Problem Set 3: solutions

Transcription

1 Univesity of Alabama Depatment of Physics and Astonomy PH 126 LeClai Fall 211 Poblem Set 3: solutions 1. A spheical volume of adius a is filled with chage of unifom density ρ. We want to know the potential enegy U of this sphee of chage, that is, the wok done in assembling it. Calculate it by building up the sphee up laye by laye, making use of the fact that the field outside a spheical distibution of chage is the same as if all the chage wee at the cente. Expess the esult in tems of the total chage Q of the sphee. Solution: Thee ae in my opinion thee easonably staightfowad ways to go about this poblem, each with its own meits. Below I will outline all thee, you only needed to have one valid solution fo full cedit. Method 1: use du=v dq We know that we can find the potential enegy fo any chage distibution by calculating the wok equied to bing in each bit of chage in the distibution in fom an infinite distance away. If we bing a bit of chage dq in fom infinitely fa away (whee the electic potential is zeo) to a point P( ), at which the electic potential is V, wok is equied. The potential V esults fom all of the othe bits of chage aleady pesent; we bing in one dq at a time, and as we build up the chage distibution, we calculate the new potential, and that new potential detemines the wok equied to bing in the next dq. In the pesent case, we will build ou sphee out of a collection of spheical shells of infinitesimal thickness d. We ll stat with a shell of adius, and wok ou way up to the last one of adius. A given shell of adius will have a thickness d, which gives it a suface aea of 4π 2 and a volume of (thickness)(suface aea) = 4π 2 d. Once we ve built the sphee up to a adius, Gauss law tells us that the potential at the suface is just that of a point chage of adius : V() = k eq() (1) Whee q() is the chage built up so fa, contained in a adius. Binging in the next spheical shell of adius + d and chage dq will then equie wok to be done, in the amount V() dq, since

2 we ae binging a chage dq fom a potential of at an infinite distance to a potential V() at a distance. Thus, the little bit of wok equied to bing in the next shell is dw = du = V() dq = k eq() dq (2) Now, fist we ll want to know what q() is. If we have built the sphee to a adius, then the chage contained so fa is just the chage density times the volume of a sphee of adius : q() = 4 3 π3 ρ (3) Next, we need to know what dq is, the chage contained in the next shell of chage we want to bing in. In this case the chage is just the volume of the shell times the chage density: dq = 4π 2 d ρ (4) Putting that all togethe: du = k e 4 3 π3 ρ 4π 2 d ρ = 16π2 k e ρ 2 4 d (5) 3 All we need to do now is integate up all the small bits of wok equied fom the stating adius = to the final adius =: 16π 2 k e ρ 2 U = du = 4 d = 16π2 k e ρ = U = 16π2 k e ρ [ 5 5 ] (6) (7) If we emembe that the total chage on the sphee is Q tot = 4 3 π3 ρ, we can ewite a bit moe simply: U = 3k eq 2 tot 5 (8) Method 2: use du= 1 2 ɛ oe 2 dv ol We have also leaned that the potential enegy of a chage distibution can be found by integating

3 its electic field (squaed) ove all space: i U = 1 2 eveywhee ɛ o E 2 dv ol = 1 eveywhee E 2 dv ol (9) If we have a sphee of chage of adius, we can beak all of space up into two egions: the space outside of the sphee, whee the field is E out () at a distance fom the cente of the sphee, and the space inside the sphee, whee the field is E in (). Theefoe, we can beak the integal above ove all space into two sepaate integals ove each of these egions, epesenting the enegy contained in the field outside the sphee, and that contained inside the sphee itself: U = U outside + U inside = 1 outside E 2 out dv ol + 1 inside E 2 in dv ol (1) Obviously, the egion outside the sphee is defined as >, and the egion inside the sphee is. Just like in the last example, we can pefom the volume integal ove spheical shells of adius and width d, letting the adius un fom inside the sphee, and outside. Ou volume elements ae thus dv ol =4π 2 d, just as in the pevious method. What about the fields? Outside the sphee, the electic field is just that of a point chage (thanks to Gauss law). Noting again that the total chage on the sphee is Q tot = 4 3 π3 ρ, E out () = k eq tot 2 ˆ = 4πk e 3 ρ 3 2 ˆ ( > ) (11) Inside the sphee at a adius, the electic field depends only on the amount of chage contained within a sphee of adius, q() = 4 3 π3 ρ as noted above, which we deived peviously fom Gauss law: E in () = k eq() 2 ˆ = 4πk eρ ˆ ( ) (12) 3 Now all we have to do is pefom the integals... i The odd notation fo volume dv ol is to avoid confusion with electic potential without having to intoduce moe andom Geek symbols. Some people will use dτ to epesent volume to this end.

4 U = U outside + U inside = 1 = 1 = 8π2 k e 6 ρ 2 9 = 8π2 k e 6 ρ 2 9 outside ( 4πke 3 ) 2 ρ 4π 2 d d + 8π2 k e ρ 2 9 [ 1 = 16π2 k e ρ 2 5 = ] + 8π2 k e ρ 2 9 kq 2 tot E 2 out dv ol + 1 inside E 2 in dv ol (13) ( ) 4πke ρ 2 4π 2 d (14) 3 4 d (15) [ 5 5 ] = 8π2 k e ρ ( ) In the last line, we used ρ=q/( 4 3 π3 ). Somehow, it is eassuing that these two diffeent methods give us the same esult... o at least it should be! (16) (17) Method 3: use du= 1 2 ρv dv ol As in the fist method, we know that we can find the potential enegy fo any chage distibution by calculating the wok equied to bing in each bit of chage in the distibution in fom an infinite distance away. Fo a collection of discete point chages, we found: U = 1 2 i j i k e q i q j ij = 1 2 q i V j = 1 q j V i (18) 2 i j i i j i ecall that the facto 1/2 is so that we don t count all pais of chages twice. If we have a continuous chage distibution - like a sphee - ou sum becomes an integal: U = 1 ρv dv ol (19) 2 eveywhee We still need the facto 1/2 to avoid double counting. Like the last example, we beak this integal ove all space up into two sepaate ones: one ove the volume inside the sphee, and one outside. Outside the sphee at a distance >, the chage density ρ is zeo, so that integal is zeo. All we have to do is integate the potential inside the sphee times the constant chage density ove the volume of the sphee! But wait... what is the potential inside the sphee? The electic field inside such a sphee at a adius we have aleady calculated, and it depended only on the amount of chage contained in a sphee of adius, q(). That is because electic field fom

5 the spheical shell of chage moe distant than canceled out. This is not tue of the potential, because the potential is a scala, not a vecto like the electic field, and thee is no diectionality that will cause the potential to cancel fom the chage moe distant than. Thus, the potential inside the sphee will actually have two components at a distance inside the sphee: a tem which looks like the potential of a point chage, due to the chage at a distance less than fom the cente, and one that looks like the potential due to the emaining shell of chage fom to. How do we calculate this mess? We can use the definition of the potential and the known electic field inside the sphee. E in () = k eq tot ˆ 3 B V B V A = E d l A (2) (21) In this case, we want to find the potential diffeence between a point at which we aleady know what it is - such as on the suface of the sphee - and a point inside the sphee. Then we can use the known potential on the suface to find the unknown potential at. We can most simply follow a adial path, d l =ˆ d. Since the electic field is consevative, the path itself doesn t matte, just the endpoints - so we always choose a vey convenient path. B V() V() = E d l = = A k e Q tot 3 k e Q tot 3 ˆ ˆ d (22) d = k eq tot 3 [ 2 2 ] (23) = k eq tot 2 3 ( 2 2) (24) We know what the potential at the suface V() is, since outside the sphee the situation is identical to that of a point chage. Thus, V() = V() + k eq tot 2 3 ( 2 2) = k eq tot + k eq tot 2 3 ( 2 2) = k eq tot 2 ) (3 2 2 (25) Now we ae eady! Once again, we integate ove ou spheical volume by taking spheical shells of adius and width d, giving a volume element dv ol = 4π 2 d. Noting that ρ is zeo outside the sphee,

6 U = 1 2 = 1 2 eveywhee k e Q tot ρ 2 = πk eq tot ρ ρv dv ol = U = 1 ρv() 4π 2 d (26) 2 ) ( π 2 d = πk eρq tot ] [ = πk eρq tot [ d (27) 2 ] = 4πk eq tot ρ 2 = 3 kq 2 tot 5 5 One poblem, thee methods, and the same answe evey time. Just how it aught to be. Which method should you use? It is a matte of taste, and the paticula poblem at hand. I tied to pesent the methods in ode of what I thought was inceasing difficulty, you opinion may diffe. 2. At the beginning of the 2 th centuy the idea that the est mass of the electon might have a puely electical oigin was vey attactive, especially when the equivalence of enegy and mass was evealed by special elativity. Imagine the electon as a ball of chage, of constant volume density ρ out to some maximum adius o. Using the esult of the pevious poblem, set the potential enegy of the system equal to mc 2 and see what you get fo o. One defect of this model is athe obvious: nothing is povided to hold the chage togethe! Solution: Using the esult of the pevious poblem, and using a= o, (28) U = 3 kq 2 tot 5 = mc 2 o = 3kQ2 5mc m (3) (29) Hee we used Q= 4 3 π3 oρ to make things simple. In tems of the chage density ρ, we have 15mc o = π 2 kρ 2 (31) 3. A spheical conducto A contains two spheical cavities. The total chage on the conducto itself is zeo. Howeve, thee is a point chage q b at the cente of one cavity and q c at the cente of the othe. A consideable distance away, outside the conducto, is a point chage q d. What foce acts on each of the fou objects, A, q b, q c, and q d? Which answes, if any, ae only appoximate, and depend on being elatively lage? Solution: The foce on the point chages q b and q c ae zeo. The field inside the spheical cavity is independent of anything outside. Since we equie the field to be zeo inside a conducto, the field

7 just outside the cavities but inside the conducto must be zeo. Fo this to be tue, chage equal and opposite to the point chages must be induced on the inside suface of the cavities, q b on cavity B, and q c on cavity C. In ode fo the spheical conducto itself to be oveall electically neutal, its suface must then have a chage q b +q c spead out unifomly on it to cancel out the induced chage on the cavity sufaces. If thee wee no chage q d, the field outside would look like that of a point chage of magnitude q b +q c, the total chage enclosed by the conducto: E= q b +q c / 2. The pesence of q d will slightly alte the distibution of chage on the conducto s suface (if q d is positive, it would push some of the peviously unifom positive suface chage to the fa side), but not the total amount of chage. Since the distibution would no longe be spheically symmetic, it would no longe be exactly like a point chage. If q d is fa enough fom the conducto, the effect is small, and the foce on q d would be appoximately F d k eq d (q b + q c ) 2 ˆ (32) ecall that by Newton s thid law that the foce on the cavity must be equal and opposite, F A = F d. Only these two foces ae appoximate and depend on being elatively lage; the net foce on q b and q c is zeo egadless. 4. We want to design a spheical vacuum capacito with a given adius a fo the oute sphee, which will be able to stoe the geatest amount of electical enegy subject to the constaint that the electic field stength on the suface of the inne sphee may not exceed E o. What adius b should be chosen fo the inne spheical conducto, and how much enegy can be stoed? Solution: Fist we need to know the capacitance of the two sphee system. Let be the distance measued fom thei common centes. Fo that, we need only find the potential between the two sphees, b < < a. Inside the oute sphee of adius a, we know that the field due to oute sphee is zeo by Gauss law. That means that the potential due to the oute sphee is constant fo a, since E = V. Thus, the potential diffeence between the two sphees depends only on the field due to the inne sphee of adius b. This we can find easily enough by integating E d l ove a adial path fom b to a, since the field of the inne sphee will be that of a point chage fo >b. Let the oute sphee have a chage Q, meaning the inne sphee has a chage Q. a a V = E d = b b k e Q 2 ( 1 d = k e Q b 1 ) = k e Q a b a ab (33)

8 The capacitance is chage divided by potential diffeence: C = Q V = 1 k e ab a b = 4πɛ ab o a b The stoed enegy can be found in tems of the capacitance and voltage: U = 1 2 C ( V)2 = 1 ab 2k e a b k2 eq 2 ( a b ab ) 2 = 1 2 k eq 2 ( ) a b ab (34) (35) Now we know that at the inne suface = b the field can be at most E o, and we know that the field is just that of a point chage of magnitude Q. This lets us ecast the enegy in tems of the maximum allowed field, since one is popotional to the othe: E o = k eq b 2 = Q = b2 E o (36) k e b 4 E 2 ( ) ) o a b U = k e k 2 = E2 o (b 3 b4 (37) e ab 2k e a Maximizing the enegy means U/ b= (and 2 U/ b 2 <), which gives us a elationship between b and a: U b = E2 o 2k e ) (3b 2 4 b3 = = b = 3 a 4 a (38) Of couse, this could be a minimum; we must apply the second deivative test (o at least gaph the function): 2 U b 2 b= 3 4 a = 6b 12 b2 a b= 3 4 a= ( ) a < (39) 4 Indeed, b= 3 4a is a maximum, giving the maximum stoed enegy if the field stength on the inne conducto is the constaint. The actual enegy stoed is then found by plugging this value back into ou expession fo U: U max = k e E 2 oa 3 (4) 5. We have two point chages connected by a igid od, foming a dipole. It is placed in an extenal electic field E( ).

9 (a) Suppose that the electic field is unifom: E= E o whee E o is a constant vecto. What will be the total foce on the dipole? (b) Now suppose the field is not unifom, but that it only changes by a small amount ove the distance d. Show that the z-component of the total foce on the dipole is appoximately F z = p z E z z (41) whee p z is the z-component of the dipole moment p. ii (c) Why is a chaged ubbe od able to attact bits of pape without touching them? Solution: efeence the figue below. We will imagine that thee is a igid od holding the two chages togethe. z P(x,y,z) q d -q θ O y (a) The total foce on the igid dipole is the sum of the foces on each chage: F net = q E o q E o = (42) (b) Fo a non-unifom field, the foce is in geneal not zeo, but depends on the diffeence between the field at the position of positive chage and the negative chage. If we woy only about a single component of the total foce, say the z component, the foce depends only on that component of the field, and we have ( ) ( d F z = q E z 2 ẑ q E z d ) [ ( ) ( d 2 ẑ = q E z 2 ẑ E z d )] 2 ẑ ( ii Fo an abitay dipole oientation, this genealizes to F= p ) E. (43)

10 In the limit of small d and slowly-vaying E(), we can appoximate the diffeence tem as ( ) ( d E z 2 ẑ E d ) 2 ẑ d E z z (44) Noting that qd is the dipole moment p, F z qd E z z (45) (c) When a chaged ubbe od is bought nea bits of pape, dipole moments ae induced in the bits of pape. The non-unifom electic field emanating fom the end of the ubbe od then inteacts with the induced dipole moments to attact the bits of pape. 6. A wie having unifom linea chage density λ is bent into the shape shown below. Find the electic potential at O. 2 O 2 Solution: The two staight bits of wie will give the same contibution to the potential at O. Take the staight segment on the ight side and beak it up into infinitesimal segments of length dx, each of which will have chage dq = λdx. The potential fom each dq is that of a point chage, we can find the potential at O by integating ove the line segment fom to 3 all such contibutions: V line = line k e dq = k e 3 λdx x = k ln 3 (46) Fo the semicicle, each infinitesimal bit of aclength ds = dθ has chage λds, and also gives a contibution to the potential k dq/. Integating ove the semicicle means unning the angle θ fom π 2 to π 2 : V semicicle = semicicle k e dq = k e pi/2 π/2 λ dθ = k e πλ (47) The total potential at O is due to two line segments and one semicicle: V O = 2V line + V semicicle = 2kλ ln 3 + kλπ = kλ (2 ln 3 + π) (48)

11 7. The two figues below show small sections of two diffeent possible sufaces of a NaCl suface. In the left aangement, the NaCl(1) suface, chages of +e and e ae aanged on a squae lattice as shown. In the ight aangement, the NaCl(11) suface, the same chages ae aanged in a ectangula lattice. What is the potential enegy of each aangement (symbolic answe)? Which is moe stable? a + - a a a 2 Solution: We need only add up the potential enegies of all possible pais of chages. In each case we have fou chages, so thee must be ( 4 2) =6 combinations. Let the uppe left chage be q1, and numbe the chages in a clockwise fashion. The combinations ae thus q 1 q 2, q 1 q 3, q 1 q 4 q 2 q 3 q 2 q 4 q 3 q 4 (49) (5) (51) Fo eithe aangement,t he enegy is then U = k eq 1 q k eq 1 q k eq 1 q k eq 2 q k eq 2 q k eq 3 q 4 34 (52) Fo the fist aangement, NaCl(1), we need only plug in the distances and chages: U 1 = k ee 2 a + k ee 2 a 2 + k ee 2 + k ee 2 + k ee 2 a a a 2 + k ee 2 a Fo the second aangement, NaCl(11), we have: = ke2 a ( 4 + ) ke2 a (53) U 11 = k ee 2 a 2 + k ee 2 a 3 + k ee 2 a + k ee 2 a + k ee 2 a 3 + k ee 2 ( 2 a 2 = ke ) ke2 a a (54) Since U 1 <U 11, the (1) suface is moe stable, in ageement with expeiments. 8. A chage Q is located h metes above a conducting plane. How much wok is equied to bing this chage out to an infinite distance above the plane? Hint: Conside the method of images.

12 Solution: Thee ae two ways to appoach this one. Fist, the pesence of the conducting plane a distance h fom the positive chage means that this poblem is equivalent to a dipole of spacing 2h (see figue below).iii Thus, we need to find the wok equied to move a chage q fom a distance 2h fom a second chage q out to infinity. Let the oigin be halfway between the eal chage q and its image chage q. The wok equied to move the positive chage away is: Z Z W= ~ F d~l = Z q~ E d = h h kq2 1 kq2 kq2 d = = 4 z h 4h (2)2 (55) P +q -q conducting plane x eal chage vitual image chage Figue 1: The field of a chage nea a conducting plane, found by the method of images. MASSACHUSETTS INSTITUTE OF TECHNOLOGY A much sneakie way is to ealize that the enegy in the electic field must just be half of that due to Depatment of Physics Physics II (8.22) - Pof. J. McGeevy - Sping 28 a eal dipole. We leaned that the enegy of a chage configuation Solutions can be found by integating the Poblem Set 4 2 electic field ove all space, U E dv. The electic field due to ou point chage above the con1. Pucell 3.5 Wok and image chages [15 points] ducting plane is identical to that of a dipole, but only fo the egion of space above the plane. Below the plane, half of all volume in space, the field is zeo. We can immediately conclude that the point chage and infinite plane have half as much enegy, since thee is no field below the conducting plane. TA - Nan Gu We can calculate the foce expeienced by the chage Q by consideing the electic field geneated by its image chage. Howeve, we must emembe that the image chage also moves when the oiginal chage is moved.!! Q2 Q2! d!s = W = Q E dh" = (2h" )2 4h h 2 The answe Q 4h must be tue because it takes no wok to move the image chage, it is simply an image of the oiginal chage. " 1 We can also look at this fom the point of view of field enegy. We leaned in chapte 1 that U = 8π dv E 2. The electic fields in the point chage and infinite plane system ae identical to the system of two point chages in the whole lane. We can immediately conclude that the system of the point chage and the infinite plane has half as much enegy because thee ae no fields in the lowe half plane U U Pucell 3.17 Designing a spheical capacito [15 points] Figue 2: The field enegy of ou single chage with a conducting plate is half that of a dipole. iii A note on calculating capacitances. It may become confusing to calculate capacitance because thee seems to be an ambiguous sign (i.e. do we take φ1 φ2 o φ2 φ1?) that might esult in a negative capacitance. One way to calculate capacitance is to choose a convention whee Q is positive and to calculate the potential fom it. Then, you choice of φ must be positive. At the level of 8.22, anothe way is just to take the absolute value of whateve capacitance answe you get, since nomal mateials neve exhibit negative capacitance. See poblem 4. 1

13 The enegy of a dipole we found aleady when we consideed point chages. If the dipole spacing is 2h, U dip = kq2 2h = U = 1 2 U dip = kq2 4h (56)