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Relations 3. The given charge distribution is shown in Fig. We choose the closed surface in the form of a flat cylinder whose endfaces are on different sides of the interface Fig. These are some ionic crystals see footnote on page 68 and ferroelectrics. 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. Here we assume that the test charge q' is sufficiently small so. In our case, we have.

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. 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. To be mOre precise, it is assumed that such transformations are negligibly small. Thus, for infinitesimal displacements we can w.

These forces appear when the diel. Ponderomotive forces appa. This phenomenon is called electrostnctwn: In the general case, the effect o a dIelectrIc on. However, in many cases these forces can. Energy Method for Calculating Forces. In this case, the charges on the conductors rem? Equation 4. This can be done as follows. Suppose that we are interested in the force acting on a given body a conductor or a dielectric.

Let us displace this body by an infinitesimal distance dx in the direction X we are interested in. It is well known that the force depends only on the position. It cannot depend on how the energy process will develope if the system starts to move under the action of forces.

And this means that in order to calculate F x by formula 4. 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.

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.. 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. 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. Irodov — With a specific end goal to accentuate the most critical laws of electroattraction, and particularly to clear up the most troublesome bestics, the creator has tried to bar the less important subjects.

While trying to portray the primary thoughts compactly, plainly and in the meantime accurately, the content has been kept free from pointless numerical recipes, and the primary pressure has been laid on the physical parts of the phenomena.

With a similar end in see, different model representations, streamlining factors, exceptional cases, symmetry considerations, and so forth have been utilized wherever possible.

SI units of estimations are utilized all through the book. In any case, considering that the Gaussian arrangement of units is still broadly utilized, we have incorporated into Appendices 3 and 4 the tables of change of the most essential amounts and equations from SI to Gaussian units. PDF Books Online: Check All Books Here.

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