PHY 8 Test Pactice Solutions Sping Q: [] A poton an an electon attact each othe electically so, when elease fom est, they will acceleate towa each othe. Which paticle will have a lage acceleation? (Neglect gavitational attaction.) a) The acceleations ae the same since the Coulomb foces ae equal. b) The electon since it is lighte. c) The poton since it is heavie. ) The paticles won t acceleate since the foces cancel each othe out. Solution: b The chages act upon each othe with foces which in the spiit of Newton s 3 law fom action-eaction pais. So, the poton an the electon will attact each othe with foces with same magnitue. Hence, in confomity with Newton s n law, thei acceleations will epen on thei elative masses: a m e p F F meae mpap ae ap a m p e That is, the acceleation of the electon is lage than the acceleation of the poton by the same facto the electon is lighte than the poton. Q: [] Two chages ae epesente in the figue with some electic fiel lines. Base on the patten, which chage is lage an by how much? a) The positive chage, thee times. b) The negative chage, thee times. c) The chages ae equal. ) The positive chage, two times. Solution: a The numbe of electic fiel lines is popotional to the intensity of the electic fiel so the positive chage is lage since it has moe lines suouning it. Also, the fiel of a chage n times lage than anothe chage shoul be epesente by n times moe fiel lines. So, since in ou case thee ae lines exiting the positive chage but only of them en on the negative chage, the positive chage must be 3 times lage. Q3: [] A poton entes with hoizontal velocity v in the vacuume space between the hoizontal plates of a chage paallel-plate capacito. Which of the fou paths epicte on the iagam will be most likely followe by the poton? a) A (staight line) b) B (ac of cicle) c) C (ellipse) ) D (paabola) A B C D v y x Solution: The electic fiel between the plates is unifom, so the foce acting on the poton is constant, so its acceleation is constant. Since the poton caies a positive chage, the foce is in the iection of the fiel, vetically upwa in the positive y-iection. Consequently, only the y-component of the paticle s motion is acceleate wheeas the x-component is not. Hence, the time epenence of the cooinates is given by Fe e y at t E t m an x vt. (Alas, PHY 7 bites us fom behin ) m Substituting the time between the two equations, we get a quaatic epenency between y an x, that is, the path is paabolic: e x ee y E y x m v mv. Q: [] An electon is place in the inteio of an imaginay sphee one thi the aius fom the ight hemispheic suface. Then the electic flux though the suface of the sphee is -e
PHY 8 Test Pactice Solutions Sping a) e b) e 3 c) e ) e 9 Solution: a The sphee can be chosen as a Gaussian. Then, by Gauss s law, the flux though its close suface is popotional to the chage within iespective of its position: e Q5: [] Which of the ajacent cuves most likely epesents the electic potential of a chage metallic sphee in electostatic equilibium as a function of the istance fom the cente of the sphee. R is the aius of the sphee. a) A b) B c) C ) D Solution: Thee is no net chage insie a metal, so the inne fiel is zeo. Theefoe, the potential must be constant thoughout the inteio of the conucto an it thee is no iscontinuity at the suface as on gaph D. A B R R C D R R Q6: [] Thee point chages with the same magnitue q =.6 μc ae locate in the -q cones of an equilateal tiangle of sie =.56 mm. What is the wok one on the positive chage to bing it to its cone fom infinity? q a) 89 J b) 56 J -q c) 8 J ) J Solution: a The wok one by the two negative chages to bing the positive chage fom infinity to the cone of the tiangle is given by the iffeence of potential between the two positions multiplie by the tanspote chage. The positive chage moves though the potential ceate by the two negative point chages. Theefoe we get q q q W q q q 8 J. Q7: [] The figue epesents the electic potential of a ipole. A positive test chage is elease in vicinity with an initial velocity oiente as in the figue. Which of the inicate tajectoies will be most likely followe by the test chage? a) A b) B c) C ) D Solution: The positive test chage spontaneously moves in the iection of the electic fiel pointing own the gaient of potential towa egions with lowe electic potentials, so it will ten to fall into the potential well coesponing to the negative chage of the ipole. Since the test chage has an initial velocity, it will fall own into the sink along a spial tajectoy much like the motion of a mass own a whilpool. C v D B A
PHY 8 Test Pactice Solutions Sping Q8: [] Suppose that you want to buil a paallel plate capacito with capacity of. F. You esign inclues cicula metallic plates sepaate by. cm of vacuum. What shoul be the aius of one of these plates? a) 5. m b) m c).5 m ) m Solution: b The capacitance of a capacito is given by its geomety an the mateial in between the plates. Thus, fo the paallel plate capacito with cicula plates of aius A C C km. So, you can see why we sai in class that Faa is a athe lage unit. Q9: [] Consie two paallel plate capacitos, with capacitances C lage than C, but the same plate sepaations =. If each of the capacitos is chage with the same amount of chage, which capacito has a lage electic fiel in between the plates? a) Capacito C. b) Capacito C. c) The electic fiel is the same. ) Depens on the plate aeas. Solution: b Capacitance C is lage than C, but they hol the same chage, so the potential iffeence acoss C must be lage: C C Q Q. Theefoe, since the potential iffeence is popotional to the constant fiel in between the plate by = E, we get E E E E. Q: [] Two ai-fille capacitos ae connecte in paallel an chage. A ielectic is insete in between the plates of one of them. What happens with the chage stoe on the othe one? a) Stays the same. b) Inceases. c) Deceases. ) Becomes zeo. Solution: c The effect of the ielectic is to incease the capacitance of the espective capacito an thus its ability to hol moe chage with a smalle potential. Since the total chage on the combination must stay the same with o without the ielectic (it cannot go anywhee), the voltage will ecease. But a smalle voltage acoss the capacito without the ielectic means that it looses chage (it goes onto the othe one). Using pime symbols fo the chages afte the ielectic is insete, we can show systematically how the chage changes on the secon capacito, as following: C KC Qnet Q Q Q Q Q Q Q Q C C since the voltage is the same, so Q /C = Q /C C KC Q C C Q Q Q Q C C Q KC C C C KC C 3
PHY 8 Test Pactice Solutions Sping P: Thee point chages = 6. μc an = 3. μc ae fixe in the cones of a vetical equilateal tiangle of sie = cm, as in the figue. A test chage q =.6 μc of mass m is place in point A whee it floats in static equilibium at height h = / cm above the chages Q. a) [] Use vecto aows to epesent the electic fiels,, an ceate by the thee chages in point A. The fiel exists in point A inepenent of the pesence of the test chage. The souce chages ae positive, such that the espective fiels ae aows pointing aially away fom each chage. b) [7] Split the vectos in x an y-components an fin a symbolical expession fo the magnitue of the net electic fiel in point A. It shoul epen on,, an constants. Then epesent the vecto net fiel on the figue an calculate the magnitue numeically. Note that the x-components of the fiels ceate by the bottom chages cancel each othe out, so we emain only with thei y-components which ae equal. Theefoe, the net fiel is given by Q Q E Esin E k k, whee an ae the istances between the espective chages an point A. Fom the geomety of the aangement, Theefoe. 3, h 3 Q Q k Q E k k Q 3 3 The negative sign means that the net fiel is oiente ownwa. Q 7. m. α = 5 Q / q A y α / B / Q x c) [3] Repesent the electic foce acting on chage q. Then use equilibium to calculate the mass m of the test chage. The test chage is negative such that the foce will be vetically upwa (against the fiel). Since the chage is in equilibium F mg qe mg m q E g.39 kg. e ) [] Suppose that the uppe chage is emove. Calculate the iffeence of electic potential ceate by the emaining chages between points A an B. Both chages contibute to the potentials equally. Theefoe Q Q Q 5 B A k k k.8. e) [] The test chage q is elease fom est an moves ownwa. Calculate its spee in point B. Since the test chage falls fom est its balance of final an initial potential enegies gives us the final kinetic enegy mg qa qb mv mv mg q B A.3 J, such that the spee is v.3.39 m s. m s. y a P: A cylinical wie of aius a is electically chage with unifom linea ensity λ = 3. - C/m. Consie the cyline infinitely long L P (,) x
PHY 8 Test Pactice Solutions Sping an aligne with y-axis, as in the figue. The questions below will lea you though the most common calculation of the electic fiel an potential in point P(,), at a istance =. cm fom the axis of the cyline. a) [] Sketch a cylinical Gaussian of aius an abitay height L aoun the wie epesente on the figue. Denote q the chage enclose by this suface an wite it own in tems of λ an L. q L b) [5] Use Gauss s Law to povie a symbolical expession fo the magnitue of the electic fiel E in point P in tems of, λ an ε. Then calculate it numeically. Inicate on the figue the iection of the vecto E in point P. Since the electic fiel must espect the aial symmety of the chage, the electic flux though the Gaussian must be nonzeo only though the cuve suface (so no flux though the top an bottom). Moeove, by symmety, the magnitue of the fiel must be constant on these cuve sufaces. Consequently, by Gauss s Law, we get E E A E L q E L L E by Gauss's Law Gaussian Since the chage on the wie is negative, the fiel vecto points aially insie as inicate by the expession fo the component E which is negative. Numeically, the fiel magnitue is E.8 m c) [5] Use the expession fo the electic fiel to fin a symbolical expession fo the iffeence of electic potential between point P an a point outsie the wie at istance fom P, locate on the x-axis. Fo any consevative fiel with aial symmety, we have the elationship between the fiel stength an the potential: E ˆ ; Hence, we get E E ln, whee the pime is a customay way to mak a ummy vaiable since is a limit of integation. The goun o suface of zeo potential can be taken on the suface of the wie, = a, an then this is an expession fo the potential at istance. ) [6] Now suppose that a poton is launche in point P moving to the ight along x-axis with initial spee v = 5 m/s. What istance oes the poton tavel befoe it stops? We can use consevation of enegy. Since thee is no nonconsevative foce acting on the poton, the consevation of enegy between the initial configuation (point P) an the point at position x whee it stops (since the poton is attacte by the negative chage on the wie) is given by: mv e e. C x Hence, we can calculate the istance = x tavele by the poton e mv ex ln mv mv mv e e ln e e.5 mm e P3: A capacitive cicuit is connecte to a battey with a potential iffeence =, as in the figue. The capacitances ae C =. nf, C =. nf, C 3 = 5.6 nf, C = nf. b a) [] Use the space below to euce the cicuit to simple foms an calculate the equivalent capacitance C of the cicuit. C 3 C 3 C C C 5 a c
PHY 8 Test Pactice Solutions Sping b c a c b) [] Calculate the net C 3 C chage Q stoe on the combination. The use it to C C calculate the potential iffeence ab between points a-b on the cicuit. a Q C nc Since the equivalent capacito C, which is connecte acoss points a-b, contibutes to the net capacitance C in a seies cicuit (see the secon cicuit iagam above), the chage Q on C is equal to the net chage Q. Theefoe, applying the efinition of capacitance to the equivalent capacito C, we get Q abc Q ab Q C. b c) [3] Now let s look close at the paallel-plate ai-fille capacitos C an C. Knowing that the aeas A of thei plates is the same, an the sepaation between the plates of C is =. mm, what is the sepaation between the plates of C? Using the epenency of capacitance on the pemittivity an geomety, we have C C C A an C A. Hence, iviing the expession to get i of the unknown aea, we obtain C C C C. mm a ) [3] Calculate the magnitues of the electic fiel E, in between the plates of the espective capacitos. Since the plate sepaations ae iffeent, the same potential iffeence ab will ceate iffeent electic fiels insie each of the two capacitos: E ab. N C an E ab. N C e) [] Calculate the chages Q, on the espective capacitos. Then, fin a way to calculate the aea A of the plates of the two capacitos (Hint: use fo instance Q an E. How can you wite E in tems of Q?) Applying the efinition of capacitance, we get Q abc 8. nc an Q abc 6 nc Notice that, inee, Q Q = Q = Q. Recall that the constant fiel between the plates is given by the chage ensity σ. Fo instance, using C, since Q A E Q A A Q E.5 m f) [] Compute the electostatic foces between the plates of each of the two capacitos. The electic foce appeas because each plate is in the constant fiel of the othe one. Because the net fiel is twice the fiel pouce by each plate, we obtain 5 F Q E 8. N 5 an F Q E 3 N 6