(1.14)
i.e. inside a uniformly charged sphere the field intensity grows linearly with the distance r from its centre. The curve representing the dependence of E on r is shown in Fig. 1. to.
General ConcllJsions. The results obtained in the above
A remarkable proper~y of electri~ theorem suggests thdat ~htls the~rb~ml'lt~s which would broa en I s POSSI 1
r~e~~s::fedsf~da ~rff:~:n?f~~: a~ an i~strument
for analysis
andI~~~~~;~~~10 (1.7~ wticg is c~~:~r:: ~hi~h~:f:bii~~:~h~l ::l~~
the differential formlo t e h agU~ensity p' and the changes in the field tion between the vo ume c ar e . ., intensity E in the vicinity of a gIven pOInt m space.
1. Electrostatic Field ia ••
Tr F
aCllllm
i':r: :
'1Vr
~. ~ E dS=(p)/Bo•
r
(1.15)
°
V·E=VxEx+VyEy..LVzEz=ax E x I
° +oy
Ey
a +0% E z•
It follows from (1. i 7) that this is just the divergence of E. Thus, the divergence of the field E can be written as div E or V·E (in both cases it is read as "the divergence of E"). We shall be using the latter, more convenient notation. Then, for example, the Gauss theorem (1.18) will have the form
The quantity which is the limit of the ratio of t. E dS to 'j' Vas V--+-O Is called the divergence of the fi ld E d ' hy defuIition. e an IS denoted by div E. Thus,
(1.20)
P Bo•
..!.- k. E dS v...o V ' j ' .
div E= lim
(1.16)
The dive!'lfence of any other v to fi . wa y• It· follows from definitioneC(1 16)eltdh ls de~rmined in a similar furi ctlon of coordinates. . at divergence is a scalar In order to obtain the ex ressi . we m~t, in accordance with f116,o~ fkr the. divergence of the field E determine the flux of E thrO~ h 't a e an Infinitely small volume Y' volu~e. and find the ratio of tfis f1~~ closed surface enveloping thi~ ohtaIned for the divergence will d the volume. The expression te sys)tem (in different systems ~~o~~. th:e c~oice of the coordierent. For example, in Cartesian coordi~~:e S'tl~ tU!D8 out to be . s 1 IS given by y div E = ~+ oE _L oE% 0:& oy I 0;-. (i.17) Thus. we have found that as Y --+-' '. ::nd,j. to plea, while the left-hand side O~~i1~5)d" Its nght.hand side e Ivergence of the field E' 1 d s IV E. Consequently lame point through the equati~~ re ate to the charge density at th~
:I
dffl
I
only in combination with a scalar or vector function by which it is symbolically multiplied. For example, if we forin the scalar product of vector V and vector E, we obtain
We now make the volume Y Je d . point we are interested in. In this c~ to zero. by c0';1tracting it to the ~iluhaneof P at the given point of the I~P) wJltohvlOusly tend to the td side of Eq. (1.15) will tend toe I' an ence the ratio on the
S:'
I
25
1.5. Circulation of Vector E. Potential
For this purpose we first rep t h e charge q in the volume V enveloped by a clo~d surface S is ~h~ volum~ charge ?ensity, averaged ~~ = I(p) V, Where {p) ~ sthl~uhte t.hlS expressIOn into Eq. (1 7) and d' e;;o bumeh Then We , w IC gives . IVI e ot Its sides by
I
f divE=ple o' (1.18) This equation expre -th----The form of man e sses .e Gauss theorem in the differential f ira~IY simplified w~P[:::;~d::~i:~ir:;P~i~fftions.can be co~:r:t n artesian coordinates, the operator Vehasrth~ ;:;:tIal operator V.
il
V=i.!-+
ox
j
3-. a oy + k 8i" '
r-.
(1.19)
where I, J, and k are the unit vecto f th operator V itself does not have any ~~anin~ X[-i and Z-axes. The . ecomes meaniD~fu J
The Gauss theorem in the differential form is a local theorem: the divergence of the field E at a given point depends only on the electric charge density p at this point. 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:dox, oEy/By, and oEz/o% as well. However. the Gauss theorem states that the sum of these derivatives, which determines the divergence of Ef"turns out to be equal to zero at all points of the field (outside the charge itself). 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.
1.5. Circulation of Vector E. Potential
Theorem on Circulation of Vector E. It is known from mechanics that any stationary field of central forces is conservative, Le. the work done by the forces of this field is independent of the path and depends only on the position of the initial and final points. This property is inherent in the electrostatic field, viz. the field created by a system oj fixed charges. 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. the elementary work of t he forces of the field done over thE distance dl is equal to E· dl, and the total work of the fiek forces over the distance between points 1 and 2 is defined a1 2
~ E dI. t
(1.21
26
This integral is taken alon '. therefore called the line inte:ra~ certam hne (path) and is ' h . We shall now show that fm integral (1.21) of the path hrot t e mdependence of line that when taken along an b\ween two points it follows gral is equal to zero. Integr:I (~r;f)Y closed path, this iute. . over a closed contour is the cIrculation of vector E d t.. called . an 'IS d.enoted by 'V' Thus we state that cireul t' f .. static field is equal to ze ro,a I.e. ~on 0 vector E in anv" electro-
~Edl=O.
1248
(1.22)
This statement is' called th h vector E. e t eorem on circulation of In order. to prove this th . eorem.. we break an arbitrary closed pa.th Into two parts la2 and 2bl 2 (FIg. 1.11). Since line integral (1.21) (we denote it by does not 1 depend on the path be~~een 'points 1
i )
c:?
(a)
f
and 2, we have = otb er h an, d it12 is
b
T
Fig. 1.11
(b)
(b)
i '. On
the. clear that
• 12 (b)
• 112 - \ ,where \ is the Integral over the same segment b b ,21 • ., 21 direction. Therefore ut taken In the opposite (n)
(b)
~ 12 L1
(a)
~
(b)
L2
-I= 11 = 0, Q.E.D. A field haVing property (1 22) . I Hence, any electrostatic field . IS ca led .the potential field. The theorem on' . lS a potentzal fleld. to draw a numbe~I~Ctl?tIon of vector E ~akes it possible sorting to calculations ~mfortant ~oncluslOns without 1'e. e us conSIder two examples.' Example f • TI Ie rIe Id I ines of an electrostatic field E cannot be closed.
I
~
27
1.5. Ctrculation of Vector E. PotenUal
1. Electrostatic Field tn a Vacuum
Indeed, if the oppollite were true and some lines of field E?lerJ! closed, then taking the circulation of vector E along this line we would immediately come to contradiction with theorem (1.22). This means that actually there are no closed lines of E in an electrostatic field: the lines emerge from positive charges F!;::::j and terminate on negative ones (or go to infinity). I • Example 2. Is the configuration of an I I electrostatic field shown in Fig. 1.12 I possible? . lr No, itisnot. 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. ---The arrows on the contour indicate the • direction of circumvention. With such a special choice of the contour, the contriFig 1t2 bution to the circulation from its ver. . tical parts is equal to zero, since in this case E .L dl and E ·dl = O. It remains for us to consider the two horizontal segments of equal length:,. The figure shows that the contributions to t.lte circulation from these regions are opposite in sign, and unequal in magnitude (the contribution from the upper segment is larger since the field lines are denser, and hence the value of E is larger). Therefore, the ci'rculation of E differs from zero, which contradicts to (1.22).
-4
'I
Potential. 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 (1.21) representing the work'of the field forces done in the displacement of a unit positive charge from point 1 to point 2 does not depend on the path allows us to!state that for electric neld there exists a certain scalar function
where
28
1. Electrostattc Field in
tJ
2U
Vacuum
called the field potential. A comparison of (1.23) with the expression for the work done by the forces of the potential field (the work being equal to the decrease in the potential energy of a particle in the field) leads to the conclusion that the potential is the quantity numerically equal to the potential energy of a unit positive charge at a given point of the field. 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 (1.23). 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 .will be"'shown later, not by the potentia] at a given point but by the·potential difference between neighbouring points of the field. The unit of potential is the volt (V). Potential of the Field of a Point Charge. Formula (1.23) contains, in addition to the definition of potential cp, the method of finding this function. For this purpose, it is sufficient to evaluate the integral ~ E dl over any path between two points and than represent the obtained result as a decrease in a certain function which is just
field of a lixed point charge: • 1
q
= -d
Ii r
IJ ,l.'l:'o
E.dt -- - (' r ·die /trrf,o ,.2
(_1_ _.'L _i-const) 4Jlco r I
r:!.
,
where we took into account that cr dt =-= 1· (dl}r= dr, sillee the projection of dl onto en ~nd hence ?ll 1', is ~qual to ~he increment of the magnitude of vector r, Le. dr. 'I he quantlty appearing in the parentheses under the diffe~enti~l is exactly i:p (1'). Since the additive constant contamed III ~he fo~' mula- does not play any physical role, it is usually omitted ill order to simplify the expression for (p. Thus, the potential of the field of a point charge is given by
I~
= __ 1 }L II _ _ 4n~_'_
(1.2:»
The absence'6f an additive constant in this expression indicates· that we conventionally assume that the potential is equal to zero at infinity (for r -+ 00). Potential of the Field ola System of Charges. Let a system consist of fixed point charges ql' q2' .... In accordance with the principle of superposition, the field intensity at any point of the field is given by E = E I + E 2 + ..., wh~re E l is the field intensity from the charge ql' etc. By llsmg formula (1.24)1 we can then write E·dl = (E l + E 2 + ...) dl = EI·dl + E 2 ·dl + ... = -dcpl - dCP2 - ... = --dcpl where cp = 2jcpf, Le. the principle of superposition t~rns out to be valid for potential as well. Thus, the potential of a system of fixed point charges is given by
.- _1_ '" !!i- -I I cp - 4nco L..J r 'I 1.
(1.26)
_i
where ri is the distance froID the point charge qj 10 the point under consideration. 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.
1.6. Relation Between Potential and Vector I!:
30 1. Electrostatic Field in a Vacuum -----------_._---.-'-'-'=";~--If the charges forming the system are distributed cUlItinuously, then, as hefore, we assume that each volume element dV contains 3 "point" charge p dV, where p is the charge . density in the volume dV. Taking this into consideration, we call write formula (1.26) in a different form: ( 1.27)
where the integration is performed either over the entire space or over its part containing the charges. If the charges are located only on the surface S, we can write
=
1 ~ adS 4neo -r-'
(1.28)
where (J is the surface charge density and dS is the element of the surface S. 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. 1.6. 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. (1.24). Let the displacement dl be parallel to the X-axis; then dl = i dx, where i is the unit vector along the X-axis and dx is the increment of the coordinate x. In this case E·dl = E·i dx = Ex dx, where Ex is the projectioll of vector E onto the ullit vector i (and not on the displacement dll). A comparison of this ex-
3t
pression with formula (1.24) gives Ex
= - orp/vx,
(1.29)
wher~ the symbol of partial.derivative emphasizes that the functIOn rp ~x, y, z) must he differentiated only with respect to x, assummg that y and z are constant in this case. . In a similar wa~, w.e can obtain the corresponding expres810.ns for the prOJectIOns E y and E z. Having determined E", Ell' and E z, we can easily fmd vector E itself:
E=
_ ( ocp ox
i
+ ~j + ~ k) iJy oz .
(1.30)
T~e quantity in the par~Iltheses is the gradient of the potentzaZ cp (grad cp or Vcp). 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. (1.30) can be represented in the form
(1.31) L~. the field inte~sity E'is equal to the potential gradient With the miflUS SIgn. This is exactly the formula that can be used for reconstructing the field E if we know the function
f
+
Let ~s derive one more useful formula. We write the righthand sl~e of (1.24) in the form E·dl c=o R I dZ, where dl = . Id1 l IS an elementary displacement and E I is the projectIOn of vector E onto the displacement dl. Hence
IE
l
= -ocpI8Z,
I
(1.32)
Le. the projection of vector E onto the direction of the dis-
1.6. llelation Between Potential and Vector E
--------------~_. ----------~
g2
On the Advantages o~ Potential. 1l was noted earlier that . electrostatic field is completely cltaracterizeu by vector function E (r). 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.. it. If we know the potential
1. Electrostatic Feeld in a Vacuum
placement dl is equal to the d' • . . p.otential (this is emphasized b;r~~~onal telflvfative .of the rivative). sym 0 0 partial deEquipotential Surfaces Let . d equipotential surface, viz. 'the s:;a~~t:~ :l~e t~etco~cept. of potentiallp has the same value. We shall sho~o~~ ~ 0 whIch each point o.f the surface is directed along thea e eqUipotentIal surface and. towards the decrease in the potential. Indeed, it follows from formula (1.32) that the projection of vector E onto any direction tangent to the eq,!ipo~entialsurface at a given pomt 18 equal to zero. This means that vector E is normal to the given surface. Further, let us take a displaC&ment dl along the normal to the surface, towards decreasing Fig. 1.13 p. Then Olp < 0, and accordmg to (1 32) E > 0 . v,ector E is. directed towards decreasing ~, ~r i~ the tIon ?pposite. to ~hat of the vector Vlp. It IS expedIent,.to draw equipotential surfaces in such a way that the potential difference between two neighbourin surfaces b~ t~e same. Then the density of e ui otentiJ s~aces WI.ll vIsually indicate the magnitudes fieYd intensttIes a~ dIfferent points. Field intensity. will be higher . t e re~lOns ~h~re equipotential surfaces are denser '"th~ pot~ntial rehef IS steeper"). \ Smce vector E ~s normal to an equipotential surface eve ry where, the field hnes are orthogonal to these surfaces.
:~
n::::.~r~
where
d;;;'
uating the integral ~ q'E dl in this case. This problem can be easily solved with the help of potential. Indeed, since all elements of the ring are at the same distance a from the centre of the ring, the potential of this point is Ipo ~= q/4Jv ou. And we know that cp = 0 at intinity. Consequently, the work A = q'cpo = q'q/4nlO oa.
of
Th
l
F~guhe df 3
shows a two-dimensional
patt~rn of
an electric field
h liDes eorres~o~d to equipoten~ial rlUrfaces, while thetiled MUd I. t. e mes of E. Such a representatIOn ean be easil v· i~~~:t:yd\:tiYghhows dthe hdirec.t!o.n of vector E, the regiorrs wi::e field er an were lower as we'l's tIl . "\h
lin:s ::
~
10 13
~~~~r steepness of the putential relief. SU(~ri a ;ai'tern ~~?':sed'to
d. t' qilal~~i~lve answers to a number of .~lU. ,.stiOJl8 sU~.h as ';In'wha.t lrec ,IOn WI "charge placed at a certain point m~veP Whe I> • iliagiutude o! the potential gradient At which 1 e orce actmg on the charge be greater~" et.c.
hi~her?
U:in~.;t.e
A
33
------
"
2. 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. Indeed, for calculating
r
I
34
1. Electrostatic Field in a Vacuum
----------'--'--'-.:-:..'-.:-:..~....:..::..:..::......:::..:......:=-:..=::.::::=-------
1.7. Electric Dipole
There are some other advantages in using potential which will be discussed later.
from the negative to the positive charge: p = ql,
1.1. Electric Dipole
where q > 0 and I is the vector directed as p. It can be seen from formula (1.34) that the dipole field depends on its electric mom~nt p.. It will be shown below that the behaviour of the dIpole III an external field .al~o depends on p. Consequently, p is an important characterIstIc ' . of the dipole. It should also be noted that the potential of the dI~)Qle field decreases with the distance r faste~ than the ~otentlal of the field of a point charge (in proportIOn to 1frz Illstead of 1fr). !In order to find the dipole field, we shall use formula (1.32) a~d calculate the projections of vector E onto two mutually perpendicular directions along the unit vectors er and ~ (Fig. 1. 14b):
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 . ,
+
p
(b)
eb +q
/ q
P
E
E
it is assumed that the dipole itself is pointlike, i.e. the distance r from the dipole to the points under consideration is assumed to be much greater than l. 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. L~t us first find the potential of the dipole field and then its illtensity. According to (1.25), the potential of the dipole field at the point P (Fig. 1.14a) is defined as 1 4itl:: o
'1) = 7..- - -,::
(lJ
1 4ne
aq>
"",' r
Fig. 1.14
<.p ,-
o
(1.35)
t)
= - --ar = = -
aq> r'~
=
2pcos~
1
4nll
o
-,-.- ,
1 p sin ~ 4neo - - , . -
(1.36)
Hence, the modulus of vector E will be z-tt. E=VE~+E6= 'tJ,80 J.: -?V-1-+-3-c-o-s r
(1.37)
In particular, for tt = 0 and tt = n/2 we obtain the expressions for the field intensity on the dipole axis (E u) and on the normal to it (E.d:
q(r_-r.) r.r_ •
Ell
= ~80 2~,
EJ.
= 4~80 ~,
(1.38)
Le. for the same r the intensity E u is twice as high as EJ.' The Force Acting on a Dipole. Let us place a dipole into a nonuniform electric field. Suppose that E+ and E_ are the intensities of the external field at the points where the positive and negative dipole charges are located. Then the resultant force F acting on the dipole is (Fig. 1.i5a): F = qE+ -qE_ = q (E+ - E_).
Since r» l, it call 1e seen fro.:n Fig. 1.14a that r__ r + = = 1 cos \} and r +r_ = r 2 , where r is the distance from the point P to the dipole (it is pointlikel). Taking this into accoulJt, we get (1.34) where p = ql is the electric moment oj the dipole. This quantity corresponds to a vector directed along the dipole axis
The difference E+ - E_ is the increment LiE of vector E on the segment equal to the dipole length 1 in the direction 3*
J
1. Electro$tatlc Field In a Vacuum f
I'
of vector I. Since the length of this segment is small, we can write ~E
~E=E+-E_=-. l
=
aE az l.
Substituting this expression into the formula for F, we find that the force acting on the d'ipole is equal to
I
F=p"*,
(1.39)
where p = ql is the dipole electric moment. 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. the direction coinciding with vector I or p. Unfortunately, the simplicity of formula (1.39) is delusive: taking the derivative fJE/fJl is a rather com~q~plicated mathematical operation. We shall not discuss this question
(.~
...
~:-q~
z:s.-
Figure 1.16 shows the directions of the force F acting on a dipole in the field of a point charge q for three different dipole orientations. We suggest that the reader prove independently that it is really so.
lf we are interested in the projection of force F onto a certain direction Xi, it is sufficient to write equation (1.39) in terms of the projections onto this direction, and we get
Fx = P
a:zx
,
37
1.7. Electric Dipole
where fJE ~/az is the derivative of the corresponding projection of vector E again onto the direction of vector I or p. 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. 1.17. Since the increqe <= p. ment of the projection E ~ in the direc~F tion of vector p will be negative, F:r < O. and hence vector F is directed to the left, i.e. towards increasing field intensity. If we rotate vector p shown in FI the figure through 90° so that the dipole q-----~ centre coincides with the symmetry axis of the field, it can be easily seen Fig. 1.16 that in this position F:II: = O. The Moment of Forces Acting on 8 Dipole. Let<4Is 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. For this purpose, we must find the moment. of external forces with respect to the dipole centre of mass*. ~By definition, the moment of forces F + = qE+ and F _ = = -qE_ with respect to the centre of mass C (Fig. 1.18) is equal to M = [r+ X F+I [r_ X FJ=[r+ X qE+l -[r_ X qEJ,
q...-----tp
+
+
where r + and r _ are the radius vectors of the charges q and - q relative to the point C. For a sufficiently small dipole length, E+ ~ E_ and M = [(r+ - r_) X qEl. It remains for us to take into account that r + - r _ = I and qI = p, which gives
IM=
[p X
EJ.!
(1.41)
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
(1.40) • We take the moment with respect to the centre of mass in order
to eliminate the moment of inertial forces.
38
1. Electroltattc Field in a Vacuum
39
Problem'
follo,,":s: under the action of the moment of force (1.41), the. dIpole tends to. get oriented along the field (p tt E), whIle under the action of the resultant force (1.39) it is
position, the moment of external forces will return the dipole to the equilibrium position.. Problems
• t.t. A very thin disc is uniformly charged with sut'fa?e char~ density lJ > O. 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.
+q
E -q
Fig. 1.17
Fig. 1.18
£.
displaced towards the region where the field E has larger magnitude. 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. qcp, where cp is the fIeld potential at the point of locatIOn of the charge q. A dipole is the system of two charges and hence its energy in an external field is ' W = q+cp+ + q_cp_ = q (cp+ _ cp_), where CP+ and cp_ are the potentials of the external field at the points of location of the charges +q and -q. To within a quantity of the second order of smallness, we can write
Solu.io1l. It is clear from symmetry considerations th.at o~ the disc axis vector E must coincide with the direction of this aXIS (FIg. 1.19). Hence, it is safWentto find the component d!£z from the charg; of the area e1emeBt tiS at the point A and th~n mtegrate the o.btamed expression over the entire surface of lthe dISC. It can be easIly seen that' 1 lJdS dE z =-4- - I cos 'It. (1' JtEo
alp
cp+-cp-= az 1, ,,:here acp/Ol is the derivative of the potential in the directIOn of the vector l. According to (1.32), acp/al = -E It and henen '" + - '"_ ~ L E ,I from which we Il"l
Ell.
IW = -p.E.
Fig. 1.20
Fig. Lt9
r
In our case (dS cos 'It)/rz = dO is the solid angle at which the area element dS is seen from the point A, and expression (1) can be written as 1 dE z = -4- (J dO. JtEo
Hence, the required quantity is
(1.42)
It follows from this formula that the dipole has the minimUt;D. energy (Wmin = -pE) in the position p tt E (the positIOn of stable equilibrium). If it is displaced from this
1
E=-4- aQ. Jtto
It should he noted that at large distances from the disc, Q = SIrS, where S is the area of the disc and E = ql4nEors just as the field of the point charge q = lJS. In the immediate vicinity of the point O,the solid angle g = 2a and E = a/2eo'
40
Problems
1. Electrostatic Field in a Vacuum
• f .2. A thin nonconducting ring of radius R is charged with a linear density A = Ao cos cp, where '-0 is a positive eoDStant and cp is the azimuth angle. Find the electric field intensity E at the centre of the ring. Solution. The given charge distribution is shown in Fig. t,20. 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
1 dq dE cos cp = -4- -R' cos cp,
ne o
This gives dEy = (A cos a da)/4ne oy and E y = A/4.tteoy·
We have obtained an interesting result: Ex = E y independently of y,
(1)
where dq = AR dlp = AoR cos cp dcp. IntegTating (1) over cp between 0 and 2n, we find the magnitude of the vector E:
"-0 R 4ne o
E
.=
a
2n
Jr cost
d
cp cp =
o
"-0
4e R • o
It should be noted that this Integral is evaluated in the most simple way if we take into account that (cost cp) = 1/2. Then 2n
~
Fig. 1.21 i.e. vector E is oriented at the angle of 45° to the filament. The modulus of vector E is E=
cos' cp dcp=(cos' cp) 2n=n.
o
• 1.3. 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! y and lying on the normal to the filament, passing through its end. Solution. The prohlem is reduced to finding Ex and E., viz. the projections of vector E (Fig. L21, where it is assumed fliat '" > 0). Let us start with Ex. The contribution to Ex from the-charge element of the segment dz Is . 1 Adx. dE x =-4- -t-sma. (1) neo r j Let us reduce this expression to the form convenient for integration. In our case, dx = r dalcos a, T = ylcos a. Then "
V E~+E~=j,. Y2i4rt8 oY •
• 1.4. The Gauss theorem. The intensity of an electric field depends only on the coordinates x and y as follows: E = a (xi yj)/(x 2 y2),
+
q=80
dE x = - A 4 sin ex da.
neoy
Integrating this expression over a between 0 and n/2, we find Ex
= "'/4neoy·
In order to find the projectiou E y , it is sufficient to recall that dEli diffeq from dE~ in that sin a"ln (1) is simply replaced by C08 lx.
+
where a is a constant, and i and j are the unit vectors of the X-and Y;-axes. Find the charge within a sphere of radius' R with the centre at the origin. Solution. In accordance with the Gauss theorem, the required charge is equal to the flux of E through this sphere, divided by &0' In our case, we can determine the flux as follows. 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. and arranged as shown in Fig. 1.22. Then
where Er
=
~ E dS=8 oE rS,
aIR and S = 2nR·2R = 4nR2. Finally, we get q = 4n8 oaR.
• 1.5. A system consists of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density p = aIr, where a is a positive constant and r is the distance from
42
43
1. Electroirtattc Field in a Vacuum
Problem.
the centre of the sphere F' d th h electric field intensity E; o~~ide ~hc arghe of. th~ dsphere for which the the value of E. e sp ere IS In ependent of r. Find
of the spheres and of the distance between their centres. 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. 1.25). • 1.7. Using the solution of the previous problem, find the field intensity E inside the sphere over which a charge is distributed with
Solution. Let the sought charge of the sphere be q. Then, Using the
,I II· 1
I
lie
Iii
Iii I
Fig. 1.22 Fig. 1.23 Gauss theorem we can writ th f II . surf~ of radi~s r (outside :he :ph~r~~i~t ~h~r~h~ir~~ ~r a spherical r
E~ ·4n r 2 - _ q .t- 1 a - \. -4nr2dr. 80
£0. R
r
After integration, we transform this equation to 2 E ·4nr = (q ~ 2naR2)leo 4nar 2/28 0 • • The intensity E does t d d theses is equal to ze~~. J!rt~~ on r when the expression in the paren-
+
=
2naR2 and E = a/280 ' ~ f .6. Find the electric fi ld . t · . sectIon of two spheres unifo In enslty .Ii: In the region of intervolume densities p and _p th c3~rged by unlike charges with the spheres is determined by v~ctor le(F::~af.r:3)~etween the centres of the Solution. Using the Gauss th electric field intensity Within eo~efm, lwe can easilY.lshow that the a UD! orm Y charged spflere is E = (p/38 0 ) r, where r is the radius vector r I t' . can consider the field in the r e.a Ive .to the c~ntre of the sphere. We sUperposition of the fields of et~on of.;n\ersecthlOn of the spheres as the an arbitrary point A (Fig 1 24)0 \lfntlh~r my.c arged spheres. Then at " 0 IS regIOn we have q
WI
E = E+
+ JL =
P (r+ -
r_V3e n = pI/3e . n
Thus, in the region of intersectio f th h form. This conclusion is valid retpardle o f etse sp ~rebs the field is uni" ss 0 he ratio etween the radii
Fig. 1.25
Fig. 1.24
the surface density CT = CTo cos {Jo, where 0"0 is a constant and {Jo is the polar angle...' Solution. 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. 1.26). Then, in accordance with the solution of the previous problem, the field in the region of intersection of these spheres will be uniform: E = (p/380 ) 1. (1)
tz
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. The thickness of the charged layer at the points determined by angle -& (Fig. 1.26) is equal to 1 cos {Jo. Hence, in this region the charge per unit area is (J = pl cos {Jo = 0"0 cos a, where °0 = pl, and expression (1) can be represented in the form E = -(0"0/380) k,
Fig. 1.26
where k is the unit vector of the Z-axis from which the angle it ii'J measured. • 1.8. Potential. Tho potential of a certain electric field has the form Ip = ex (xy - z2). Find the projection of vector E onto thp. direction of the vector a = i 3k at the point M (2, 1, -3). Solution. Let us first find vector E: E = -Vip = -a (yl xj - 2zk).
+
+
45
2.1. Field in a Substance
1. Electrodatlc Field In a Vacuam
where dq is the charge contained between the spheres of radii rand r + dr. Hence
The sought projection is E a =E . ..!..= -0: (yl-zj- 2lk)(i+3k). . a }f1+31 At the point M we have
E.
-0:~1t18)
1 , d r, r:dE r +2rE.r dr=-pr
-0: (y-61)
Eo
JfiO
p
• f.9. Find the potential lp at the edge of a thin df!e of l'Jdius R with a charge uniformly distributed over olle of its sides with the .... face density o. . S~lut~on. ~y definition, the potential in the case of a surface charge dl8trl~ution IS defined by integral (1.28). In order to simplify in. tegratlOD, we shall choose the area element dS in the fonn of a part of the ring of radius rand width!,dr (Fig. 1.27). Then dS = 2'&r dr, r = 2R cos '6-, and dr = -2R sin '6- t:H}. After substituting these expressions into integral (1.28), we obtain the expression for cp at &be point 0:
neo
Jo ~sin'6'd{;. n/2
We integrate by parts, denoting and sin t}- dt}- = dv:
~
=•
) 'It sin t}- dt}- = - '6- cos '6-
+ \ cos'6-dtJo=-'6-cos{}+sinil'.
Fig. 1.27
01
which gives :-1 after substituting the limits of integration. As a result, we obtam ~/ lp = oRfneo'
• 1.10. T.he potential of the field inside a charged sphere depends ?nly on the dI.stance r from its. Cl'ntre to the point under consideration I~ the foll?wI~g 'Yay: lp = ar' + b, where a and b are constants. Fmd the dIstrIbution of the volume charge p (r) within the 'sphere. Solution. Let us first find the field intensity. According to (1.32) we have • E r = -oq>/ar = -2ar. (1) I
Ii
I
T~en we US? th~ Gauss theorem: 4nr'Er = q/eo • The differential of thiS expressIOn IS 1 i 4n d(rIE r) = - dq = - p·4nr' dr, 66
eJ
ar
+2. E_' r r
__
...£... eo
Substituting (1) into the last equation, we obtain
J~o 0:.
lp=- oR
aE r
=-"
-flEoa,
i.e. the charge is distributed uniformly within the sphere. • 10ft. Dipole. 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. Solution. According to (1.39), we have F = PI I aElal I• where E is the field intensity from the dipole P2' determined by the first of formulas (1.38): E=_1_ 2P2 ...' 4nEo za Taking the derivative of this expression with respect to I and substituting it into the formull\. for fl, we obtain 1 _ 6PIP2 F __ - 41tEo l4' It should be noted that the dipoles will be attracted when PI t t P2 and repulsed when PI H P2'
2. A Conductor in an Electrostatic Field 2.1. 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. electrons and nuclei. The solution of this problem is obviously not feasible. In any case, the result would be so complicated
2.2. Fields Inside and Outside a Conductor
46
2. A
Co..a~ctor·tn
an Electrodatic Pield
that i~ would ~e impossibl~ to use it. Moreover, the knowledge of tillS field IS not reqUIred for the solution of macroscopic problems. 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:roscopic field) we shall understand the microscopic field averaged over space (in this case time averaging is superfluous). This averaging is performed over what is called a physically infinitesimal volume, viz. the volume containing a large number of atoms and having the dimensions that are many times smaller than the dli!tances over which the macroscopic field noticeably changes. The averaging over such vo~umes smooth~ns all i.rregular and rapidly varying fluctuatlOns of the mIcroscopIC field over the distances of the order of atomic ones, but retains smooth variations of the macroscopic field over macroscopic distances. Thus, the field in the substance is E=Emacro=(Emlcro). (2.f) The Inftuence of a Substance on a Field. 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.al separation· of these charges. In certain regions of the suhstance, uncompensated charges of different signs appear. This phenomenon is called the electrostatic . / induction, while the charges appearing as a result of separation are called induced charges. Induced charges create an additional electric field which in combination with the initial (external) field forms the resultant field. Knowing the external field and the distribution of induced charges, we can forget about the presence of the substance itself while calculating the reSultant iield since the role of the substance has already been taken int~ account with the help of induced dlarges Thus, the resultant field in the presence of a substance is determined simply as the superposition of the external Bald and the field of .wauced charges. 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-
41
bution of induced charges is mainly determined by the properties of the substance, i.e. its physical nature and the shape of the bodies. We shall have to consider these questions in greater detail. 2.2. Fields Inside and Outside a Conductor Inside a Conductor E = O. 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.l.'ges of the conductor, and as a result all the negative charges (electrons) will be displaced in the direction against the field. 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.eld at a~l the points inside the conductor vanishes. Thus, in the statIC case the electric field inside a conductor is absent (E = 0). Fu.rther, siw;e E = 0 everyWhere in the conductor, the densIty of excess (uncompensated) charges inside the conductor is· also equal to zero at all points (p = 0). This can be easily explained with the help of the Gauss theorem. Indeed since inside the conductor E = 0, the flux of E through an; closed surface inside the conductor is also equal to zero. 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. It should he noted that the excess s~face charge is located in a very thin surface layer (whose thIckness amounts to one or two interatomic distances). The absence of a field inside a conductor indicates in accordance with (1.31), that potential cp in the cond~r has the same value for all its points, i.e. any conductor in an electrostatic field is an equipotential region, its surface being an equipotential surface. 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
2. A Condllctor in an Electrostatic Field
-------
49
conductor, Le. charge equilibrium would be impossible.
2.3. Forces Acting on the Surface of a Conductor
Example. Find the potential of an undlarged condudiug sphere provided that a point charge q is located at a distance r from its centre (Fig. 2.1). 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:
shall take a small cylinder and arrange it as is shown in Fig. 2.3. Then the flux of E through this surface will be equal only to the flux through the "outer" endface of the cylinder- (the fluxes through the lateral surf~ce and the inner endface are equal to zero). Thus we obtam En /18 =
Ip=
1
q.,
4nco' r~t-fjl·
(1)
where the tJrst term is the potential of the charge q, while the second is the potential of the charges induced on the surface of the sphere.Dut since aU induced charges are at the same distance a from the point 0 and the total induced charge is equal to zero, q;'=O as well. Thus, in this case the putential \if the sphere will be determined only by the Erst term ill (1). Fig. 2.1
l"igure 2.2 shows the field and the charge distributions for a system consisting of two conducting spheres one of which (left) is charged. As a result of electric induction, the ch?rges of the opposite sign appear on the surface of the right uncharged sphere. The field _of these.eharges will in turn cause a redistribution of charges on the surface of the left sphere, and their surface distribution will become nonuniform. The solid lines in the figure are the lines of E, while the dashed lines show the intersection of tlquipotential surfa~s with the plane of the figure. As we move away from this...system, the equipotential. surfaces become closer_and closer to spherical, and the field lines become closer to radial. The field itself in this case resembles more and more the field of a point charge q, viz. the total charge of the given system. 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
n
4'
Fig. 2.2 Fig. 2.3 = 0/18/80' where En' is ..the projection of (vector E onto the outward normal n (with respect to the conductor), /18 is"9the cross-sectional area of the cylinder and 0 is the local s~face charge density of the conductor. Cancelling both sides of this expression by /18. we get
I
En = a/so·
I
(2.2)
If 0 > 0, then En > 0, Le. vector E is directed from the conductor surface (coincides in direction with the normal n). If 0 < 0, then En < 0, and vector E is directed towards the conductor surface. Relation (2.2) may lead to the erroneous conclusion that the field E in the vicinity of a conductor depends only on the local charge density 0'. 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. 2.3. 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. The force acting on a small area· ~S of the conductor surface is ~F = at.S .E o, (2.3)
\ 50
2.4,., Properties
2. A Conductor in an Electrostatic Field
where 0 f!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. It should b~ noted at the very outset that Eo is not equal to the field intensity E in the vicinity of the given surface element of the cojductor, although there exists ~a certain relation between th~m. Let us find this relation, Le. express Eo through Be' Let E a be the intensity of the field created by the charge on. the area element f!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 (1.10), E a = = 0/ 28 0' The resultant field both inside and outside the conductor (near the area element f!S) is the superposition of the fields Eo and Eo. On both sides of the area element f!S the field E is p~actically ~he same, while the field Eo has opposite di~ ractIOns (see FIg. 2.4 where (for the sake of definiteness it is assumed that C1 > 0). From 'the condition E = 0 inside the conductor, it follows that Eo = Eo, and then outside the conductor, near its surface, E = Eo Eo = 2Eo. Thus, Eo = E/2, (2.4) and Eq. (2.3) becomes
+ 1
AF=T aAS·E. AS
(2.5)
Dividing both sides of this equation by f!S, we obtain the expression for the force acting on unit surface of a conductor: 1 Fu-T.aE.
(2.6)
Fig. 2.4 :-
We can write this expression in a different form since the quantities a and E appearing in it are interconnected. Indeed, in accordance with (2.2), En = a/eo, or E = (a/eo) n, where n is the outward normal to the surface element at a given point of the conductor. Hence
(2.7)
of a Closed Conducting Shell
5i 2
where we took into account that 0 = eoEn \ and E; = E • The quantity F u is called the surface density of force. Equation (2.7) shows that regardless of the sign of 0, and hence of the direction of E, the force Fu is always directed outside the conductor, tending to stretch it: Example. 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 (2.7) by dS, we obtain the expression for the force dF acting on the surface element dS:
dF=+ eo£2 dS, where dS = n dS. The resultant force acting on the entire conductor can be found by integrating this equation over the entire conductor lIUliace:
2.4. Properties of a Closed Conduding Shell
It was sho.wn that in equilibrium there.are no excess charges inside a conductor, viz. the material inside the conductor is electrically neutral. Consequently ~ if the substance is removed from a certain volume inside a conductor (a closed cavity is created), this does not change the field anywhere, Le. does not affect the equilibrium distribution of charges. This means that the excess charge is distributed on a conductor with a cavity in the same way as on a uniform conductor, viz. on its outer surface. 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. the screening of bodies, e.g. measuring instruments, from the influence of external electrostatic fields. 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. Let us take a closed surface S enveloping the cavity and lying completely in the material 4*
52
2. A Conductor in an Electrostatic Field
2.4. Properties Df a Closed Conducting Shell
of the conductor. 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 2.5, when the surf/tge
space outside the cavity, the field created by the charges located inside the cavity. 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.ng only a conducting shell around the cavity, the field WIll not be ~hanged anywhere and will remain equal to zero beyond thIS shell. Thus the field of the charges surrounded by a conducting shell a~d of the charges induced on the surface of the cavity (on the inner surface of the shell) is equal to zero in the entire outer space. We arrive at the following important conclusion: a closed conducting shell divides the entire space into the inner .and outer parts which are completely independent oj one another in respect of electric fields. This must be i~te:preted as follows: any arbitrary displacement of charges InSIde the shell does not introduce any change in the field of the outer space, and hence the ch.arge distribution on the outer surface of the shell remains unchanged. The same refers to the field inside the cavity (if it contains cbarges) and to the ~istribution ?f charges induced on the cavity walls. 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.
p
r
Fig. 2.5
Fig. 2.6
of the cavity itself contains equal quantities of positive and negative charges. However, this assumption is prohibited by another theorem, viz. the theorem on circulation· of vector E. 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. Since E = 0 inside the conductor, the field flux through a closed surface surrounding the cavity is also equal to zero. 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. In equilibrium the charges induced on the surface of the cavity are\arranged so as to compensate compleLely, in the
53
Example. A point charge q is within an electrically neutral sh~ll whose outer surface has spherical shape (Fig. 2.6). 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 . the outer spherical surface since, a~ was shown ab«;lve, the field of the point charge q and of the charges mdu.ced on the. mner surf~ee ?f the sphere is equal to zero everywhere outsIde the caVIty. Next, III VIew of symmetry, the charge on the outer surface of the shell is distributed uniformly, and hence 1 q qJ =
4n: eo -,:-.
An infinite conducting plane is a special case of a closed conducting shell. The -space on one side of this plane is .electrically independent of the space on its other side. We shall repeatedly use this property of a closed conducting shell.
2.5. General Problem of Electrostatilcs
2.5. General Problem of Electrostatics.
static case the charge is distributed over thlle surface of a conductor in a unique way as well. Indeed, th'Iere is a one-toone correspondence (2.2) between the charges; on the conductor and the electric field in the vicinity of itts surface: (J =
Image Method
d . ~r~~ue~tlY~ we must solve problems in which the charge t~:.rl ~tJon IS . unknown but the potentials of conductors Jr s ape an d relative arrangement are given We must find th~ pot~ntial cp (r) at any point of the field between the ~~a~d uc ors. t should be recalled that if we know the potend th (r),. the field E (r) itself can be easily reconstructed an f en Its value in the immediate vicinity of the condu~tor s'!r a~es .can be used for determining the surface ch dlstrlbutlOll for the conductors. arge The Poisson and I aplac t· L . equation for the functi e equa .lOns. et ~s derIve the differential into the left-hand sid~~( /r~t)n~hal). For t~IS p¥rpose: we substitute i.e. E = -v
.
ell
e
I
Olsson equation:
V2
7thh:~ r~eifo~ Lapl~ce
(2.8)
operator (Laplacian). In Cartesian coordinates
V2=~ 2
1:\
55
2. A Conductor in an El~ctrostatic Field
ax
a +~ ay + az' 2
2
'
i.e. the scalar product v·v [see (1.19)). is tran~f~:ma:d ~~tcharg~s beltween th!! cond.uctors (p = 0), Eq. (2.8) o a SImp er equatIOn, VIZ. the Laplace equation:
I
V'
O.
I
(2.9)
Eq.S:(2~h8)t~~(4~~ f~\h~t~~ii;;:s~a~~t ~:t~:e~u~h~i~~n~~~ :~~sfies
qUIres t e given values
It can ~e prove? theoretically that this problem has a unique solutIOn. ThiS statement is called the uniqueness theorerr;. Fro~ the physical point of view, this conclusion is q'!Ite obvIOUS: If there are more than one solution there wIll be .several potential "reliefs", and hence the field E at each pom~ generally has not a single value. Thus we arrive at a phYSically absurd conclusion. Usm~ the uniqueness theorem, we can state that in a
(a)
(b)
(e) l
Fig. 2.7 = EllEn. 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. (2.8) aM (2.9) in the generail case is a complicated and laborious ·problem. The analytic solutions i of these equations were obtained only for a few particular cases. As ffor the uniqueness theorem, it simplifies the solution of a number of «electrostatic problems. 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). Example. Prove that in an empty cavity of a cconductor the field is absent. The potential d that makes it possible to calculate in a simple way the ,electric field in some (unfortunately few) cases. Let us consiider this method by using a simple example of a point charge ~q near an infinite conducting plane (Fig. 2.7a). , The idea of this method lies in that we mmst find another problem which can be easily solved and whwse solution or a part of it can be used in our problem. In ourr case such a simple problem is the problem about two changes: q and -g.
---56
2. A Conductor in an Electrostatic Field
2.6. C.pacUance. Capacitors
57
The field of this system is well known (its equipotential surfaces and field lines are shown in Fig. 2.7b). Let us make the conducting plane coincide with the middle equipotential surface (its potential qJ = 0) and remove the charge --g. According to the uniqueness theorem, the field in the upper half-space will remain unchanged. Indeed, qJ = 0 on the conducting plane and everywhere at infinity. 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. 2.7c). It should he noted that we can arrive at this conclusion proceeding from the properties of a closed conducting shell [see Sec.2.1J. since hoth half-spaces separated by the conducting plane are electrically independent of one another, and the removal of the charge -g will not affect the field in the upper half-space. Thus, in the case under consideration the field differs from zero only in the upper half-space. In order to calculate this field, it is sufficient to introduce a fictitious image charge g' = -g, opposite in sign to the charge q, by placing it on the other side of the conducting plane at the same distance as the distance from g to the plane. 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.e. we strive to find another problem (eonfignration of charges) in which the field configuration in the region of space we are interested in is the same. The image method proves to be very effective if this·can be done with the help of sufficiently simple configurations. Let liS consider one more example.
charges whose action on the charge q,is eq~ivalent to the action of all
Example. A point charge q is placed between two mutually perpendicular half-planes (Fig. 2.8a). Find the location of fictitious point
It can he seen from formula (2.10) that for this jlurpose. we must mentally chargre the conductor by a charge q and calculate Its potell-
cha~~\\~~deu~~dii~3:~~~etc~;l~l~~r::'chargesfor which the eqnipoten-
tial ~ul'face~ with cp = 0 wo~J!d (a) (~) coinCide With the cond~c!Illg half-planes. One or two fI~tltIO~StL'q f charges are insuflicient III tillS I case' there should be three of -q I q the~ (Fig. 2.8b). Only with such a configuration of the ~:ystem of fOl~r I charges we can realize the ft.--l--quired "trimming"', i.e. ensure 7, j that the potential on the conI dueling" half-plane~ ~e equal to q b---lf---<>- q zero. These three hctl tlOllS cl.Jarges create just the same held within the "right angle" as the Fig. 2.8 field of the charges induced on
con~ucting plah.es. n alion of point chal'gl's {anolher proble~n), HavIll\r.found I IS con 19l1b f tlH'r questions, for example, lmd ili'e ~~~e~t~~{a~ldsfi~f/i~~~~~i~; ~n ~ny point within the "right angle" or determine the fw-ce acting Oil the charge q. the
2.6. CapaCitance. Capacitors Capacitance of an Isolated Conductor. Let I~S fconSid~r a solitarv conductor, i.e. the conductor .remove( r~m 0 101' . 1)0 d'Ies". and charge" show that the conductors, , , , . Expenments . . 1 't cl·r.r e of this conductor is directly propor~lOJ.la to I S p~~e~tia\
58
2. A Conductor in an Electro8tattc Field
2.6. Capacitance. Capacitors
tial <po In accordance with (1.23), the potential of a sphere is
J'
00
1 q 1 q Erdr=---2 dr = - - - . 4nfo . r 4n8 o R R R
I
Substituting this result into (2.10), we find
C
=
4m oR.
. the charges on the plates must be equal in magnitude I.e. ( d ) and opposite in sign ~ :.n ~apacitor is its capacitance. The basic charactens lC 0 . d onductor the capaciUnlike the capacitance of an Isolate c t' f i'ts charge to 'tor is denned as t lle ra 10 0 • tance 0 f a. cap~cl b t n the plates (lhis difference 18 the potentIal difference e wee called the voltage):
f!
f
00
59
(2.U)
The unit 'ofcapacitance is the capacitance of a conductor whose potential changes by 1 V when a charge of 1 C is supplied to it. This unit of capacitance is called the farad (F). The farad is a very large quantity. It corresponds to the capacitance of an isolated sphere 9 X 106 km in radius, which is 1500 times the radius of the Earth (the capacitance of the Earth is 0.7 mF). In actual practice, we encounter capacitances between 1 ~F and 1 pF. Capacitors. If a conductor is not isolated, its capacitance will considerably increase as other bodies approach it. This is due to the fact Jhat the field of the given conductor causes a redistribution of charges on the surrounding bodies, Le. induces charges on them. Let the charge of the conductor be g > O. Then negative induced charges will be nearer to the conductor than the positive charges. For this reason, the potential of the conductor, which is the algebraic sum of the potentials of its own charge and of the charges induced on other bodies will decrease when other uncharged bodies approach it. This means that its capacitance increases. This circumstance made it possible to create the system of conductors, which has a considerably higher capacitance than that of an isolated conductor. Moreover, the capacitance of this system does not depend on surrounding bodies. Such a system is called a capacitor. The simplest capacitor consists' of two eonductors (plates) separated by a small distance. In order to exclude the effect of external bodies on the capacitance of a eapacitor, its plates are arranged with respeet to one another in such a way that the field created by the charges accumulated on them is concentrated almost completely inside the capaeHor. This means that the lines of E emerging on one plate must terminate on the other,
(2.12) C=qlU. \ \ The charge g ofa capacitor is the charge of its positively charged' plate. ., f a capacitor is also measured Naturally, the capaCitance 0 , in farads. " t depends on its O'eometry The capacitance of a capaCl or a between th~ plates, (size and shape. of i~S pl~\~s)th~leC;p~citor. Let us derive and the m~teqalf t ~~e capacitances of some capacitors the expreSSIOns or. cuum between their plates. assuming that there is a I I te Capacitor. This capacitor Capacitance of a Para e -p a d b a ap of width h. consists of two parallel pl~ttes ~epar~~:n :cco~ding to (1.11), If the charge of the capaCi or IS g, . '1 . E _ ale . h fi Id between Its pates IS 0' the intensity of t e .e h f each plate. Consequentwhere a = glS and S IS t e area 0 . ly, the voltage between the plates IS
rr
U
=
Eh
=
ghlfoS.
Substituting this expression into (2.12), we obtain (2.13) C = foS Ih . d 'thout taking into account This calculation was rna e WI lates (edge effects). field disto~tions n~ar thel e~~~e °1a~~~lor is determined ~Y The capaCItance 0 a rea p tely the smaller the gap h III this formula the more. accura. nsions of the plates. compari~on with th; l~ne~r t~~acitor. Let the radii of thE CapaCItance of a p. enca es be a and b respectively. II inner and outer capaclt?r pl.at field intensity between thE the charge of the capaCItor. IS g, . plates is determined by the Gauss theorem. 1
Er = into
q
rt·
\'
60
2. A Conductor in an Electrostatic Field Problems
The voltage of the capacitor is charges on the outer surface. It follows from this expression that 1 _ _q_ cP - 4n80 r a b
u = J"r E r dr = _q_ ( ..!.. __ 1 ). 4n80 a b a
(..i.. __ +..!..).
It can be easily seen that the capacitance of a spherical capacitor is given by C=4Jt£o ~ b-a .
(2.14)
It is interesting that when the gap between the plates is small, Le. when b - a «a (or b), this expression is reduced to (2.13), viz. the expression for the capacitance of a parallel-plate capacitor. Capacitance of a Cylindrical Capacitor. By using the same line of reasoning as in the case of a spherical capacitor, we obtain ol c-·- In2n8 (b/a)
(2.15)
,
where l is the capacitor's length, a and b are the radii of the inner and outer cylindrical plates. Like in the previous case, the obtained expression is reduced to (2.13) when the gap between the plates is small. The influence of the medium on the capacitance of a capacitor will be discussed in Sec. 3.7. Problems • 2.1. On the determination or potential. A point charge q is at a distance r from the centre 0 of an uncharged spherical conducting layer, whose inner and outer radii are equal to a and b respectively. Find the potential at the point 0 if r < a. Solution. As a result of electrostatic induction, say, negative charges will be induced on the inner surface of the layer and positive charges on its out{~r surface (Fig. 2.9). According to the principle of superposition, the sought potential at the point 0 can be represented in the form q-J - -1 -4rtE o
(q- - -t- \.~' - -dST, '1) - -dS- ) 0_
r
•
0+
a
:.
b
'
where the first integ-ral is taken over all the charges induced on the inner surface of the layer, while the second integral, over all the
a
It should be noted that the potential.can. be found in such ~f~e form only at the poin~ 0 s~nce all the. like mduced charges are a e same distance from this pomt and theIr distribution (which is unknown to us) does .not play any role.
• 2.2. A system consists of two concentric spheres, the inner sphere of radius R l having a charge ql' What charge q must be placed onto the outer spher~ of radius R 2 to make the potential of the inner sphere equal to zero? What will be the dependence of potential cP on the distance r from the centre of the system? Plot schematica~ly the graph of this dependence, assummg that ql < o. 4.' 'Fig. 2.9 Solution. We write the expressions. . h for potentials outside the slstem (CPn) and In the regIOn between t e spheres (CPI): 1 ql +q2 1 ql CPn=-- - - - , CPI=-4-- - - CPo, 4n80 r n80 r
+
where CPo is a c~r~ain constant. Its val~ can be easily found from the boundary conditIOn: for r = R 2 , CPn - CPl, Hence lfu = q2/4ncoR2' From the condition
ql
1-R 2/R J r
,CPr
=.3..L 4n80
(_1r
=
-ql R 2 IR l •
~).
R1
• 2.3. The force acting on a surface charge. An. uncharged metallic sphere of radius R is placed into an external umform. field'Js a result of which an induced charge appears on the sphere Wit~ su ice density = 00 cos 'ft, where 00 is a positive consta.nt an~ ~ IS ~ po ar angle. Find the magnitude of the resultant electnc force actmg on like charges. Solution. According to (2.5), the force acting on the area element dS is
°
dF=-taEdS. .
'"
(1)
It follows from symmetry considerations that t~e resultant force f is directed along the Z-axis (Fig. 2.11), and henc0 it can be represented
2. A Conductor in an Electrostatic field
Problem.
as the sum (integral) of the projections of elementary forces (1) onto the Z-axis: dF z = dF cos {}. (2)
where the minus sign indicates that the induced charge is opposite to sign to the point charge q.
It is expedient to take for the area element dS a spherical zone dS =
• 2.5. A point charge q is at a distance 1 from an infinite conducting plane. Find the work of the electric force acting on the charge q done upon its slow removal to a very large distance from the plane. f
1
o F
-q
Eq
I
,
o Fig. 2.10
= 2nR sin t}·R dt}. Considering that E = a/eo, we transform (2) as follows: clFz = (na I R2/ eo ) sin t} cos {} d{} = - (na~R2/po) cos3 {} d (cos t}). Integrating this expression over the half-sphere (i.e. with respect to cos if between 1 and 0), we obtain F = na8R2/4eo' • 2.4. Image method. A point charge q is at a distance 1 from an infinite conducting plane. Find the density of surface charges induced on the plane as a function of the distance r from the base of the perpendicular dropped from the charge q onto the plane. ~olutton. Accord.ing to (2.2), the :surface charge:density on a conductor IS connected .wIth the electric field near its surface (in vacuum) through the relatIOn a = eoEn' Consequently, the problem is reduced to determining the field E in the vicinity of the conducting plane. Using the image method, we find that the field at the point P (Fig. 2.12) which is at a distance r from the point 0 is
Hence
dx
x
Fig. 2.13
Fig. 2.12 Fig. 2.11
q 1 E=2Eqcosa=2 - - - - 4neox2 x '
x
q
Solution. By definition, the work of this force done upon an elementary displacement dx (F;.ig. 2.13) is given by ql 6A = F % dx = - 4neo (2%)1 dx,
where the expression for the force is obtained with the help of the image method. Integrating this equation over x between land 00, we find
r
00
ql
A= - 16neo
dx
J 7= I
ql
16ne ol •
Remark. An attempt to solve this problem in a different way (through potential) leads to an erroneous result ~h!ch differs from w~at was obtained by us by a factor of two. ThIS IS because the re.latIOn A = q (IJ>l -
• 2.6. A thin conducting ring of radius R, having a charge q, is arranged 80 that it is parallel to an infinite conducting plane at a ~is tance I, from it. Find (1) the surface charge density at a point of the plane, which is symmetric with respect to the ring and (2) the electric field potential at the centre of the ring. . Solutton. It can be easily seen that in accordance with the image method, a fictitious charge -q must be located on a similar ring but
65
Problems
2. A Conductor in an Electrostatic Field
on the other side of the conducting plane (Fig. 2.14). Indeed, only in this case the potential of the midplane between these rings is equal to zero, Le. it coincides with the potential of the conducting plane. Let us now use the formulas we already know. (1) In order to find {J at the point 0, we must, according to (2.2), find the field E at this point (Fig. 2.14). The expression for E on the R
the required force (see Fig. 2.15b) 2
V 2-
1
q2
F =c F 2 - F 1= ---'74- 2a 2 neo
,
and answer the second question. • 2.8. Capacitance of parallel wir~s. :rwo long straight wires with the same cross section are arranged In aIr parallel to one another.
1 2
1I
I
A
-qC---l;---~
........ _----
(8)
B
(b) Fig. 2.15
Fig. 2.14
Fig. 2.17
Fig. 2.16 axis of a ring was obtained in Example 1 (see p. 14). In our case, this expression must he doubled. As a result, we obtain ql {J=
21t (RI+12)'/1
(2) The potential at the centre of the ring is equal to the algebraic
sum of the potentials at this point created by the charges q and -q: ljl=
1 41t8o
(q if
q) yR2+411 •
• 2.7. Three unlike point charges are arranged as shown in Fig. 2.15a, where AOB is the right angle formed by two conducting half-planes. The magnitude of each of the charges is I q I· and the distances between them are shown in the figure. Find (1) the total charge induced on the conducting half-planes and (2) the force acting on the charge -q. Solution. The half-planeD forming the angle AGE go to infinity, and hence their potential q; = O. It can be easily seen that a ·~Ylltem haVing equipotential surfaces with ljl = 0 coinciding with the conducting half-planes has the form sho1inl in Fig. 2.15b. 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: -qe By reducing the syst0m to iour point ch:trges, we can easily find
The distance between the wires is 11 times larger th~n the radi~s of the wires' cross section. Find the capacitance of the WIres per umt length provided that 11:;;}> 1. Solution. Let us mentally charge the two wires by charg~s of the same magnitude and opposite signs so that t~e charge per um.t length is equal to "'. Then, by defmition, the reqUIred capaCitance is Cu
=
",IU,
(1)
and it remains for us to find the potential difference between the wi~es. It follows from Fig. 2.16 showing the dependences of the potentl~ls lp + and Ip_ on the distance between the plates that the sought potential difference is
u= ILlcp+1 + ILlCP-l =2jLl(jl+I·
(2)
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: E = "'/2ne ox. Then b-a
I 8(jl+ I =
J\
'" b-a E dx = 2neo In --a- ,
(3 )
a
where a is the radius of the wires' cross section and b is the separation between the axes of the wires. 5-0181
.,
!i 'I'
A Conductur ------tili
'I' 1,1., .
'I' i I
!
!
2.
3.1. Polarization
in an Electroltatlc Field
difference between them is equal to zero. Consequently,
It follows from (1), (2) and (3) .nat
E 1X l 1
Cu = neo/ln 1').
where we took into account that b :it> a. • 2.9. 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. 2.17). Solution. 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. 2.18). 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. By defini tion, the capacitance of the system in this case is C = qo/V
E = 2E'.
E 2x (l -
= 0,
11)
where E 1x and E x are the projections of ve.ctor E onto the X-axis to the left and to tte right of the plane P (Fig. 2.19b). On the other hand, it is clear that a
=
-(a1
+ a.),
where, in accordance with (2.2), a 1 = eOE1n1 = eOE 1X and a2 = e E 2 = -eOE 2X (the minus sign indicates that the normal o n, . .) is directed oppositely to the umt vector 0 f t.he X,-axIs. . Eliminating E!X and E 2x from these equatIOns, we obtam a1
=
-a (I -
11)/1,
a2
=
= U2
-al 1/l·
The formulas for charges q1 and q2 in terms ~f q have a si~ilar form. It would be difficult, however, to solve thiS problem WIth the help p
-
r------
r-r-A
-=01" _02
B
x
~
r-1 r-
a
;r
(2)
And since E ex a, we can state that according to (2) the charge qo on plate 1 is divided into two parts: qo/3 on the left side of the plate 1 and 2qo/3 on its right side. Hence V = Eh = ahlE o = 2qoh/3e oS, and the capacitance of the system (between points 1 and 2) is 2h
+
. (1)
where U is the required potential difference between the points 1 and 2. Figure 2.18 shows that the putential difference V between the inner plates is twice as large as the potential difference between the outside pair of plates (both on the right and on the left). This also refers to the field intensity:
C= 3eoS
67
0/ Dielectrics
(b)
(a)
Fig. 2.1st=
Fig. 2.19
of the image method, since it would require an infmite series of fIctitious charges arranged on both sides of the charge q, and to find the field of such a system is a complicated problem.
•
• 2.10. 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. 2.19a). Solution. Let us use the superposition principle. We mentally place somewhere on a plane P the same charge q. Clearly, this will double the surface charge on ~ach plate. If we now distribute uniformly on the surface P a certain cnarge with surface density a, the electric field can be easily calculated (Fig. 2.19b). The plates are connected by the wire, and hence the potential
3. Electric Field in Dielectrics 3.1. Polarization of Dielectrics Dielectrics. Dielectrics (or insulators) are substances that practically do not conduct electric current.. This ~neans that in contrast, for example, to conductors dielectriCS do not contain charcrcs that can move over considerable distances t> and create elcctric current.
~
'
,"~i)
69
3.1. Polarization of Dielectrics
68
3. Electric Field in Dielectrics
When even a neutral dielectric is introduced into an external electric field, appreciable changes are observed in the field and in the dielectric itself. This is because the dielec~ric is acted up?n. by a force, the capacitance of a capacitor Increases when It IS filled by a dielectric, and so on. 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 . of a crystal lattice (ionic crystals, for example, of the NaCI type). 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. Nonpolar molecules do not have intrinsic dipole mo~ents, since the "centres of mass" of the positive and negatIve charges in them coincide. Polarization. Under the action of an external electric field, dielectric is polarized. This phenomenon consists in the following. If a dielectric is made up by nonpolar molecules, the positive cll~rge. in each molecule is shifted along the field and the negatIve, In the opposite direction. If a dielectric consists of polar molecules, then in the absence of the field their d~pole moments are oriented at random (due to thermal motIOn). 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 ?f the NaCI ty~e, an external field displaces all the positive IOns along the field and the negative ions, against the field. * Thus, the m~chanism of polarization depends on the structure of a dielectric. For further discussion it is only important that regardless of the polarization mechanism all the positive ~harges duri~g this process are displaced aiong the field, whIle the negatIve charges, against the field. It should be noted that under normal conditions the displacements of charges are very small even in comparison with the dimensions of the molecules. This is due to the fact that the intensity of the external field acting on the dielectric is consider* There exist ionic crystals polarized even in the absence of an external field. This property is inherent in dielectrics which are called electrets (they resemble permanent magnets).
ably lower than the intensities of internal electric fields in the molecules. Bulk and Surface Bound Charges. As a result of polarization, uncompensated charges appear on the dielectric surf!"ce as well as in its bulk. To understand better the mechamsm £=0 p~
7~
.!.-
~I
I
I
I I I
I I
p~
,
It I p,+ I
I:
I
I
I. I.
X
X
X
(a)
p~
(b)
(c)
Fig. 3.1 A.~
of emergence of these charges (and especially bulk charges), let us consider the following model: Suppose that we have a plate made of a neutral inhomogeneous dielectric (Fig. 3.~a) whose density increases with the coordinate x accordIng to a certain law. We denote by p~ and p~ the magnitudes of the volume densities of the positive and negative charges in the material (these charges are associated with:nuclei and electrons). In the absence of an external field, p~ = p~ at each point of the dielectric, since the dielectric is electrically neutral. However, p~ as well as p~ increase with x due to inhomog~ neity of the dielectric (Fig. 3.1b). This figure shows that III , the absence of external field, these two distributions exactly coincide (the distribution of p~(x) is shown by the solid line, while that of p~ (x), by the dashed line). 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. 3.1c). As a result, uncompensated charges will appear on the dielectric surface as well as in the bulk (in Fig. 3.1 an uncompensated negative charge appears in the bulk). It should be noted that the re-
3.2. Polarization 70
3. Electric Field in Dielectric,
version of the field direction changes the sign of all these charges. It can be easily seen that in the case of a plate made of a homogeneous dielectric, the distributions p' (x) and p~ (x) would be IT-shaped, and only uncompensated surface charges would appear upon their relative displacement in the field E. 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. They can move only within electrically n~utr~l m?lecules. We shall denote bound charges by a prime (q , p , (J). Thus, in the general case the polarization of a dielectric !ea~s to the appearance of surface and bulk bound charges III It.
We shall call the charges that do not constitute dielectric molecules the extraneous charges. * These charges may be located both inside and outside the dielectric. 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: 0 E = Eo + E', (3.1) where Eo and E' are macroscopic fields, Le. the microscopic fields of extraneous and bound charges, averaged over a physically infinitesimal volume. Clearly, the field E in the dielectric defined in this way is also a macroscopic field. 3.2. Polarization
Definition. It is natural to describe polarization of a dielectric with the help of the dipole moment of a unit volume. If an .ext~rnal field or a dielectric (or both) are nonuniform, polarl.zatlOn. turns out to be different at different points of t~e dlel~ctrlc. In order to characterize the polarization at a given POIllt, we must mentally isolate an infinitesimal volume !'1 V containing this point and then find the vector sum of the • ~xtraneous c~arg~s are frequently called free charges, but this term IS not convement In some cases since extraneous charges may be not free.
,
I
'1 1,1
'1'
i i!
','II
7f
dipole moments of the molecules in this volume and write the ratio
(3.2) Vector P defined in this way is called the polarization of a dielectric. 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 !'1 V contain !'1N dipoles. We multiply and divide the right-hand side of (3.2) by !'1N. Then we can write P = n (p), (3.3) where n = !'1NI!'1V is the concentration of molecules (their number in a unit volume) and (p) = ('T.PI)!!'1N is the mean dipole moment of a molecule. 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 !'1 V inside the dielectric. Upon polarization, the positive charge p+!'1 V contai~ed in this volume will he displaced relative to the negative charge by a distance I, and these charges will acquire the dipole moment !'1p = p~ !'1 V·1. Dividing both sides of this formula by !'1 V, we obtain the expression for the dipole moment of a unit volume, i.e. vector P: P = pi.!. (3.4) The unit of polarization P is the coulomb per square meter (elm'). 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. For an isotropic dielectric and for not very large E, there exists a relation P = 'X8 oE J (3.5) where 'X is a dimensionless quantity called the dielectric susceptibility of a substance. This quantity is independent of E and characterizes the properties of the dielectric itself. 'X is alwavs greater than zero. Hencef~rth, if the opposite is not stipulated, we shall
73
3. Electric Field in Dielectrics
3.3. Properties 0/ the Field 0/ P
consider only isotropic dielectrics for which relation (3.5) is valid. H~wever, there exist dielectrics for which (3.5) is not applIcable. These are some ionic crystals (see footnote on page oR) and ferroeZectrics. The relation between P and E for ferr~electr~cs .is nonlinear and depends on the history of the dIeI?ctnc, l.e. on the previous values of (this'phenomenon IS called hysteresIs).
of the positive and negative bound charges as a result of polarization. Then it is clear that the positive charge p:Z+dS cos a inclosed in the "inner"part of the oblique cylinder will pass through the area element dS from the surface S outwards (Fig. 3.2b). Besides, the negative charge p:tdS cos a enclosed in the "outer" part of the oblique cylinder will enter the surface S through the area element dS. 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. area element dS of the surfa~e S in the outward direction: dq' = p:Z+ dS cos a + Ip:IL dS cos a.
72
E;
3.3. Properties of the Field of P The Gauss Theorem for th£' Field of P. 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
Since
p
I r: I =
dq'
=
p~ we have
P+ (l+
+ L) dS
cos a
=
p+l dS cos a,
(3.7) where l = l+ L is the relative displacement of positive and negative bound chaJ;ges in the dielectric during polarization. Next, according to (3.4), p:l = P and dq' = P dS cos at or
"+
(a)
( b) Fig. 3.2
ar~itrary closed sur~ace S is equal to the excess bound charge
(wIth the reverse sIgn) of the dielectric in the volume en. closed by the surface S, i.e.
dq'
=
P n dS
=
P·dS.
(3.8)
Integrating this expression over the entire closed surface
S, we find the total charge that left the volume enclosed by the surface S upon polarization. This charge is equal to
~ PdS=
-qint.
(3.6)
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. 3.2a, the dielectric is hatched). When an external electric field is switched on the ~ielectric is polarized-its positive charges are displaced relative to the negative charges. Let us find the charge which passes through an element dS of the closed surface S in the outward direction (Fig. 3.2b). Let 1+ and C be vectors characterizing the displacement
~ P.dS. As a result, a certain excess bound charge
?'
will be left inside the surface S. 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 (3.6). Differential form of Eq. (3.6). Equation (3.6), viz. the Gauss theorem for the field of vector P, can be written in the differential form as follows:
I
V·P=-p',
(3.9)
c:
74
3.3. Properties of the Field of P 3. Electric Field in Dielectrics
i.e. the divergence of the field of vector P is equal to the volume density of the excess bound charge at the same point, but taken with the oppOsite sign. This equation can be obtained from (3.6) in the same manner as the similar expression for vector E was obtained (see p. 24). 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: (1) the dielectric is homogeneous and (2) there are no extraneous charges within it (p = 0). Indeed, it follows from the main property (3.6) of the field of vector P that in the case of a homogeneous dielectric, we can substitute xeoE for P in accordance with (3.5), take x out of the integral, and write x
&eoE·dS .. -q'. oJ
The remaining integral is just the algebraic sum of all the charges-extraneous and bound-inside the closed surface S under consideration, Le. it is equal to q + q'. Hence, x (q + q') = - q', from which we obtain q' =
- 1~X
q.
(3.10)
This relation between the excess bound charge q' and the extraneous charge q is valid for any volume inside the dielectric, in particular, for a physically infinitesimal volume, when q' __ dq' = p'dV and q -+- dq = pl1V. Then, after cancelling out dV, Eq. (3.10) becomes , p =
-
x 1+x p.
• (J.H)
Hence it follows that in a homogeneous dielectric p' = 0 when p = O. 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
75
such a dielectric there is no excess bound bulk1ch~rge tn~ only a surface bound charge appears as a resu t 0 po arl-
zat~~·us
find the relation between polarizatio? P a nd ~he 'ty a' of the bound charge at the lOterf ace ed surf ace ensl . 1 t s use property tween the dielectrics. For this purpose, e u ' (3.6) of the field of vector~. We D choose the closed surface lO the form of a flat cylinder whose end2~J~~~t--faces are on different sides of the 1 interface (Fig. 3.3). We shall asI sume that the height of the cyl• n' inder is negligibly small and the F' 3 3 area !1S of each endface is so small Ig. . that vector P is the same at all . d' . ts of each endface (thiS also refers to the surface ensl~Y ~?I~f the bound charge). Let 11 be the common normal to t e interface at jl. given point. We shall always draw vector n froDm. dielecd~riC ~h~o g~~e~}ri~ ~hrOugh the lateral surface d 'th (3 6)' Isregar Illg ., of the cylinder, we ca'h write, in accor ance WI P !1S + Pin' t1S = - a' t1S, 2n h P and P , are the projections of vector P in d ielecw. er; on2to the n~~al n and in dielectric 1 onto the normal t~I(p' 3 3) Considering that the projection of. vector Ponto n Ig. . 1 ~, is equal to the projection of thIS vector o~to . t h e norma. ( m n) normal n taken with the opposIte t~e o~pos~te ~m ~ we can ~ite the previous equation sIgn, l.e. tn' - In' 11' t1S)' in the following form (after cance mg . P2n - Pin = - a'. (3.12) This means that at the interface bet~een ~iel~ctrics the normal component of vector P ha~ a dlsc.ofntIDdu!ty, ;~ose . d d d a' In partIcular, I me mID IS a magmtu e epen s on . d d't' (3 12) acquires a vacuum, then P 2n = 0, an con I Ion . simpler form: (3.13) where P n is the projection of vector ~ onto th~ outward normal to the surface of a given dielectrIC. The SIgn of the pro-
('
76
3. Electric Field in Dielectrics
jection P n determines the sign of the surface bound charge a' at a given point. Formula (3.13) can be written in a diHerent form. In accordance with (3.5), we can write (3.14) where En is the projection of vector E (inside the dielectric and in the vicinity of its surface) onto the outward normal. Here, too, the sign of En determines the sign of a'. A Remark about th(' Field of Vector P. Relations (3.6) and (3.13) may lead to the erroneous conclusion that the field of vector P depends only on the hound charge. Actually. this is not true. The field of vector P, as well as the field of E, depends on all the charges, both hou~d and extraneous. This can he proved if only by the fact that vectors P and E are connected through the relation P = xeoE. 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.
3.4. Vedor D The Gauss Theorem for Field D. 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:
~ eoE dS =
(q + q')lnt,
(3.15)
where q and q' are the extraneous and hound charges enclosed by the surface S. The appearance of the hound charge q' complicates the analysis, and formula (3.15) turns out to be of little use for finding the field E in a dielectric even in the case of a "sufficiently good" symmetry. 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 (3.6). Then
a.4.
77
Vector D
expression (3.15) C an he transformed as follows:,
~ (eoE + P) dS = qlnt. The quantity in the parentheses in t~~ integrand by D. Thus, we have defined an auxIlIary vector
ID=eoE+P, I
(3.1.6) ~
denoted (3.17)
through an arbitrary dosed surface is equal.to the h fl w ose uxsum of extraneous c1larges enclosed hy this sur'algebraic face:
I~ DdS =
(3.18)
qlnt.\
.- This statement is called the ~auss thearer- for fiel~e~ I t should h~. n ote? ~h~t ve~or D ISpth~~~~~ist~Za~~~P it is d ly different quantities. eo a~. have an' deep inde~d an aux~liary veft~~rW~~: ;::~e~t~ of the leld of physICal meamng. Howe 't. (3 18) justifies the introvdectt~r Do'f etxhf:~:~~or~~ne~ua~~o~ase~ it ~onsiderably simpliuc IOn . d' I t ' * fies the ~naly(s3iS107f) thedfle(idH~n) a:: ~can~sfor any dielectric, '. , RelatIOns . an
both isotr?pic (~nf7)anS\~~~~P;~~t the dimensions o.f vect~r D as those of vector P. The 2quantlty D IS measured in coulombs per square metre (elm ).
~~~r~:~lOs~me'
Differential form of Eq. (3.18) is
I V·D=p, I
(3.19)
. equal to the volume densi ty of an Le. the divergence 0 f th e field D is oint extran~ous ch~rge .at the obat~i~fJ fro~ (3.18) in the same way as it can Ee,\see p. 24\" It suffices to replace E by D and wasThis doneequatiOn for the field take into account only extraneous charges. . '. ftPJ] called dielectric displacement! or electro* The ql;'antity this D term, however, III order to static inductton. We Dh~~llo s a no i b e u~ing ~ .
emphasize the auxiliary nature of
\~etor
.
78
79
3.4. Vector lJ
3. Electric field in Dielectrics
At the points where the divergence of vector D is positive we have the sources of the field D (p > 0), while at the points where the divergence is negative, the sinks of the lield D (p < 0).
Relation Between Vectors D and .E. In the case of isotropic dielectrics, polarization P = xeoE. Substituting this expression into (3.17), we obtain D = 8 0 (1 x) E, or
Let us illustrate what was said above by several examples. Example t. An extrallPUUS 'poil~t charge '1. is loc.atc(~ at th~ cenl~e of a sphere of radius a, made 01 an IsotropIC dIelectrrc WIth a dlelectrrc
+
I
D=eoeE,
I
(3.20)
where e is the dielectric constant of a substance:
8=1+x.
(3.21)
The dielectric constant e (as well as x) is the basic electric characteristic of a dielectric. For materials 8> 1, while for vacuum 8 = 1. 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). The value of e for water is rather high (8 = 81). Formula (3.20) shows that in isotropic dielectrics vector D is collinear to vector E. 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). This follows if only from the relation D = 8 oeE. .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 (3.18) and (a.1!J). These laws express only a certain property of field D but do not determine this field proper.
r
Fig.
Fig. ;LS
~L4
constant e. Find"'ihe projectiun 1,', of Iield intensity E as a functiun of the distance r frolll the centre of this sphere. The syIllllletry uf the system allows us to use the Gauss th~orem for vector D for sol ving the 'problem (we cannut usc here the srn.l1lar theorem for vector E, since the bound char~e is unlwo.wn to us). l' or a sphere of radius r with the centre at the POlllt of locatIOn of the charge q we can write the following relation: 4;r.r"lJ r . = q. Hence we can lind lj, and then, usillg formula (3.20), the required quantity t r : 1
E r (r < a) = 4ne";
'/ <:/:2'
E
r (r > a
)
1
q
=="4ne o ~.
Figure 3.4 shows the curves lJ (r) and E (r). Example 2. Suppose that a system consists of apoint ~harg.e q >.0 and an arbitrary sample 01 a homogelleou~. IsotropIc dielectrrc (Fig. 3.5), where S isa certain closed surface. h~d ?ut what Will happen to the ileIds of vectors E and D (and to theIr fluxes through the surface .) if the dielectric is removed. The field E at any point of space is determined by the charge q and by bound charges of the polarized diele~tr.ic. Since in o.ur case D = e eE this refers to the lield D as well: It IS also determIned by the exrra~eous charge q and by the bound char~es of the dielectric. The removal of the dielectric will change the held E, and hence the lield D. The flux' of vector E through the surface S will also cha~ge, since negative bound charges will vanish from inside. this s~rface. However the flux of vector D through the surface SWill remalll the same in ;pite of the change in the lield D. . . . Example 3. Let us consider a system C?ntallllllg no extraneo.us charges and having only bound charges. :::;uc~ a ~ystem can, for example, be a sphere made 01 an electret (see p. 6b). 1, I~ure 3.?a shows the field E of this sy::;tem. What can we say about the fIeld Dt
RO
3, Electric Field in Dielectrics
First of all, t~e abs~nee of extraneous charges means that the field has ~o sou:(;~s, the Imes of D d.o IIOt emerge or terminate anywhE're. lJowever, the 11I~ld J) PXISts 'md.IS.sllown I'n I"I'g . 3 .u. "b 'fl}' _ ( Ie Illes 0 l' LL' J)
81
3.5. Boundary Conditions
where the projections of vector E are taken on the direction of the circumvention of the contour, shown by arrows in the figure. If in the lower region of the contour we take the n n ..
Lines ofE
Lines of D
(a)
Cb)
Fig, 3.G and ~ coi~cid~ outsi?e the sphere, but inside the sphere they have 01.PdPoslte dlfectlOns, smce here the relation D = 2 eE is no longer val and D = FoE P. 0 '
+
!------I ---:' I
c
2 [ ~ ~ '"'r'
I~
J
•
Fig. 3.7
Fig. 3.8
projection of vector E not onto the unit vector 't' bu~ onto the common unit vector 't, then En' = - E H , and It follows from the above equation that Ii:
(3.22)
3.5. Boundary Conditions Let ~s first consider the behaviour of vectors E and D at. the mterface between two homogeneous isotropic dielectrics. Suppose that, for greater generality, an extraneous su:face charge exists at tl16 interface between these dielectrICS. The required conditions cari be easily obtained with the help of two theorems: the theorem on circulation of vector E and the Gauss theorem for vector D:
~E dI = 0
and
~ n dS =--= glnt-
. Boundary Condition for Veclor E. Let the field near the lDterface be E 1 in dielectric 1 and E 2 in dielectric 2. We choose ~ sm~ll ?l(~_ngateu rectangular contour and orient it as ~hown In Fig, .L'. The sides of the contour parallel to the lllterfac~ must have slIch a length that the fwld E over this !,eng.th ,,1Tl each dielectric can be assumed constant. The height of th~ contour must be negligibly small. Then, in accordance With the tllf'orem on circulation of vector E we IJave '
I.e. the tangential component of vector E turns out to be the same on both sides of the interface (it does no.t have a discontinuity). . Boundary Condition for Vector D. ~et us ta~e a cylInder of a very small height and arrange it at the lll~erface between two dielectrics (Fig. 3.8). 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 .!1S DIn' ·!1S = (JAS,
+
where (J is the surface density of the extraneous charge at the interface. Taking both projections of vector D onto the common normal n (which is directed from dielectric 1 to dielectric 2), we obtain Din' = - DIn' and the previous equation can be reduced to the form \
D 2n - Din = a.
(3.23)
It follows from this relation that the normal component of vector D generally has a discontinuity when passing 6-0181
82
throngh tllO inlt'l'[a(;('. If, however, tltel'O aro charges at the interface (u = 0), wo ohtain
I
Din
/10
extraneous
E H /E 2n E u /E l1l
tan Cl2 tan (;(,
LJ 1
E2
Et
FieldE Fig 3.9
FieldD
Fig. 3.1U
We see that in the case under consideration, the lines of E are
refracted and undergo discontinuities (due to the presence of bound
charges), whileihe lines of D are only refracted, WI ~hou t thscon ttnuities (since there are no extraneous charges at the Illterlace).
Boundary Condition 'on the Conductor-Dielectric Interface. If medium 1 is a conductor and medium 2 is a dielectric (see Fig. 3.8), it follows from formula (3.23) that (3.26)
(;3.2!'i)
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. ;3.9, 8 2 > 8 1), Example. Let us represent graphically the fields E and D at the i II ["ri'ace hetween two homogeneous dielectrics 1 and 2, assuming that
awl llli~re is 110 extraneous charge on this surface. Since 1'2 > ;'n in accordance with (3.25) Cl 2 > at (Fig. 3.10). Considering that the tangential component of vector E remains unchanged anrl using Fig. ;·1.9, We can easily show that F 2 < Hl in magnitude, i.e. the lines of E ill dielectric 1 must be denser than in dielectric 2, as is shown in Fig. 3.10. The fact t~at. the normal corn:.'2> 1-'1
>
•
Taking into aceount the above conditions, we obtain the law of refraction of lines E, and hence of lines D: . ~
pouents o[ vectors D are equal leads to the conclusion that LJ 2 in magnitude, Le. the lines of D must be denser in dielectric 2.
fl~1l" I
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. are refracted (Fig. 3.9). Let us flIld the relation between the angles a l and a 2 • In the absence of extraneous charges at the interface, we have, in accordancp- with (3.22) and (3.24), E 2 , = E l , and 8 2 E 2n = 8 r E ln • Figure 3.9 shows that tan Cl2 tan Cll
83
9.5. Boundary Conditions
3. Electric Field in Dielectrics
where n is the conductor's outward normal (we omitted the subscript 2 :;ince it is inessential in the given case). Let us verify formula (~).26). In equilibriulll, the electric fteld inside a conductor is E =,c 0, and hence the polarization P = O. This means, according to (;3.17), that vector D =~ 0 inside the conductor, Le. in the notations of formula (:L~3) D1 = 0 and DIn = O. IIenceD 2n = a. 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:;ity a' appeal' at the conductor-dielectric interface (recall that the volume density of bound charges p' = 0 for a homogeneous dielectric). Let us now apply the Gauss theorem to vector E in the same way as it was done while deriving formula (2.2). Considering that there are both bound and extraneous charges (a 0*
3.6. Field in 84
3. Electric Field in Dielectrics
' I
I
(3.27)
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. 3.6. Field in a Homogeneous Dielectric
III II' I :1
II
1
I'
Homogeneous Dielectric
85
a' are unambiguously connected with the extraneous charges
and G') at the conductor-dielectric interface we arrive at the following expression: E" = (G + a')le o' On the other hantl, according to (3.2G) E" = Dn/ee,o = alee o' Combining these two equations, we obtain Gle = G + a', whence 0 = - -8-1 8- 0 .
/l
It was noted in Sec. 2.1 that the determination of the resultant field E in a substance is associated with considerable difficulties, since the distribution of induced charges in the substance is not known beforehand. 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: determination of the macroscopic field E' of bound charges in each specific case is generally a complicated independent problem, since unfortunately there is no universal formula for finding E'. 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 (3.27) the charges
a on the surface of the conductor. As before, the,re will be no field inside the conductor (E = = 0). 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. In accordance with the Gauss theorem, a + G' = eDEn, where En = Dnlee o = a/eeo' and hence G + a' = Gle. (3.28) But if the charges creating the field have decreased by a factor of e everywhere at the interface, the field E itself has become less than the field Eo by the same factor: (3.29) E = Eo/e. Multiplying both sides of this equation by eeo, we obtain • D = Do, (3.30) Le. the field of vector D does not change in this case. It turns out that formulas (3.29) and (3.30) are also valid in a more general case when a homogeneous dielectric fills the volume enclosed between the equipotential sur1aces of the field Eo of extraneous charges (or of an external field). In this caBe also E = Eo/e and D = Do inside the dielectric. 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'
=
-Ple{}.
(3.3t)
This relation can be easily obtained from the formula E = = Eo + E' ifl we take into account diat Eo = eE and
P = xeoE.
.
As was mentioned above, in other cases the situation is much more complicated, and forml1las~ (3.29)-(3.3t) are inapplicable. CornU.Fie. Thus, if a homogeneous dielectric fills the entire space oeeupied by a field, the intensity E of the field
87
Problems 9. Electric Field in Dtl!lectrics
will be lower than the intensity Eo of the field of the same extraneous charges, but in the absence of dielectric, by a factor of e. Hence it follows that potential cp at all points will also decrease by a factor of e: cp = cpo/e, (3.32) where CPo is the field potential in the ahrnce 01. the dielectric. The same applies to the potential difference: U = Uoh, (3.33) where U0 is the potential difference in a vacuum, in the absence of dielectric. 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). And since it is so, the capacitance e = ql U of the capacitor filled by dielectric will increase e times e' = ee, (3.34) where e is the capacitance of the capacitor in. the absence of dielectric. It should be noted that this formula is valid when the entire space between the plates is filled and edge effects are ignored.
Problems
. .3.1. Polarization of a dielectric and the bound charge. An extraneous point charge q is at the centre of a spherical layer of a heterogeneous isotropic dielectric whose dielectric constant varies only in the radial direction as 8 = air, where a is a constant and r is the distance from the centre of the system. Find the volume density p' of a bound charge as a fuuction of r within the layer. Solution. We shall use Eq. (3.6), taking a sphere of radius r as the closed surface, the centre of the sphere coinciding with the centre of the system. Then 4Jtr2 • P r = -q' (r),
dq'.
+
(1)
between the spheres with a;:erp'4Jtr2 dr, we transform (1) , 2
r 2 dP r +2rP r dr=-pr
d
r,
whence (2)
In the case under consideration we have e-l 8-1 Pr=xeoE r =
-e D r =-;-
q 4n:r 2
. (2) will have the form and after certain transformation expressIOn 1 q p'=~
?"
which is just the required result.
f vector D An infmitely large pla~e • 3.2. The Gauss thedo~i t~~c with th~ dielectric constant e IS made of a hBtnogeneous Ie ee 1 uniformly charged by an. extraneous Ex I{J charge with volume densIt~ P > O. The thickness of the platt! IS 2a. 1~h1lii-l~ Find the magnitude of vector E an. l;; E?, the potential cp as functions of the dIstance 1 from the middl~ of. the pl a~e (assume that the poten~Ial IS ~ero n the middle of the plate), ChOOSlll I' X-axis perpendicular to the pal'. Plot schematic curves for the pro. . n E (x) of vector E and the po~~~t:fal ~(x). (2) Find the surface and volume densities of the bound charge. / Solution. (1) From symmetry co~. siderations it is clear that ~'I= ~;ii FIg. 3.11 the middle of the plate, w 1 I' a d-' other points vectors E f a~h pef P:: In order to determine E, w,e s~bIl icular to the surface 0 I' P a . D (since we know the dIstrr u'use the Gauss theorem for vector W take for the closed su~face a tion of only extran~ous charges)! hose endfaces coincid~s WIth the right cylinder of helghthl, one 0 ~'onal area of this cylInder be S. midplane (x = 0). Let t I' cross-sec 1 Then
(ll
h
y\
DS
where q' (r) is the bound charge inside the sphere. Let us take the differential of this expression: 4Jt d(rl·p r) = -
. h' I Here dq' is the bound c~arg~ ill th \ ~n, radii rand r dr. Considermg a q as follows:
DS
=
pSI,
=
pSa,
D
D
= =
pi, pu,
E
=
pl/ee:o (1 ~ a),
E = paleo (l ~ a).
88
The graphs of the funtions Ex (x) and
,
89
Problems
3. Electric Field in Dielectrics
where P is the volume density of the extraneous charge. Hence IJ p r3 -a' E-- -=-- --,- eEo 38 08 r The corresponding curves for E (r) and cp (r) are shown in Fig.· 3.12b.
e-1 =Pn="X8 0 E n =(e-1)pa/e=--. pa>O. e
This result is valid for Il0th sides of the plate. Thus, if the extraneous charge p > 0, the bound charges appearing on both surfaces of the plate are also positive. In order to find the volume density of the bound charge, we use Eq. (3.9) which in our case will have a simpler form:
p'=_ ap x ax
=_.!.( ax
o\---'---!---~, a b
e-1 px)=-- e-1 p. e e
(b)
(a)
It can be seen that the hound charge is uniformly distributed over
Fig. 3.12
the bulk and has the sign opposite to that of the extraneous charge.
• 3.3. A homogeneous dielectric has the shape of a spherical layer whose inner and outer radii are a and b. Plot schematically the curves of intensity E and potential a) = q/4nr 2 •
The required intensity is E (r < a) = 0,
E (r
>
a) = D/eeo.
The curve for E (r) is shown in Fig. 3.12a. The curve for
a
0
. ·f I d· tributed with the volumc .3.4. Extraneous ~argerUd·lu~r:\:ad~of a homogeneous dielecdeMity p > 0 over ~ sp ~:e 0 ~~nd (1) the magnitude of vector E as tric . with the pet:nntttvI y 8. he centre of the sphere and plot the function of the dl!~tanceEr(f)ro~d tm (r). (2) the surface and volume dencurves of the functions raT ' sitie5 of bound charge. d . E we shall use the Gauss theaSolution. (1) In .order to ketcr~h~e di~tribution of only extraneous rem for vector D smce we now
a
c~:
r