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Feynman Physics Lectures V2 Ch06 1962-10-18 Electric Field in Var Circumstances

Feynman Physics Lectures V2 Ch06 1962-10-18 Electric Field in Var Circumstances

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Unformatted text preview: 6 The Electric Field in Various Circumstances 6—1 Equations of the electrostatic potential This chapter will describe the behavror of the electric field in a number of different circumstances. It Will prov1de some experience With the way the electric field behaves, and will describe some of the mathematical methods which are used to find this field. We begin by pointing out that the Whole mathematical problem is the solution of two equations, the Maxwell equations for electrostatics: v~E = L (6.1) 60 v x E = 0. (6.2) In fact, the two can be combined into a Single equation. From the second equation, we know at once that we can describe the field as the gradient of a scalar (see Section 3—7): E = — W). (6.3) We may, if we wish, completely describe any particular electric field in terms of its potential <13. We obtain the differential equation that dz must obey by sub- stituting Eq. (6.3) into (6.1), to get v V¢— _ —i’—. (6.4) V'V¢=V¢=~i (6.5) so we write Eq. (6.4) as (6.6) The operator V2 is called the LaplaCian, and Eq (6 6) is called the Porsson equa- tion. The entire subject of electrostatics, from a mathematical pomt of view, is merely a study of the solutions of the Single equation (6.6). Once 451s obtained by solvmg Eq. (6.6) we can find E immediately from Eq. (6.3). We take up first the speCial class of problems in which p is given as a function of x, y, z. In that case the problem IS almost trivial, for we already know the solution of Eq. (6.6) for the general case. We have shown that if p is known at every point, the potential at pornt (l) is _ (2) (IV M1) ‘ / 47mm: where p(2) is the charge den51ty, (1V2 is the volume element at pomt (2), and r12 is the distance between points (I) and (2). The solution of the diflerentiul equation (6.6) is reduced to an integration over space. The solution (6.7) should be especially noted, because there are many situations in physics that lead to equations like (6.7) V2 (something) = (something else), and Eq. (6.7) is a prototype of the solution for any of these problems. The solution of electrostatic field problems is thus completely straightforward when the p05itions of all the charges are known. Let’s see how it works in a few examples. 6—1 6—1 6—2 6—3 6—4 6—5 6-6 6—7 6—8 6—9 Equations of the electrostatic potential The electric dipole Remarks on vector equations The dipole potential as a gradient The dipole approximation for an arbitrary distribution The fields of charged conductors The method of images A point charge near a conducting plane A point charge near a conducting sphere 6—10 Condensers; parallel plates 6—11 High-voltage breakdown 6—12 The field emission microscope Revzew. Chapter 23, Vol. I, Resonance Fig. 6—1. A dipole: two charges +q and —q the distance d apart. +0 ¢+ Fig. 6—2. The water molecule H20. The hydrogen atoms have slightly less than their share of the electron cloud; the oxygen, slightly more. 6—2 The electric dipole First, take two point charges, +q and —q, separated by the distance d. Let the z-axis go through the charges, and pick the origin halfway between, as shown in Fig. 6—1. Then, using (4.24), the potential from the two charges is given by ¢(x.y. 2) l q —q _ 47reo lx/[z — (ti/2)]2 + x2 + y2 + [z + (d/2)]2 + x2 + y2]' (6-8) We are not going to write out the formula for the electric field, but we can always calculate it once we have the potential. So we have solved the problem of two charges. There is an important special case in which the two charges are very close together—which is to say that we are interested in the fields only at distances from the charges large in comparison with their separation. We call such a close pair of charges a dipole. Dipoles are very common. A “dipole” antenna can often be approximated by two charges separated by a small distance—if we don’t ask about the field too close to the antenna. (We are usually interested in antennas with moving charges; then the equations of statics do not really apply, but for some purposes they are an adequate approximation.) More important perhaps, are atomic dipoles. If there is an electric field in any material, the electrons and protons feel opposite forces and are displaced relative to each other. In a conductor, you remember, some of the electrons move to the surfaces, so that the field inside becomes zero. In an insulator the electrons cannot move very far; they are pulled back by the attraction of the nu- cleus. They do, however, shift a little bit. So although an atom, or molecule, remains neutral in an external electric field, there is a very tiny separation of its positive and negative charges and it becomes a microscopic dipole. If we are interested in the fields of these atomic dipoles in the neighborhood of ordinary- sized objects, we are normally dealing with distances large compared with the separations of the pairs of charges. In some molecules the charges are somewhat separated even in the absence of external fields, because of the form of the molecule. In a water molecule, for example, there is a net negative charge on the oxygen atom and a net positive charge on each of the two hydrogen atoms, which are not placed symmetrically but as in Fig. 6—2. Although the charge of the whole molecule is zero, there is a charge distribution with a little more negative charge on one side and a little more positive charge on the other. This arrangement is certainly not as simple as two point charges, but when seen from far away the system acts like a dipole. As we shall see a little later, the field at large distances is not sensitive to the fine details. Let’s look, then, at the field of two opposite charges with a small separation d. If d becomes zero, the two charges are on top of each other, the two potentials cancel, and there is no field. But if they are not exactly on top of each other, we can get a good approximation to the potential by expanding the terms of (6.8) in a power series in the small quantity d (using the binomial expansion). Keeping terms only to first order in d, we can write (—‘2—1)2~22—zd. x2+y2+zz=r2. 2 (2—521)+x2+y2zr2—zd=r2(l—E(—1)s It is convenient to write Then and 1 z 1 z 1 (1 — 2:1)—1I2. [2 — (av/2)? + x2 + y2 r2i1 — (zd/r2)i r '2 6—2 Using the binomial expansion again for [1 — (zd/r2)]_“2—and throwing away terms with higher powers than the square of d—~we get 1 lzd 70%;?) ____1___~1(1_ 12d). [2+ (d/2)]2 + x2 + y2 r 272 The difference of these two terms gives for the potential Similarly, —— qd. (6.9) ¢>(x, y, 2) = The potential, and hence the field, which is its derivative, is proportional to qd, the product of the charge and the separation. This product is defined as the dipole moment of the two charges, for which we W111 use the symbol p (do not confuse with momentum!): p = qd. (6.10) Equation (6.9) can also be written as l pcosO 47reo r2 d>(x, y, Z) = ’ (6-11) since z/r = cos 0, where 0 is the angle between the axis of the d1pole and the radius vector to the point (x, y, z}——see F ig. 6—1. The potential of a dipole decreases as l/r2 for a given direction from the axis (whereas for a point charge it goes as l/r). The electric field E of the dipole wrll then decrease as 1/r3. We can put our formula into a vector form if we define p as a vector whose magnitude is p and whose direction is along the axis of the dipole, pointing from q_ toward q+. Then cos 0 = p-er, (6.12) where e, rs the unit radial vector (Fig. 6—3). We can also represent the point (x, y, z) by r. Then D' o] t t' l: - ‘ 1p epoenla (W) = 1 p er: “Lu. (6.13) This formula 1s valld for a dipole with any orientation and position if r represents the vector from the dipole to the point of interest. If we want the electric field of the dipole we can get it by taking the gradient of ¢. For example, the z-component of the field 1s —6¢/az. For a dipole oriented along the z-axis we can use (6.9): _§2__P33__P 1-22 62 ' 41reo 62 r3 — 41reo r3 r5 5 or 3 2o 1 _ p cos — _ E2 — 4760 ————————r3 (6.14) The x- and y-components are p 32x p 3zy_ __ , E = _,_ ._ 471'60 r5 1’ 41reo r5 Ex: These two can be combined to give one component directed perpendicular to the z-axis, which we will call the transverse component E1: E1=VE3+E2= p 3~2«/x2+y2 1/ 47r60 r5 or 3 o ‘ 0 E; p cos s1n . — 411'60 r3 (6.15) 6—3 p Fig. 6—3. Vector dipole. notation for 0 Fig. 6—4. dipole. The electric field of a g. The transverse component E i is in the x-y plane and points directly away from the axis of the dipole. The total field, of course, is E: x/E§+Ei. The dipole field varies inversely as the cube of the distance from the dipole. On the axis, at 6 = 0, it is tWice as strong as at 0 = 90°. At both of these special angles the electric field has only a z—component, but of opposite sign at the two places (Fig. 6—4). 6—3 Remarks on vector equations This is a good place to make a general remark about vector analysis. The fundamental proofs can be expressed by elegant equations in a general form, but in making various calculations and analyses it is always a good idea to choose the axes in some convenient way. Notice that when we were finding the potential of a dipole we chose the z—axis along the direction of the dipole, rather than at some arbitrary angle. This made the work much eaSier. But then we wrote the equations in vector form so that they would no longer depend on any particular coordinate system. After that, we are allowed to choose any coordinate system we wish, knowing that the relation is, in general, true. It clearly doesn’t make any sense to bother with an arbitrary coordinate system at some complicated angle when you can choose a neat system for the particular problem—provided that the result can finally be expressed as a vector equation. So by all means take advantage of the fact that vector equations are independent of any coordinate system. On the other hand, if you are trying to calculate the divergence of a vector, instead of just looking at V - E and wondering what it is, don’t forget that it can always be spread out as 6E, 6E, 6E, 6x W + a: ‘ If you can then work out the x-, y-, and z-components of the electric field and differentiate them, you will have the divergence. There often seems to be a feeling that there is something inelegant—some kind of defeat involved—in writing out the components; that somehow there ought always to be a way to do everything with the vector operators. There is often no advantage to it. The first time we encounter a particular kind of problem, it usually helps to write out the components to be sure we understand what is going on. There is nothing inelegant about put- ting numbers into equations, and nothing inelegant about substituting the deriva- tives for the fancy symbols. In fact, there is often a certain cleverness in doing just that. Of course when you publish a paper in a professional journal it will look better—and be more easily understood—if you can write everything in vector form. Besides, it saves print. 6—4 The dipole potential as a gradient We would like to point out a rather amusing thing about the dipole formula, Eq. (6.13). The potential can also be written as (,2 1 1,.V<l). (6.16) 47reo r If you calculate the gradient of l/r, you get and Eq. (6.16) is the same as Eq. (6.13). How did we think of that? We Just remembered that e./r2 appeared in the formula for the field of a point charge, and that the field was the gradient of a potential which has a 1/r dependence. 6—4 There is a physical reason for being able to write the dipole potential in the form of Eq. (6.16). Suppose we have a point charge q at the origin. The potential at the point P at (x, y, z) is _ 9.. ¢0 — r (Let’s leave off the l/41reo while we make these arguments; we can stick it in at the end.) Now if we move the charge +q up a distance Az, the potential at P will change a little, by, say, A¢+. How much is A44? Well, it is just the amount that the potential would change if we were to leave the charge at the origin and move P downward by the same distance Az (Fig. 6—5). That is, Aqfi = —%A2, where by A2 we mean the same as (1/2. So, using ¢ = q/r, we have that the po— tential from the positive charge is a d 45+ = g ‘ age) 5' {6-17) Applying the same reasoning for the potential from the negative charge, we can write _ :1 i :1 é . ¢—— r+az<r>2 (6.18) The total potential is the sum of (6.17) and (6.18): 6 ¢ = ¢+ + ¢_ = ‘a (13).; (6.19) ll 6 1 —6—2 (7) 4d- For other orientation of the dipole, we could represent the displacement of the positive charge by the vector Ar+. We should then write Eq. (6.17) as A¢+ = _V¢0 'Ar+, where Ar is then to be replaced by 11/2. Completing the derivation as before, Eq. (6.19) would then become 1 . This is the same as Eq. (6.16), if we replace qd = p, and put back the 1/471'60. Looking at it another way, we see that the dipole potential, Eq. (6.13), can be interpreted as 4; = —p-VcI>0, (6.20) where (£0 = l/47reor is the potential of a unit point charge. Although we can always find the potential of a known charge distribution by an integration, it is sometimes possible to save time by getting the answer with a clever trick. For example, one can often make use of the superposition principle. If we are given a charge distribution that can be made up of the sum of two dis- tributions for which the potentials are already known, it is easy to find the de- sired potential by just adding the two known ones. One example of this is our derivation of (6.20), another is the following. Suppose we have a spherical surface with a distribution of surface charge that varies as the cosine of the polar angle. The integration for this distribution is fairly messy. But, surprisrngly, such a distribution can be analyzed by super- position. For imagine a sphere with a uniform volume density of positive charge, and another sphere with an equal uniform volume density of negative charge, 6—5 Fig. 6—5. The potential at P from a point charge at A2 above the origin is the same as the potential at P’(Az below P) from the same charge at the origin. Fig. 6—6. Two uniformly charged spheres, superposed w ment, are equivalent to a nonuniform — —- —— _. _. _. _ _. distribution of surface Fig. 6—7. Computation of the po- tential at a point P at a large distance from a set of charges. ith a slight displace- charge. (0) 'l' (b) = (c) originally superposed to make a neutral—that is, uncharged—sphere. If the positive sphere is then displaced slightly with respect to the negative sphere, the body of the uncharged sphere would remain neutral, but a little positive charge will appear on one side, and some negative charg. Wlll appear on the opposite side, as illustrated in Fig. 6—6. If the relative displacement of the two spheres is small, the net charge is equivalent to a surface charge (on a spherical surface), and the surface charge density will be proportional to the cosme of the polar angle. Now if we want the potential from this distribution, we do not need to do an integral. We know that the potential from each of the spheres of charge is—for points outside the sphere—the same as from a point charge. The two displaced spheres are like two point charges; the potential is just that of a dipole. In this way you can show that a charge distribution on a sphere of radius a with a surface charge density a = 0'0 cos 9 produces a field outside the sphere which is just that of a dipole whose moment is 47r00a3 p = 3 u It can also be shown that inside the sphere the field is constant, with the value 0'0 E = _ . 360 If 0 is the angle from the positive z-axis, the electric field inside the sphere is in the negative z—direction. The example we have just conSIdered is not as artificial as it may appear; we will encounter it again in the theory of dielectrics. 6—5 The dipole approximation for an arbitrary distribution The dipole field appears in another circumstance both interesting and im- portant. Suppose that we have an ObjCCt that has a complicated distribution of charge—like the water molecule (Fig. 6—2)—and we are interested only in the fields far away. We will show that it is possible to find a relatively simple expression for the fields which is appropriate for distances large compared with the size of the object. We can think of our ObJCCt as an assembly of point charges qz in a certain limited region, as shown in Fig. 6—7. (We can, later, replace q, by p dV if we wish.) Let each charge q, be located at the displacement d, from an origin chosen somewhere 6—6 in the middle of the group of charges. What is the potential at the point P, located at R, where R is much larger than the maximum 11,? The potential from the whole collection is given by _ 1 fl ¢ _ 4m 1 n, (6.21) where r, is the distance from P to the charge q, (the length of the vector R ~ d,). Now if the distance from the charges to P, the point of observation, is enormous, each of the r,’s can be approximated by R. Each term becomes q,/R, and we can take l/R out as a factor in front of the summation. This gives us the simple result _ 1 l _ Q ¢ — 47TEO R 2 qt — 47r€0R5 (622) where Q is just the total charge of the whole object. Thus we find that for points far enough from any lump of charge, the lump looks like a point charge. The result is not too surprising. But what if there are equal numbers of positive and negative charges? Then the total charge Q of the object is zero. This is not an unusual case; in fact, as we know, objects are usually neutral. The water molecule is neutral, but the charges are not all at one point, so if we are close enough we should be able to see some efiects of the separate charges. We need a better approximation than (6.22) for the potential from an arbitrary distribution of charge in a neutral object. Equation (6.21) is still precrse, but we can no longer just set r, = R. We need a more accu- rate expression for r,. If the point P is at a large distance, r, will differ from R to an excellent approximation by the projection of d on R, as can be seen from Fig. 6-7. (You should imagine that P is really farther away than is shown in the figure.) In other words, if e, is the unit vector in the direction of R, then our next approximation to r, is r, z R ~ 11,- e,. (6.23) What we really want is l /r,, which, since d, << R, can be written to our approxima- tion as 1 1 d ~ _ ‘L ' er . 71 ~ R (1 + R ) (6.24) Substituting this in (6.21), we get that the potential is _ 1 Q dye, ) ¢‘4fi(§+;q‘ R2 +.... (6.25) The three dots indicate the terms of higher order in d/R that we have neglected. These, as well as the ones we have already obtained, are successive terms in a Taylor expansnon of Ur, about UK in powers of d1/R. The first term in (6.25) is what we got before; it drops out if the object is neutral. The second term depends on l/RZ, just as for a dipole. In fact, if we define P = 2 qt d1 (6-26) as a property of the charge distribution, the second term of the potential (6.25) is __ 1 p-er (bu—47760 R2 , (6.27) precisely a dipole potential. The quantity p is called the dipole moment of the distribution. It is a generalization of our earlier definition, and reduces to it for the special case of two point charges. Our result is that, far enough away from any mess of charges that 15 as a whole neutral, the potential is a dipole potential. It decreases as l/R2 and varies as cos B—and its strength depends on the dipole moment of the distribution of charge. It is for these reasons that dipole fields are important, since the Simple case of a pair of pomt charges is quite rare. 6—7 Fig. 6—8. The field lines and equipo- tentials for two point charges. CON DUCTOR / Fig. 6—9. The field outside a con- ductor shaped like the equipotential A of Fig. 6—8. The water molecule, for example, has a rather strong dipole moment. The electric fields that result from this moment are responsible for some of the im- portant properties of water. For many molecules, for example C02, the dipole moment vanishes because of the symmetry of the molecule. For them we should expand still more accurately, obtai...
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