PH6 Exam I Solutions q Q Q q. Fou positively chage boies, two with chage Q an two with chage q, ae connecte by fou unstetchable stings of equal length. In the absence of extenal foces they assume the equilibium configuation shown in the iagam. Show that tan 3 =q /Q. Note: This can be one in two ways. You coul show that this elation must hol if the total foce on each boy, the vecto sum of sting tension an electical epulsion, is zeo. O you coul wite out the expession fo the enegy U of the assembly an minimize it. Enegy-base appoach. Let the length of the stings connecting ajacent Q an q chages be. Call the istance between the two Q chages hoizontally l, an the vetical istance between the two q chages h. Using tigonomety, then: cos = l/ = l sin = h/ = h The total potential enegy of this system can be foun by aing the potential enegies of all unique pais of chages, ecalling that fo a pai of point chages q an q sepaate by a istance the potential enegy is k e q q /. We also note that thee ae fou equivalent paiings of the q an Q chages, all sepaate by a istance. U = k eq l + k eq h + 4k eqq = k eq cos + k eq sin + 4k eqq Now we have the potential enegy of the entie system as a function of the angle. Notice that the last tem - the potential enegy ue to the chages fixe by the sting - oes not epen on, since the istance between ajacent q an Q chages is fixe by the length of the sting.
At equilibium, this potential enegy shoul be at a minimum with espect to any angula vaiation. If U() shoul be at a minimum, we must have U/ = : Solving the equation above, U = [ ke Q cos + k eq ] sin + 4k eqq = U = k eq sin cos k eq k e Q sin cos = k eq Q sin cos cos sin = cos sin = q cos sin q Q = sin3 cos 3 = tan3 Now, we have fogotten to be caeful about one thing: is this a maximum, a minimum, o an inflection point? Setting U/ = only ensues we have foun one of the thee; ecall fom Calculus I which one it is epens on the sign of U/. One can ague on physical gouns that it must be a minimum, but mathematically one must show that U/ > to be cetain. Fining the secon eivative of U() is athe messy; you shoul fin something like this once you gin though it: U = U = k ( ) e Q cos + q sin Fo the pesent poblem, the angle can only be between an 9 without beaking the stings. The equation above is positive ove that entie ange of angles (though singula at the enpoints an 9 ), which means that U/ > fo any physically possible choice of, an we have inee foun a minimum of potential enegy, athe than a maximum o inflection point. Thus, ou conition epesents a stable situation. Foce-base appoach. Fist, efe to the figue below, whee we have awn a simple fee-boy iagam about one of the q chages, an one of the Q chages. We will call the foce between ajacent Q an q chages F qq, the foce between two q chages F qq, the foce between two Q chages F QQ, an finally, the tension in the stings is T. All fou stings must have the same tension, base on the symmety of the system an Newton s thi law. Since we know the istances between the chages (see above), we aleay know the electostatic foces involve:
F qq T F qq q F qq T y x T T Q F qq F QQ F qq Figue : Poblem 3: fee-boy iagam F QQ = k eq 4 cos F qq = k eq 4 sin F qq = k eqq Next, focus on one of the q chages. We will pick the uppemost one just to be concete. As inicate in the fee boy iagam above, thee will be two epulsive F qq foces fom the two ajacent Q chages, an these foces will be iecte at an angle above the inicate x axis. The sting tensions will act opposite these two epulsive foces. At equilibium, all foces must sum to zeo. Summing the foces along the x an y axes, we have: on q chage: Fx = F qq cos F qq cos + T cos T cos = Fy = F qq sin T sin + F qq = The foces in the x iection give us nothing useful, but those in the y iection o. Plugging in ou expessions fo the foces: k e Qq sin T sin + k eq 4 sin = () This looks useful, but it is not enough. We must eliminate the tension T, an the only way to get enough equations to o so is to also pefom a foce balance aoun one of the Q chages. Pick the ightmost one: on Q chage: Fx = F QQ + F qq cos T cos Fy = F qq sin F qq sin + T sin T sin = This time, the y foce balance is useless, but the x foce balance gives us anothe inteesting equation. 3
Again, plugging in ou expessions fo the foces: k e Q 4 cos + k eqq cos T cos = () Now: compae equations () an (). We can solve both equations fo T, an eliminate the tensions entiely: = T = k eqq + k eq 4 sin 3 T = k eqq + k eq 4 cos 3 k e q 4 sin 3 = k eq 4 cos 3 q Q = sin3 cos 3 = tan3 fom () fom () Thus, as it must, the foce-base appoach yiels the same answe as the enegy-base appoach. 4
-q a +q E. An electic ipole in a unifom electic fiel E is isplace slightly fom its equilibium position, as shown above. The angle between the ipole axis an the electic fiel is (you may assume is small). The sepaation of the chages is a, an the moment of inetia of the ipole is I. Assuming the ipole is elease fom this position, show that its angula oientation exhibits simple hamonic motion with a fequency f = qae π I Define the positive x axis to be in the iection of the electic fiel, an the positive y axis pepenicula to it in the upwa iection. This means the z axis points out of the page fo a ight-hane cooinate system. On the +q chage, thee will be foce F + = qe along ˆx, an on the q chage, a foce F = qe (along ŷ). Both of these foces will ty to cause the ipole to otate an oient itself along the electic fiel; that is, both will esult in a clockwise toque about the cente of the ipole. The sum of these toques must equal the ipole s moment of inetia I times the esulting angula acceleation α. Let the position of the +q chage be efine by a vecto + whose oigin is at the ipole cente, an similaly will give the position of the q chage. We also efine a unit vecto ˆ pointing fom the q to the +q chage. Finally, emembe that a clockwise otation efines a negative toque - this is the vesion of the ight-han ule fo toques. i ) ) τ = ( + F + + ( F = qea ˆ ˆx + ( qe) ( ˆ ˆx) = qea ( ẑ sin ) + qea ( ẑ sin ) = qae sin ẑ = I α τ = qea sin = I α We coul have avoie the vecto baggage ight off the bat, if we just chose the esultant toque to be negative base on the ight-han ule. In oe to show simple hamonic motion, we nee to show in this case that α = ω. ecalling the efinition of α, we have: i We have to pick the sign because τ, esulting fom a coss-pouct, is a pseuovecto. 5
Iα = I = qea sin t If the angle is small, we can appoximate sin by the fist tem in its Taylo expansion (a fist-oe appoximation): Taylo expansion sin = small : sin n= ( ) n (n + )! n+ = 3 3! + 5 5!... Using this appoximation, I t qea o t qea I This is ou belove iffeential equation fo simple hamonic motion, viz., t small, = ω, an thus fo ω qae I o f qae π I z P( ) Q +Q x 3. The chage istibution shown above is not quite a ipole, but may be consiee to be the supeposition of a ipole an a monopole. (a) Fin an appoximate fom fo the potential at a point P( ) fa fom the chages ( x, z) in tems of the aial istance an angle. You may teat the poblem in two imensions if you wish. (b) Fin an appoximate fom fo the electic fiel at P. Note: you may fin the following appoximation useful: ( + x) n + nx. See the last exam sheet fo fomulas elating to spheical cooinates... 6
One thing to ecognize ight off the bat is that this chage istibution is equivalent to a ipole plus one exta negative chage at the oigin: = + Q +Q Q +Q Q Figue : Ou chage istibution is equivalent to a ipole plus a point chage. Thus, the solution to ou poblem is ou usual ipole potential plus the potential of a point chage. Fist, let s consie the ipole alone, an we can a the point chage in late. We can eaily wite own the potential fo the ipole at the point P it is just a supeposition of the potential ue to each of the chages alone. We ll wok in two imensions, so long as we have the option. V ipole (x, z) = k e q q + (3) (x ) + z x + z Since we ae assuming, we simplify the enominato in the fist tem a bit, emembeing that =x + y : (x ) + y + z = x x + + z = x + z x = x/ + / x/ x +z + ( + x x +z ) (4) Hee we use the given (binomial) appoximation once again in the vey last step. Substituting this in to ou expession fo the ipole potential above, V ipole (x, z) k eq ( + x ) k eq = k eq x = k eq cos (5) In the last step, we note that x/=cos, using the angle as given in the figue. This is the potential ue to the ipole alone; fo the full poblem, we nee only a in the potential ue to a chage q at the oigin: V tot (x, z) V ipole (x, z) + V q (x, z) = k eq cos k eq = k eq ( cos ) The fiel is no poblem at all, emembeing that E = V (an that we ae in spheical cooinates). Fist, the aial pat: (6) 7
Next, the angula pat: E = V ˆ k eq cos 3 ˆ k eq ˆ = k ( ) eq cos ˆ (7) In total, E k eq E = V ˆ k eq sin ˆ 3 (8) ( ) cos ˆ ˆ + sin ˆ = k ( ) eq 3 cos ˆ + sin ˆ k eq ˆ (9) Just like the potential, the fiel is the supeposition of a ipole an a single point chage q. 4. A sphee of aius caies a chage ensity ρ()=c, whee c is a constant. (a) Fin the total chage Q containe in the sphee. (b) Fin the electic fiel eveywhee. (c) Fin the enegy of the configuation. Note: thee ae two staightfowa ways fo the last pat: fom the enegy in the electic fiel eveywhee, an fom the potential ove the chage istibution. The total chage is foun by integating the chage ensity though the volume of the sphee. emembeing to use the iffeential volume element in spheical cooinates, Q tot = π π π ϕ ρ() sin ϕ = = 4πc 3 = π ϕ c sin ϕ [ ] 4πc 4 = πc 4 () 4 The chage istibution is spheically symmetic (ρ oes not epen on o ϕ), so fo > the fiel looks like that of a point chage of magnitue Q tot : E = kq tot ˆ = kπc4 ˆ = c4 4ɛ o ( > ) () Fo points insie the sphee, <, we nee only woy about the chage containe within a sphee of aius, which can be foun fom the integal above if we eplace the uppe limit with instea of. E = kq() ˆ = kπc4 ˆ = kπc ˆ = c ˆ ( ) () 4ɛ o 8
Once we have the electic fiel eveywhee, the easiest way to fin the enegy is to integate the squae of the electic fiel eveywhee. We ll have to beak this up into two integals: one fo aii less than an one fo aii geate than, since the fiel is iffeent in these two egions. U fiel = ɛ o = ɛ o = ɛ o E τ (3) π + ɛ o π π ϕ π π sin π ( ) ( ) c c ˆ ˆ sin ϕ (4) 4ɛ o 4ɛ o ϕ ϕ ( ) ( ) c 4 c 4 4ɛ o ˆ 4ɛ o ˆ sin ϕ (5) c 6 6ɛ + ɛ π o o sin π ϕ c 8 6ɛ (6) o [ ] = c π 7 + c π 8 [ ] (7) 8ɛ o 7 8ɛ o ( ) = c π 7 8ɛ o 7 + 7 = c π 7 (8) 7ɛ We coul also fin the enegy of the system by integating the potential times chage ensity though the volume of the sphee: U fiel = ρv τ (9) Since the integan is non-zeo only in the egion whee we have chage ensity i.e., fo < we only nee the potential ove that egion as well. We can get the potential V fom E eaily by integation. In oe to fin the potential at a istance fom the cente of the sphee (still with < ), we ll nee to integate E l fom infinity own to, as if we ae bining in the chage to buil up the sphee bit by bit. Since E is consevative, we can integate ove any path we like, so we may as well make it a nice aial path, ˆ. As with ou pevious calculation, we ll have to beak the integal up into two egions, one outsie the sphee, an one within the sphee, since the fiels ae iffeent in those two egions. V() = = c4 4ɛ o E l = [ ] c 4ɛ o E ˆ = [ ] 3 3 c 4 4ɛ o c 4ɛ o = c3 c3 + c3 = 4c3 c3 = 3ɛ o ɛ o ɛ o ɛ o ɛ o c 3ɛ o ( ) 3 3 4 () Once we have the potential as a function of, we can integate ρv though the volume of the sphee to fin the enegy: 9
U fiel = ρv τ = π = 4πc 6ɛ o [ 3 4 4 7 8 π ] ϕ = πc 3ɛ o ( ) ( ) c c 3 3 sin ϕ () 3ɛ o 4 [ 7 ] 4 7 = πc 7 () 8 7ɛ o As it must be, the potential an fiel methos yiel the same esult. In my opinion, the fiel metho is somewhat easie in this case, paticulaly since you wee aleay aske to fin the fiel in the pevious pat. Still, thee is always moe than one way to o a poblem.