Physics 2212 GH Quiz #2 Solutions Sping 216 I. 17 points) Thee point chages, each caying a chage Q = +6. nc, ae placed on an equilateal tiangle of side length = 3. mm. An additional point chage, caying a chage 3Q, is placed on one side of the tiangle as dawn below. Calculate the electic potential enegy of this system, elative to zeo when the fou chages ae infinitely fa away fom each othe............... The potential enegy of a system of two chages, elative to zeo fo infinite sepaation, is U = K q 1q 2 The potential enegy of the system shown, is the sum of the potential enegies of all possible pais of chages within it. Thee pais can be fomed fom the thee chages on the vetices: U vetices = 3 K QQ ) = 3K Q2 The chage centeed on the bottom side makes thee pais, one with each of the chages on a vetex. Note that this chage is a distance /2 fom two of the vetices, and a distance 3/2 fom the thid. The potential enegy of the system is U side = 2 K 3Q) Q ) + K 3Q) Q /2 3/2 = 12K Q2 Q2 6K 3 U total = U vetices + U side = 3K Q2 = 12K Q2 8.988 1 9 N m 2 /C 2) 6. 1 9 C ) 2 3. 1 3 m Q2 6K 3 = KQ2 9 6 ) 3 9 6 3 ) = 1.3 1 3 J = 1.3 mj Quiz #2 Solutions Page 1 of 7
II. 16 points) A thin non-unifom chaged od of length is bent into a quate-cicle. When placed as shown, it has a linea chage density λ that vaies with angle θ accoding to λ = λ sin θ whee λ is a positive constant and θ is measued as usual, fom the +x axis towad the +y axis. What is the magnitude of the esulting electic potential at the cente of the ac the oigin), with espect to zeo at infinite distance?............... An element of chage dq, with an element of ac length ds, geneates an element of potential dv. Since the od is thin, this element of chage is point-like, so V = dv = K dq = K λ ds = π/2 K λ sin θ dθ = Kλ π/2 sin θ dθ [ ] π/2 [ ] [ ] = Kλ cos θ = Kλ cos π/2) cos ) = Kλ 1 = Kλ 1. 5 points) In the poblem above, in what quadant, if any, is the diection of the potential at the cente of the ac the oigin)?.................. Potential is a scala! The potential has no diection. Quiz #2 Solutions Page 2 of 7
III. 17 points) Calculate the magnitude of the electic field inside a solid, non-unifomly chaged sphee of adius R. The chage density inside the sphee is ρ) = ρ 2 R 2 fo < R, whee ρ is a constant and is the distance fom the cente of the sphee........................ Use Gauss aw, ϵ Φ = q in. To find the flux, choose a Gaussian Suface with the symmety of the chage distibution, and that passes though the point at which the field is to be found. This is a sphee of adius. Φ = E da = E cos θ da On that Gaussian Suface, the electic field has a unifom magnitude E at evey point. The electic field vecto is pependicula to the suface, and so paallel to the vecto epesenting the element of aea, at evey point. Φ = E cos da = E da = EA = E 4π 2 The chage inside this Gaussian Suface can be detemined fom the volume chage density. ρ = dq dv q in = ρ dv Since the volume chage density vaies with adius, choose volume elements that ae small in the adial diection. Thee ae thin spheical shells of adius and thickness d. Add up integate) the chage in all the shells fom the cente to the adius of Gaussian Suface,. q in = 2 ρ R 2 4π2 d = 4πρ R 2 4 d = 4πρ 5 R 2 5 = 4πρ 5R 2 5 Relate the flux to the chage with Gauss aw, and solve fo the field magnitude. ϵ E 4π 2 = 4πρ 5R 2 5 E = ρ 3 5ϵ R 2 2. 5 points) Which sketch shows the electic field outside the sphee of the pevious question? The electic field inside the sphee is not shown accuately. It is endeed as a staight dashed line)........................ Outside a spheically symmetic chage distibution, the field must be the same as the field that would be poduced if the entie chage wee concentated at a point in the cente. The field due to a point chage follows an invese squae law with distance. So, the field at 2R must have one-fouth the magnitude of the field at the suface. Quiz #2 Solutions Page 3 of 7
3. 5 points) An insulating spheical bead of adius R is chaged with an unifom positive volume chage density ρ. The bead is placed in the cente of a hollow conducting sphee of inne adius 2R and oute adius 3R. The hollow conducting sphee is initially neutal. What is suface chage density η on the exteio suface of the hollow sphee?................ Stat by consideing a Gaussian Suface entiely within the thickness of the hollow conducting sphee. The field in the conducting mateial is zeo at equilibium, so the flux though the Gaussian Suface is zeo, so the chage within the Gaussian Suface is zeo. But we know thee is a chage Q bead = ρv = ρ 4 3 πr3 on the inne bead. Theefoe, thee must be the opposite chage Q inne = ρv = ρ 4 3 πr3 on the inne suface of the conducting sphee. Since this conducting sphee is neutal, thee must be the opposite chage Q oute = ρv = ρ 4 3 πr3 on the oute suface of the conducting sphee. The aea chage density on this oute suface is η = Q oute A = ρ 4/3) πr 3 4π 3R) 2 = ρ R 27 4. 5 points) A poton is sent with a velocity v = 3. 1 6 m/s 1% of the speed of light) towad a mecuy nucleus containing Z = 8 potons, initially located infinitely fa away. The adius of a mecuy nucleus is so small 1 fm = 1 14 m), that the nucleus can descibed as a point chage. How close to the nucleus will the poton get? Reminde: 1 pico-mete pm) = 1 12 m........................ Use the Wok-Enegy Theoem. Choose a system consisting of the poton and the mecuy nucleus. No extenal foces do wok on this system, and thee ae no non-consevative foces within it to change its themal enegy. W ext = K + U + E th = K + U + The mass of a mecuy nucleus is so much geate than that of a poton, that the mecuy nucleus can be consideed fixed in place, so the only kinetic enegy change in the system is due to the poton. The potential enegy change in the system is due to the intenal electic foce. = 1 2 m pvf 2 1 2 m pvi 2 ) + K Q pq Hg K Q ) pq Hg = 1 f 2 m pv 2 ) + K e Ze) ) i f whee v f, the speed of the poton at the instant of closest appoach, is zeo. As the poton stats infinitely fa away, the initial potential enegy of the system is zeo, too. Solve fo f. f = 2KZe2 m p v 2 = 2 8.988 1 9 N m 2 /C 2) 8) 1.62 1 19 C ) 2 1.673 1 27 kg) 3. 1 6 m/s) 2 = 2.45 pm Quiz #2 Solutions Page 4 of 7
5. 5 points) Twelve identical point chages, each caying a chage Q, ae oganized on the kagome patten sketched below. Each tiangle in the patten is equilateal with a side length a. The electic potential enegy of the system is U. What is the electic potential enegy U 1 of the system when the distance a is multiplied by 3? Kagome is the Japanese wod fo taditional bamboo baskets having this geomety.)................ Since the potential enegy of each pai of chages in the system is invesely popotional to the distance between them, the potential enegy of the entie system must be invesely popotional to a. U 1 = U /3 6. 5 points) Conside two scenaios inside identical capacitos. In each scenaio, the capacito is chaged identically, with chage Q on the left-hand plate and +Q on the ight-hand plate. In scenaio 1, a negatively chaged paticle is moved fom the left ) to the ight R). In scenaio 2, a positively chaged paticle of equal magnitude) is moved fom to R. In each case, the potential diffeence is defined as V = V R V, and the potential enegy diffeence is defined as U = U R U. Which of the statements below is tue?......... In situation 2, the paticle would not move spontaneously fom left to ight. The potential enegy of this system inceases when the paticle does this U 2 > ). Since the potential change is the potential enegy change pe unit chage, and the paticle is situation 2 is positive, this paticle is moving though a positive potential diffeence V 2 > ). Since the capacitos in the two situations ae chaged identically, and potential diffeences do not depend on the natue of the pobe chage used to detemine them, the potential change in situation 1 must be the same as that in situation 2 V 1 > ). Howeve, the paticle in situation 1 would move spontaneously fom left to ight. The potential enegy of this system deceases when the paticle does this U 1 < ). V 1 > ; U 1 < ; V 2 > ; U 2 > Quiz #2 Solutions Page 5 of 7
7. 5 points) The plates in an ideal paallel plate capacito ae 5. mm apat, as shown. The potential diffeence between these plates is 15 V. Point 1 is 1. mm fom the negative plate, and point 2 is 3. mm fom the negative plate. If it can be detemined, what is the potential diffeence fom point 1 to point 2?.............. As it would equie extenal wok to foce a positive test chage fom point 1 to point 2, the potential diffeence fom point 1 to point 2 must be positive. As the potential diffeence within a paallel-plate capacito depends linealy on distance, this potential diffeence must be s 1 2 2. mm V 1 2 = V total = 15 V) s total 5. mm = +6. V 8. 5 points) Two positive chages +q and a negative chage 2q ae placed at the vetices of an equilateal tiangle, as shown. Use the convention that V = at infinity. Which statement about the point p, at the cente of the tiangle, is tue?............ Each chaged paticle is the same distance fom the point p. The 2q chage contibutes a potential of 2Kq/ at the point p, and each of the q chages contibute +Kq/. The potential at the point p must be zeo. If a positive pobe chage wee placed at the point p, thee would be an attactive foce towad the 2q chage, and a epulsive foce fom the +q chages. The foce, and thus the field, would be towad the West. V = ; E points West. Quiz #2 Solutions Page 6 of 7
9. 5 points) A positively chaged paticle and a negatively chaged paticle, each having chage magnitude q, lie on the axis of a cylindical suface, equidistant fom the ends, as shown. Rank the flux though the top, side, and bottom of the cylinde, fom geatest to least. Remembe that electic flux can be positive o negative................... Field lines stat on positive chages and end on negative chages. Because the positive chage is neae the top suface than the negative chage is, the net flux though the top suface is outwad. Because the negative chage is neae the bottom suface than the positive chage is, the net flux though the bottom suface is outwad. The aea vectos fo the cylinde sufaces point outwad, so the angle between the field vectos and aea vectos is between and 9 on the top suface, and between 9 and 18 on the bottom suface. As flux is Φ E = E da = E cos θ da the flux though the top suface is positive, while that though the bottom suface is negative. Because the chaged paticles ae symmetically placed, any field line that passes outwad though the side of the cylinde as it heads away fom the positive chage, must also pass inwad though the side of the cylinde as it heads towad the negative chage. The flux though the side is zeo. Φ top > Φ side > Φ bottom 1. 5 points) A positively chaged paticle lies in the plane of a cylindical suface s top, a distance d fom the axis, as shown. The cylinde has height h. What is the sign, if any, of the flux though the cylinde s cuved side?............... Since the field fom the chaged paticle is paallel to the top suface of they cylinde pependicula to the aea vecto), the flux though the top is zeo. At the bottom suface, field only exits the cylinde. The flux though the bottom is positive. As thee is no chage within the cylinde, the net flux though it must be zeo. Theefoe, Φ side is negative. Quiz #2 Solutions Page 7 of 7