! E da = 4πkQ enc, has E under the integral sign, so it is not ordinarily an

Transcription

1 Physics 142 Electostatics 2 Page 1 Electostatics 2 Electicity is just oganized lightning. Geoge Calin A tick that sometimes woks: calculating E fom Gauss s law Gauss s law,! E da = 4πkQ enc, has E unde the integal sign, so it is not odinaily an equation one can use easily to get a fomula fo the E-field. But in a few cases, if the geometic symmety of the chage distibution is simple, one can in fact use Gauss's law to find the E-field. The tick is to make a cleve choice of the closed Gaussian suface though which the flux is calculated; if this suface eflects the symmety appopiately, making E da simple enough to allow E to be taken outside the integal, then the total flux becomes a simple poduct of E and a known suface aea. Dividing by the suface aea, one can obtain E. Thee ae only a few cases whee this woks; we examine two. Spheical symmety. In this case thee is a point (the cente of symmety ) elative to which the physical situation is the same in all diections. This is an impotant case because isolated atoms ae spheically symmetic with the nucleus at the cente. If a chage distibution is spheically symmetic its volume chage density ρ can vay only with the distance fom the cente of symmety, not with the diection. The E-field poduced by this chage must eflect this symmety, so its diection is adial elative to the cente of symmety outwad fo positive net chage, inwad fo negative net chage and its magnitude depends only on the distance fom that cente. To find the field at distance we take fo ou Gaussian suface a sphee of adius about the cente of symmety. Then da will point adially outwad, so E da = E da, whee E, the adial component of the field, is positive if the field lines go outwad though the suface, negative if they go inwad. By the symmety E has the same value at all points on ou Gaussian suface, so fo the integation it is a constant. The total flux becomes! E da = E! da = E 4π 2. On the ight side of Gauss's law we have 4πk times the net chage inside ou sphee of adius, which we call Q(). Then Gauss s law gives E 4π 2 = 4πk Q(), o E () = k Q() 2. If Q() is negative, E is negative because E is diected inwad, towad the cente of symmety. This simple fomula gives the E-field fo any situation of spheical symmety. In analyzing a situation with spheical symmety, unless you ae instucted to do othewise, you can simply stat fom this fomula.

2 Physics 142 Electostatics 2 Page 2 Note that the fomula has exactly the same fom as that fo the E-field of a point chage of amount Q() placed at the cente of the distibution. All spheically symmetic situations have this vey simple geneal popety: The E-field at distance fom the cente of a spheically symmetic distibution is the same as that of a point chage located at the cente, with chage Q(), the total chage contained in a sphee of adius about the cente. The fist dawing shows the situation when the field point is outside the entie chage distibution. The dotted cicle indicates the spheical Gaussian suface. The chage is taken to be positive, so E is adially outwad ( E is positive). In this case Q() is simply the total chage Q tot, which is a numbe independent of. This means that the E-field at points outside a spheical chage distibution is given by E () = kq tot / 2, which is the same E-field as that of a point chage Q tot located at the In this case the field point P is outside all of the chage, so Q() is cente of the distibution. the total chage of the distibution. This is not an appoximation. If the total chage is zeo, the field outside the distibution is exactly zeo. Outside a single isolated atom thee is no E-field due to the chages in the atom. (This is not tue fo molecules, which do not have spheically symmetic chage distibutions.) The second dawing shows the situation when the field point is inside the chage distibution, again fo positive chage. The quantity Q() is the total chage within the Gaussian suface, which depends on how the chage is distibuted. As goes to zeo that amount of chage will vanish unless thee is a point chage at the cente so geneally the E-field at the cente of any continuous spheical distibution is zeo. These conclusions allow fo calculation of the E-field fo any spheically symmetic chage distibution. The difficulty, if any, lies in finding the chage Q(). Plane symmety. Let the chage be spead unifomly ove a lage flat suface. If one consides only points fa fom the edges and nea the suface, one can appoximate the suface by an infinite plane and again use symmety aguments. E P da Any point on an infinite plane can be egaded as its cente, so thee is no distinction among diffeent diections paallel to the suface. Any non-zeo component of E paallel to the suface would violate this symmety, so E must be pependicula to the suface. To apply Gauss's law we constuct a pillbox cylindical Gaussian suface, with sides pependicula to the chaged plane. E is paallel to the cuved sides of the pillbox (thus E P da In this case the field point P is inside the chage distibution, so Q() is the total chage only inside the Gaussian sphee of adius.

3 Physics 142 Electostatics 2 Page 3 pependicula to da) so thee is no flux though those sides. At the ends E and da ae paallel (antipaallel) if the chage on the sheet is positive (negative). da E A da In the case shown the chage on the sheet is positive. E is diected away fom the sheet both above and below, so at both the top and the bottom of the pillbox E da = EdA. E is also constant acoss both sufaces. The total flux is thus 2AE. The chage enclosed is σ A. Gauss's law gives 2AE = 4πkσA, o E = 2πkσ = σ /2ε 0. This is the fomula we found ealie by supeposition, but this method is obviously much simple. Besides these cases, one can use Gauss s law in cases of axial symmety, whee otation about some staight line makes no change in the physical situation. An example is a unifom staight line distibution of chage, such as a chaged staight nylon thead. The appopiate Gaussian suface is a cicula cylinde with axis along the symmety axis. This is woked out in the assignments. Effects of the field on chages: the foce and toque on a dipole Now we tun fom methods of calculating the E-field to consideation of the effects a given E-field has on chages that ae placed in it. The fundamental fomula is that fo the foce on a point chage: F = qe. In pinciple one can find the total electic foce on any object by adding up the foces on all the chages in it. One might think that an object of zeo total chage placed in an electic field would show no electic effects, but this is not usually tue. Odinay neutal matte contains, in its atoms and molecules, electons and potons of both signs of the chage; these can give ise to electic dipole moments, which ceate E-fields of thei own and also inteact with E-fields due to extenal souces. These inteactions ae impotant in pactice, epesenting the main effects an extenal E-field has on unchaged matte. Conside a small dipole in an extenal E-field, as shown. Its dipole moment p makes angle θ with the diection of the field. The field exets a foce on +q paallel to E, and it exets a foce on q opposite to E. These tend to cancel, and if E is the same at the location of both chages, they will cancel exactly, giving no net foce on the dipole. This shows that: Thee is no net foce on an electic dipole in a unifom E-field. E p q +q θ E

4 Physics 142 Electostatics 2 Page 4 But if the dipole moment p is not along the line of E then these two foces do not act along the same line, so they ceate a toque about the dipole's CM. It is elatively easy to show that this toque is given by Toque on an electic dipole τ = p E This toque twists the dipole towad alignment with E. If p is aleady paallel to E, the toque is zeo and we have stable otational equilibium. Fo p anti-paallel to E the equilibium is unstable. Even if a neutal object has (on aveage) no dipole moments in it befoe it was placed in the field, the field will often induce dipole moments. The foces exeted by the field on the positive atomic nuclei and the negative atomic electons ae in opposite diections. These foces cause a small displacement of these chages fom thei equilibium aangement, ceating dipoles automatically aligned with the field. The enegy associated with these pocesses is vey impotant, and will be discussed late. If the E-field is not unifom, thee can also be a net foce on the dipole, because the foces on the positive and negative chages of the dipole may not cancel. Suppose that p is aligned along E, as it would be fo an induced dipole moment o if the toque has aleady aligned a pemanent dipole with the field. Then thee will be a net foce towad the egion whee E has lage magnitude. On the othe hand, if fo some eason the dipole wee aligned opposite to the field, the foce would be in the othe diection. In a non-unifom E-field, the foce on a dipole aligned with the field is towad the egion whee E is stonge; on a dipole aligned opposite to the field the foce is away fom that egion. It is this foce that causes small bits of matte to be attacted towad electified objects the vey fist electical phenomenon discoveed by the ancients. This is a case (thee ae many othes) whee the fist phenomenon discoveed in a field of science is fa fom the easiest to explain. How intoducing conductos esticts the behavio of the field It was found in the 18th centuy that some mateials notably metals but also some liquids and gases conduct electification (chage) fom one place to anothe, while othe mateials keep it in place. The fome we call conductos, the latte nonconductos. Non-conductos ae sometimes called insulatos, o dielectics. We now know that all mateials ae made of atoms containing chaged paticles, including electons. In some mateials the binding fo one o moe electons pe atom is weak enough that when the atoms ae close togethe (as in a solid) some of the electons become fee to move about within the mateial. It is these conduction electons that cay electic effects though the solid, making it a conducto. Mateials having vey few o no conduction electons ae non-conductos. The distinction is not totally shap: except fo some mateials at low tempeatues no mateial is a pefect conducto, and fo vey high applied E-fields insulatos can become conductos. In liquids and gases conduction takes place by motion of ionized atoms o molecules. Ou discussion hee mostly concens solids.

5 Physics 142 Electostatics 2 Page 5 In a good conducto even a feeble applied E-field within the mateial o paallel to its suface will cause the conduction electons to move. This means that if the chages in a conducto ae known to be at est (as in electostatics) thee must be no E-field eithe within it o along its suface. (Thee can be a field pependicula to the suface, since electons ae not fee to leave the suface; but usually it takes a stong E-field pependicula to the suface to pull them fee.) Shown is a conducto of non-spheical shape, caying a net positive total chage. The chages aanges themselves on the suface, but not unifomly ove it. Whee the suface is moe shaply cuved the chage density is geate, and hence the stength of the E-field at the suface is also geate. This non-unifomity is equied to make the E- field within the conducto zeo, as it must be if the system is in electostatic equilibium. If the adius of cuvatue at some pat the suface is vey small (like a needle point), the E-field thee can be quite intense even if the total chage is not vey lage. Thee ae many applications of this fact. Using Gauss s law one can pove two impotant facts about electostatic equilibium: Any non-zeo net chage density on a conducto must eside entiely on the suface. Any such suface chage density at a point is popotional to the E-field at that point. When a conducto is bought into a egion whee thee is aleady an E-field due to othe chages, at fist the conduction electons move about apidly until the total field the extenal field plus the one ceated by the edistibuted chages on the conducto itself is zeo within the conducto and has no component paallel to its suface. The chages then etun to est, binging the system into electostatic equilibium. The elaxation time fo this equilibium to be established is usually quite shot, a few ms o less. The dawing shows the situation when an unchaged conducting sphee is placed in a peviously unifom E-field. Chages ae induced as shown they should be on the suface, of couse until the total E-field (oiginal field plus those of the induced chages) becomes zeo within the conducting sphee. Then all chages can emain at est, in electostatic equilibium. The following table summaizes all these popeties: E is zeo within the conducto. E has no component paallel to the suface. Popeties of conductos in electostatic equilibium Any net chage esides on the suface. If thee is chage on the suface, its suface density is elated to the outwad nomal component of E at the suface by E n = 4πkσ = σ /ε 0.

6 Physics 142 Electostatics 2 Page 6 A useful scala field: electostatic potential To each consevative foce coesponds a potential enegy, a scala field that epesents the effects of the foce in consideations of enegy. The Coulomb foce is consevative, so thee is a potential enegy epesenting the electostatic inteaction. Just as we found it convenient to intoduce the vecto field E epesenting the electic foce pe unit chage at a point in space, it is convenient to intoduce a coesponding scala field V, called electostatic potential, epesenting the electostatic potential enegy pe unit chage at a point in space. The defining connection is this: Moving a point chage q fom a point in space whee the potential (due to othe chages) is V 1 to a point whee it is V 2 changes the electostatic potential enegy of the system by the amount ΔU = q(v 2 V 1 ). In tems of the E-field, V is given by Electostatic potential V(2) V(1) = E d Hee 1 and 2 epesent points in space, and the integal is the same type of line integal used to calculate the wok done by a foce. In SI units potential is measued in volts (V), whee 1 V = 1 J/C. The stength of an E-field is usually given in V/m instead of N/C. Like potential enegy, potential is ambiguous in that only the diffeence between its values at two sepaate points is uniquely defined. The actual value at any point depends on the (abitay) choice of whee the potential is zeo. In dealing with objects nea the eath, one fequently chooses the eath itself to have potential zeo. In cases whee the chages ae confined to a finite egion of space, one usually chooses the potential to be zeo at infinite distance fom all the chages. Equipotentials and the effects of conductos Fo chages at est, E is zeo within a conducto, and at the suface it is eithe zeo o pependicula to the suface. If we calculate the diffeence V(2) V(1) between two points in o on a conducto, we see fom the definition that it is zeo. All points in a conducto in electostatic equilibium ae at the same potential. Regions consisting of points at the same potential ae called equipotential egions. Fo chages at est, the space occupied by a conducto is necessaily an equipotential egion. As we follow a line of the E-field ( E! d ) we see fom the elation between V and E that the potential deceases, so we have a useful ule: The potential deceases as one moves in the diection of the E-field. This means that as one moves away fom positive chages (o towad negative chages) the potential deceases, while as one moves towad positive chages (o away fom negative chages) the potential inceases. 1 2

7 Physics 142 Electostatics 2 Page 7 Shown is a pai of oppositely chaged paallel conducting plates. The E-field lines in the egion between the plates ae as shown, and lines of equal potential ae also shown, at 25 V intevals. Except nea the ends of the plates, the E-field is appoximately unifom, so the equipotential lines ae equally spaced. (In ou applications we will often neglect the non-unifomity nea the ends of the plates fo simplicity.) If the field is taken to be unifom, then the potential at hoizontal distance x fom the positive plate is given by V(x) 100 = 0 x Edx = E x. If the sepaation of the plates is d, we find (setting V (d) = 0 ) that E = 100 / d, so the fomula fo the potential can be witten as V (x) = 100(1 x / d). Calculating V fo given souces Fo a given distibution of chages, thee ae two appoaches to calculation of V: Find E (pehaps using Gauss's law) and use its elation to V, given above. Use supeposition: decompose the chage distibution into a set of point chages and add the contibutions of these to the potential. To use the second appoach we need to know the potential of a single point chage. Fo simplicity we assume the point chage is at the coodinate oigin. Then its E-field is E() = kq 2 ˆ. Fom the elation between V and E, we find (using the fact that ˆ d = d ): This gives us a simple esult: V(2) V(1) = kq d = kq 2 kq 1. etal plates. Potential of point chage V() = kq To get this fomula with no additive constant we have chosen V = 0 at infinite distance. This is the usual choice when dealing with chage distibutions confined to a finite egion of space. To find the potential of a system of point chages we add thei contibutions: V(P) = k q i. i i Hee i is the distance (a positive numbe) fom the ith chage q i to the field point P. If the chage distibution is continuous, we teat infinitesimal bits dq like point chages, and the sum becomes an integal: V(P) = k dq. Both of these fomulas ae useful in solving poblems.

8 Physics 142 Electostatics 2 Page 8 Example: the potential of an electic dipole As an example, we will find the potential field fo an electic dipole. y In the case shown the dipole has length d and lies along the x axis, centeed at the oigin. The field point P is chosen to lie in the x-y plane. (Because of the axial symmety, the potential will be the same at all q +q points on a cicle aound the x-axis and passing though P.) p We denote the distances fom the two chages to point P by + = (x d/2) 2 + y 2 = (x + d/2) 2 + y 2 P x In tems of these the total potential at P is V(x, y) = kq This is the exact answe. If >> d, we obtain fom the binomial appoximation: V kqd x 3 = k p 3. The last fomula gives the answe in a fom independent of the coodinate system. Note that V falls off with distance as the invese squae. This is a geneal popety of the dipole potential. Shown ae the field lines (blue) and the equipotential lines (ed) fo a dipole, close to the dipole. Note that the field lines ae always pependicula to the equipotential lines. Whee the equipotential lines ae close togethe, as between the chages, the E-field is stong. The vetical equipotential line unning halfway between the chages will epesent V = 0 if the potential is taken to vanish an infinite distance, as in the fomulas we have been using. To the left of that line the potential will be positive, to the ight negative. (Diections ae not indicated on the field lines. They emege fom the positive chage and teminate on the negative chage, of couse.) The components of E can be obtained mathematically fom the fomula fo V by the pocedue given next. Calculating E fom known V Since V is obtained by integating E, it is to be expected that E can be obtained by diffeentiating V. This is indeed the case. But E is a vecto, with thee components, so thee ae thee deivatives to calculate. In addition, V will geneally depend on the thee vaiables (x,y,z) specifying the location of field point. The fomula elating these fields is E = i V x + j V y + k V. z The deivatives hee ae patial deivatives, in which all othe vaiables ae teated as constants.

9 Physics 142 Electostatics 2 Page 9 The quantity in [ ] is called the gadient of V; one says that the electostatic field E is the negative of the gadient of the potential field V. In the simple case whee V depends only on one vaiable, say x, the gadient becomes an odinay deivative: E x = dv /dx. Ou cases whee we calculate E fom V will be of this simple type. Example: V and E fo a unifom spheical shell Spheically symmetic chage distibutions ae impotant in pactice, and elatively simple to analyze. We will teat as an example the case of a thin spheical shell of adius R, with chage Q spead unifomly ove its suface. We ask fo V and E at points outside and inside the shell. Any spheically symmetic distibution can be consideed to be the supeposition of such shells, like layes of an onion. Using ou geneal esult fom Gauss's law fo spheical symmety, we find that E is adially outwad (fo positive Q) with adial component Note that inside the shell E is zeo. kq E () = 2 fo > R 0 fo < R This is like a simila popety fo the gavitational field g of a spheical shell of mass. It is not tue if the shell is not spheical, o the chage is not unifomly distibuted ove its suface. We find V by integating (assuming V = 0 at infinite distances): kq fo > R V() = E d = kq R fo < R Inside the shell V is constant, but not zeo. (It is zeo only at infinite distance.) These fomulas ae useful fo many applications. Potential enegy of a system of fixed chages If the inteaction foces among paticles in a system ae consevative, we ascibe to the system a potential enegy. This is a popety of the system as a whole, not of any individual paticle. When we take a chage q, oiginally at infinite distance, and place it at a point whee the potential (due to othe chages) is V we incease the potential enegy of the whole system by qv. This additional potential enegy is the negative of the wok done by the Coulomb inteaction foces as the new chage is moved to its location. Conside the simplest such system, of two point chages: q 1 is at point 1 and q 2 is at point 2. We stat with both chages infinitely fa away, choosing the potential enegy in that configuation to be zeo, and assemble the system in steps: Q R P

10 Physics 142 Electostatics 2 Page Bing in q 1 and place it at 1. The inteaction foce is zeo because the chages ae still infinitely fa apat, so no wok is done. The potential enegy is still zeo. 2. Now bing in q 2 and place it at 2. The wok done by the inteaction foce is equal to q 2 V 2, whee V 2 is the potential at 2 due to q 1. Thus the potential enegy is: U = q 2 V 2 = k q 1 q Of couse we could put q 2 in place fist and then bing in q 1. The esult in tems of potential enegy would be the same. (Intechanging the labels 1 and 2 leaves U unchanged.) But we would have witten it as U = q 1 V 1, whee V 1 is the potential at 1 due to q 2. To emphasize this symmety, we can also wite U = 1 2 (q 1 V 1 + q 2 V 2 ) This indicates how to genealize to a system of moe than two chages: Electostatic potential enegy of a system of point chages U = 1 2 i q i V i Hee V i is the potential at the location of the ith chage due to all the othe chages. Fo a system of a few point chages, an altenative (and often simple) way to calculate U is to calculate it fo each pai and add the esults. Cae must be taken to count a given pai exactly once. The sign of the potential enegy indicates whethe the system is bound o unbound. If the total potential enegy is positive, it equies extenal (non-electostatic) foces to hold the chages in place at est. If those extenal foces wee emoved, the Coulomb foces would push the chages away fom each othe. The system is unbound. On the othe hand, if the total potential enegy is negative, the chages attact each othe and it equies extenal foces to keep them apat. The system is bound. We ae talking hee about chages at est, with no kinetic enegy. In dynamic systems with moving pats the question of bound o unbound is detemined by the sign of the total enegy. Fo a system of point chages inteacting only though thei mutual Coulomb foces, thee is no stable static equilibium, egadless of the sign of the total enegy. Fo a paticula chage thee may be a point at which the total foce fom the othes is zeo, but a slight displacement of that chage in some diection will esult in a net foce away fom the point, not back towad it, so the system will fall apat. This fact, poved by examining the potential enegy function, is called Eanshaw s theoem.