.

IS ca led. The theorem on'. Indeed, if the oppollite were true and some lines of field E? This means that actually there are no closed lines of E in an electrostatic field: Is the configuration of an I I electrostatic field shown in Fig.

This immediately becomes t clear if we apply the theorem on circulaI lr I tion of vector E to the closed contour L -. J shown in the figure by the dashed line. With such a special choice of the contour, the contriFig 1t2 bution to the circulation from its ver. It remains for us to consider the two horizontal segments of equal length: The figure shows that the contributions to t.

Therefore, the ci'rculation of E differs from zero, which contradicts to 1. Till now we considered the description of electric field with the help of vector E. However, there exists another adequate way of describing it by using potential cp it should be noted at the very outset that there is a one-toone correspondence between the two methods. It will be shown that the second method has a number of significant advantages. The fact that line integral 1. A comparison of 1.

Basic laws of electromagnetism. Irodov I.E.

We can conditionally ascribe to an arbitrary point 0 of the field any value CPo of the potential. Then the potentials of all other points of the field will be unambiguously determined by formula 1. If we change CPo by a certain value Acp, the potentials of all other points of the field will change by the same value. Thus, potential cp is determined to within an arbitrary additive constant. The value of this constant does not play any role, since all electric phenomena depend only on the eleetric field strength.

It is determined, as. The unit of potential is the volt V. Potential of the Field of a Point Charge. Formula 1. We can make it even simpler. Let us use the fact that' formula 1. This function willbo the potential cpo Let us apply this method for finding the potential of the. Thus, the potential of the field of a point charge is given by. Potential of the Field ola System of Charges.

Let a system consist of fixed point charges ql' q2' By llsmg formula 1. Thus, the potential of a system of fixed point charges is given by. Here we also omitted an arbitrary constant. This is in complete agreement witll the faet that any real system of charges is Lounded in space, and honce its potential can bo taken equal to zero at inflIlity. Taking this into consideration, we call write formula 1. A similar expression corresponds to the case when the charges have a linear distribution.

Thus, if we know the charge distribution discrete or continuous , we can, in principle, fmd the potential of any system. Relation Between Potential and Vector E. It is known that electric fwld is completely described by vector function E r. Knowing this function, we can find the force acting on a charge under investigation at any point of the neld, calculate the work of fwld forces for any displacement of the charge, and so on.

And what do we get by introducing potential? First of all, it turns out that if we know the potential cp r of a given electric fIeld, we can reconstruct the fwld E r quite easily. Let us consider this question in greater detail. The relation between rp and E can be established with tho help of Eq. A comparison of this ex-. Having determined E", Ell' and E z, we can easily fmd vector E itself:.

We shall be using the latter more convenient notation and will formally consider VCP 'as the product of a symbolic vector V and the scalar cpo Then Eq. Find the field intensity. E if the field potential has the orm. Then with the help of formthil! It can be seen that in.. Thea what is the use of introducing potential? There are several sound reasons for doing that. The concept of potential is indeed very useful, and it is not by chance that this concept is widely used not only in physics but in engineering as well..

Indeed, it follows from formula 1. This means that vector E is normal to the given surface. It IS expedIent,. Field intensity. 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.

Calculation of the work of the lield forces with the help of formula 1. Find the work of the lield forces dune in the displacement of a point charge q' from the centre of tpe ring to i: Since the distributiun o[ the charge q over the ring is unknown, we cannot say anythiug detinite about the intensity E of the field created by this charge. This means that we cannot calculate the work by eval-.

This problem can be easily solved with the help of potential. It turns out in many cases that in order to find electric field intensity E, it is easier lirst to calculate the potential lp and then take its gradient than to calculate the value of E directly.

This is a considerable advantage of potential. It can be seen from formula 1. It will be shown below that the behaviour of the dIpole III an external field. Consequently, p is an important characterIstIc '.

In order to find the dipole field, we shall use formula 1. The Field of a Dipole. The electric dipole is a system of two g and - g, separated equal in magnitude unlike charges by a certain distance l.

When the dipole field is considered. The dipole field is axisymmetric. Therefore, in any plane passing through the dipole axis the pattern of the field is the same, vector E lying in this plane. According to 1. Let us place a dipole into a nonuniform electric field. Then the resultant force F acting on the dipole is Fig. Taking this into accoulJt, we get 1. This quantity corresponds to a vector directed along the dipole axis. Substituting this expression into the formula for F, we find that the force acting on the d'ipole is equal to.

The derivative appearing in this expression is called the directional derivative of the vector. The symbol of partial derivative indicates that it is taken with respect to a certain direction, viz. Unfortunately, the simplicity of formula 1. First of all, note that in a uniform Fig. This means that generally the force acts on a dipole only in a nonuniform field. Next, i:: Vector F coincides in direction only with the elementary increment of vector E, taken along the direction of I or p Fig.

Figure 1. We suggest that the reader prove independently that it is really so. Let a dipole with moment p be oriented along the symmetry axis of a certain nonuniform field E. We take the positive direction of the X-axis, for example, F as shown in Fig.

The Moment of Forces Acting on 8 Dipole. For this purpose, we must find the moment. This moment of force tends to rotate the dipole so that its electric moment p is oriented along the external field E. Such a position of the dipole is stable. Thus, in a nonuniform electric field a dipole behaves as.

A very thin disc is uniformly charged with sut'fa? Find the electric field intensity E on the axiS of thIS disc at the point from which the disc is seen at an angle.

Both these motions are simultaneous. The Energy of a Dipole in an External Field. We know that the energy of a point charge q in an external field is W. To within a quantity of the second order of smallness, we can write. It is clear from symmetry considerations th.

Hence, it is safWentto find the component d! It follows from this formula that the dipole has the minimUt;D. If it is displaced from this. Find the electric field intensity E at the centre of the ring. The given charge distribution is shown in Fig. The symmetry of this distribution implies that vector E at the point 0 is directed to the right, and its magnitude is equal to the sum of the projections onto the direction of E of vectors dE from elementary charges dq.

The projection of vector dE onto vector B Is. IntegTating 1 over cp between 0 and 2n, we find the magnitude of the vector E:. Then 2n.

A semi-infinite straight uniformly charged filament has a charge'" per unit length. Find the magnitude and the direction of the field Intensity at the point separated from the filament by a distanC'-t! The prohlem is reduced to finding Ex and E. Let us start with Ex. The contribution to Ex from the-charge element of the segment dz Is. Then ". The Gauss theorem. The intensity of an electric field depends only on the coordinates x and y as follows: 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 , we arrive at the conclusion that the flux through the sphere of radim R is equal to the flux through the lateral surface of a cylinder having the same radius and the height 2R. In particular, it is valid when one sphere is completely within the other or, in other words, when there is a spherical cavity in a sphere Fig.

Using the solution of the previous problem, find the field intensity E inside the sphere over which a charge is distributed with. Find the electric fi ld. Then at " 0 IS regIOn we have q. Thus, in the region of intersectio f th h form. Let us consider two spheres of the same radius, having uniformly distributed volume charges with the densities p and -po Suppose that the centres of the spheres are separated by the distance 1 Fig.

Then, in accordance with the solution of the previous problem, the field in the region of intersection of these spheres will be uniform: In our case, the volume charge differs from zero only in the surface layer. 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.

Basic Laws Of Electromagnetism

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 intrinsic energy of each charged body is an essentially positive quantity. Let us now apply the Gauss theorem to vector E in the same way as it was done while deriving formula 2.

Before Download Kindly Login and download Login. This is an artificial method that makes it possible to calculate in a simple way the electric field in some unfortunately few cases. Charge Distribution. December 28, Spread the love As I have mentioned in an earlier post, it is not a widely known fact that Prof. Find the electric field intensity E at the centre of the ring.

Electric Field in Dielectrics consider only isotropic dielectrics for which relation 3. The problems are closely related to the text and often complement it. Currently there is reprint still in stock from CBS Publishers.

STACEE from Round Lake Beach
I enjoy reading books upward. Look through my other articles. I have always been a very creative person and find it relaxing to indulge in legends car racing.