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Juniper McDonald
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2 Vetri Velan GSI, Physics 7B Miterm 2: Problem Solution. Outsie sphere, E looks like a point charge. E = The total charge on the sphere is Q sphere = ρ 4 3 πr3 Thus, outsie the sphere, E = ρr3 3ǫ 0 ˆr r 2 Q sphere 4πǫ 0 r 2 Ro of charge is oriente in the raial irection, with constant linear charge ensity λ = Q. We split the ro into ifferential elements q. By Coulomb s Law, the force on each element is F = qe = q ρr3 ˆr 3ǫ 0 r. Since the ro is in the raial irection, 2 we can rewrite this: F = ρr3 λr 3ǫ 0 r ˆr 2 Integrate from R to R to fin the total force on the ro. F = F = ρr3 λ 3ǫ 0 ˆr R r R r 2 F = ρr3 λ 3ǫ 0 ( R R )ˆr = ρr3 Q 3ǫ ( 0 R R )ˆr Solution: F = ρr 3 Q 3ǫ 0R(R)ˆr 4 for getting the right form of the Efiel outsie the sphere: 2 for unerstaning the Efiel is like a point charge 2 for the correct Efiel (i.e. correct istance an total charge) 8 for setting up the integral properly: 2 for splitting ro up into elements for correct charge of each element for correct position of each element 2 for correct form of the integral for correct integral lower limit for correct integral upper limit 3 for correct result ˆr

3 Vetri Velan GSI, Physics 7B Miterm 2: Problem 2 Solution 2. The charges in the cavities are shiele, so the net effect is that it appears to an outsie observer that Q an Q 2 are at the center of the sphere. Solution: E = Q Q 2 4πǫ 0 r 2 ˆr 3 for unerstaning that charges are shiele, so E = 0 everywhere 4 for unerstaning that charges are transferre to the surface of the conuctor 4 for unerstaning that the total charge on the surface of the conuctor acts like a point at its center 2 for using Coulomb s Law (or Gauss s Law, if use correctly) 2 for correct answer

4 Physics 7B Fall 205 Miterm 2 Birgeneau Solutions Problem 3 a) By symmetry, the potential on the yaxis is zero. b) First, assume the point is locate on the positive xaxis. V = k e q x 0 x = k e λx x 0 x = k ea ( xx x 0 x = ak e 2x 0 log ( )) x0 x 0 where k e = 4πǫ 0 is Coulomb s constant. Now, assume the point is locate on the negative xaxis: q V = k e x 0 x = k λx ( ( )) e x 0 x = k xx ea x 0 x = ak x0 e 2 x 0 log, (2) x 0 note that this is just the negative of the answer we ha before, which coul also have been argue base on symmetry. ()

5 MT2 Problem 4 Rubric (Date: November 9, 205) PART A 2 points for a formula for the energy store in a capacitor:u = 2 CV 2 2 points for a formula for the capacitance: C = ǫ0a 0.25 points for the initial capacitance: C i = ǫ0a x 0.25 points for the final capacitance: C f = ǫ0a 2x 0.25 points for the initial energy store in the capacitor: U i = ǫ0av 2 2x 0.25 points for the final energy store in the capacitor: U f = ǫ0av 2 4x PART B 0.25 points for the capacitance as a function the istance between plates: C(y) = ǫ0a y 0.25 points for the charge as a function of the istance: Q(y) = C(y)V = ǫ0av y 0.5 points for the surface charge ensity on the secon plate: σ(y) = Q(y) A = ǫ0v y point for the electric fiel create by plate on plate 2: E(y) = σ(y) 2ǫ 0 = V 2y point for the force acting on the secon plate: F(y) = Q(y)E(y) = ǫ0av 2 2y 2 point for the work: W = 2x x F(y)y = ǫ 0AV 2 2 y 2 y 2x x point for the final answer: W pull apart = ǫ0av 2 PART C point for the change in the energy store in the capacitor: U capacitor = U f U i = ǫ0av 2 4x 2 points for the conservation of energy: U capacitor = W pull apart W battery point for the expression: W battery = ǫ0av 2 ǫ 0AV 2 4x point for the final answer: W battery = ǫ0av 2 4x 4x 2x

6 PROBLEM 5 ^ x L Q Q L x Figure. The physical setup of the problem (a) For the first part, we can assume that we know the formula for the capacity of a parallel plate capacitor with a vacuum between the plates, it s C = ǫ 0A. () If we have a capacitor where the ielectric (with ielectric constant K) is fully inserte, the strength of the electric fiel between the plates is reuce, which increases the capacitance to C = ǫ 0KA. (2) In this case, we have neither of these situations. But we can clearly split up the capacitor into two parts: The right part, where the ielectric is fully inserte, which has surface area x L, an the right part with no ielectric, which has surface area (L x) L. We shoul then notice that these two capacitors are connecte to each other in parallel: The positively charge plate of the left capacitor is irectly connecte to the positively charge plate of the right capacitor, an the same thing hols for the negative plates. So we can a these capacitances: C total = C left C right. (3) Stating (an explaining!) this gives you 2 points. Then writing own the iniviual capacities gives you point each: C left = ǫ 0L(L x) Then you get point for writing own the sum, C right = ǫ 0KLx. (4) C(x) = ǫ 0L (L(K )x). (5)

7 2 PROBLEM 5 (b) Here we have to use the formula for the electrostatic energy store in a capacitor U = 2 C(x)V 2 0. (6) ThisisequivalentU = Q2 2C,ifwethenplugintherightvalueforQ,Q = CV 0. This formula gives you 3 points. We then have to plug in C to get U = V 0 2 ǫ 0 L (L(K )x), (7) 2 which gives you 2 points if you wrote it own in the right variables. (c) This part is more involve than the others, an unfortunately, it s very easy to get the right magnitue by completely wrong arguments. So if you in t get points for this part, try to unerstan the actual solution. Let s first unerstan what s physically going on: Figure 2. A microscopic picture of the ielectric an the plates There are two things we shoul observe in figure 2: The first one is that the ipoles which are just outsie of the plates feel the fringe electric fiels at the bounary. From the picture, it shoul be clear that the ipole I painte in green feels a force pulling it to the upper plate, an the ipole in blue feels a force to the lower plate. In total, the vertical components cancel out, an there is a net force which pulls the ielectric slab to the left. For fining this irection, either by using this argument or via the actual calculation we o below, you get 5 points. The secon thing in the picture is that the charge ensity on the plates is not constant when we insert the ielectric. If we in t keep the voltage ifference between the plates constant, this woul simply mean that some charges are pulle from the left parts of the plates to the right parts uring the insertion, which reuces the potential ifference. (Alternatively, note that Q = CV, if C increases an Q is constant, V has to ecrease). In this case, however, the voltage is constant, so there can be no charges leaving the left parts of the plates. So there have to be some charges ae to the metal plates uring the insertion. Alternatively, observe that if Q = CV an C increases while V is constant, Q has to increase. So in summary there has to be some voltage source connecte to the plates that pushes an aitional charge Q(x) = (C(x) C(0))V 0 onto them for example, a battery. To fin the magnitue of the force on the ielectric, we note that x escribes the position of the slab in a coorinate system where the xaxis goes to the left (cf. figure ). So we can fin the force on the slab by the formula F = U total, (8) where U total is the total potential energy of the system. (Note that actually, x woul have to be the centerofmass coorinate of the slab, but because

8 PROBLEM 5 3 this is just a constant shift, this oesn t matter). So let s first fin the total energy of the system (which consists of the battery an the capacitor): U battery = U 0 work = U 0 V 0 Q(x) = U 0 V 2 0 (C(x) C(0)), (9) where U 0 is the energy of the battery before the insertion. So U total (x) = 2 C(x)V 2 0 const. C(x)V 2 0 = const. 2 C(x)V 2 0, (0) where the terms not epening on x are summe in the const. term. Plugging this into 8 gives F = V 0 2 ǫ 0 L (K )ˆx, () 2 so since K > we can also see here that the force goes to the left (note the irection of ˆx in figure ). Getting the right magnitue (in the right way, so by incluing the battery) gives you 5 points. Note that if we ha forgotten the battery, we woul have, by sheer coincience, gotten the same magnitue (an the wrong irection). So if you i the same erivation, but just forgot the battery an i everything else right, you got 2 points. As a last point, note that there s something obviously wrong about this answer: WhenxapproachesL, sotheslabisalmostfullyinserte, wewoul still have a nonzero force! In the physical picture, we can see that there are almost no ipoles left to be pulle in, so the force shoul go to zero. But when eriving the capacitance C = ǫ0a, you assume that the plates are so large that you can neglect any bounary effects, which you woul have to take into account here. Fortunately, this only changes the result when the ielectric is not inserte at all or when it s pulle in almost all the way.

9 Physics 7B Fall 205 Miterm II Solutions Part (a) In the first case the equivalent capacitance is C e = C C 2 = C, while in the secon case the C C 2 2 equivalent capacitance is C e = C C 2 = 2C. Part 2 Thus, in the first case, the charge store on each capacitor is Q = C e V = CV 2. In the secon case, the charge on each capacitor is half the charge on the equivalent capacitor, which is Q = 2 C ev = CV. So, the parallel capacitors hol more charge. 5 marks (a) In branch, the equivalent resistance is ( R e = R 2R ) R 2R = 2RR In branch 2, the equivalent resistance is R e = = 3R 2.5 Marks ( R R ) ( R R ) R R = R 3 R 2 R = R Marks (b) The potential ifference across the two branches will be equal. Ohm s Law tells us that V = I R = I 2 R 2, where R an R 2 refer to the equivalent resistances of branches an 2, respectively. So, I R I 2 R 2 = I I 2 = R 2 R I I 2 = 8. 5 Marks (c) Assume we have the capacitors in parallel The equivalent capacitance is 2C an the total charge is Q 0 = 2Q ( Mark). The equivalent resistance is R e = ( 3R 6 ) R = 33 R. Mark 29

10 Physics 7B Fall 205 Miterm II Solutions The potential ifference is V = Q 0 2C, an Ohm s Law tells us that V = IR = Q t R. Thus, Q 0 2C = Q t R Q Q = 2RC t ( Q(t) = Q 0 exp t ) 2RC The total charge on the equivalent capacitor is equal to twice that of one of the capacitors, so we have ( Q(t) = 2CV exp 29t ). 3 Marks 66RC 2

Where A is the plate area and d is the plate separation.

Where A is the plate area and d is the plate separation. DIELECTRICS Dielectrics an the parallel plate capacitor When a ielectric is place between the plates of a capacitor is larger for the same value of voltage. From the relation C = /V it can be seen that