Exam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B

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1 PHY 249, Fll 216 Exm 1 Solutions nswer 1 is correct for ll problems. 1. Two uniformly chrged spheres, nd B, re plced t lrge distnce from ech other, with their centers on the x xis. The chrge on sphere is q, wheres tht on sphere B is 2q. If the electrosttic force on sphere is Fî, tht on sphere B is (1) Fî (2) Fî (3) 2Fî (4) 2Fî (5) 1 2 Fî Recll Newton s third lw. 2. In ech of the four configurtions lbeled, B, C, D in the figure, five chrges re eqully spced long the x xis. ll chrges hve the sme mgnitude, e, but some of them re +e nd some e. Rnk the mgnitude of the force on the middle chrge for the different configurtions, with the lrgest first nd the smllest lst. B C D (1) C, D,, B (2) C,, D, B (3) C, B, D, (4), C, D, B (5) D, C,, B In C, the forces by the four chrges dd up, wheres they completely cncel out in B. Therefore, C is the first, nd B the lst. Of the remining two configurtions, D comes before, becuse the two outermost chrges do not contribute to the net force in D, wheres in they compete with the forces by the two inner chrges. 3. prticle with positive chrge Q is on the y xis distnce from the origin, nd prticle with positive chrge q is on the x xis distnce d from the origin. Wht is the vlue of d for which the x component of the electrosttic force on the second prticle is the lrgest? (1) / 2 (2) (3) 2 (4) 2 (5) 2 2 For convenience, let s cll the second distnce x insted of d. The mgnitude of the force is F = kqq/( 2 +x 2 ). The similrityofthetwotringles,onedefinedbythetwoprticlesndtheoriginndthe otherdefinedbytheforcevectornd its x nd y components, leds to F x = Fx/( 2 +x 2 ) 1/2 for the x component of the force. Hence F x = kqqx/( 2 +x 2 ) 3/2. To mximize this, set df x /dx =, which gives 1/( 2 +x 2 ) 3/2 3 2 x(2x)/(2 +x 2 ) 5/2 =. Solve this for x. 4. Three prticles, ech chrged with 1. nc, re plced on three corners of squre of edge lengths = 1. m, s shown. Wht is the mgnitude of the electric field t the empty corner, lbeled? (1) 17 N/C (2) 27 N/C (3) 12 N/C (4) 24 N/C (5) 31 N/C Find the x nd y components of ech force, then dd the components of the three forces seprtely to obtin F x nd F y of the net force F. Finlly, find F by computing (F 2 x +F 2 y) 1/2. lterntively, notice tht the sum of the forces by the two nerest chrges, 2(kq 2 / 2 ), is in exctly the sme direction s the force by the third chrge on the digonl corner, kq 2 /(2 2 ). So just dd the two expressions nd plug in numbers.

2 5. line of chrge of length L nd chrge per unit length, λ, is locted on the x xis, s shown in the figure. Which expression below gives the x component of the electric field, E x, t distnce D from the left end of the line? D x x= x=l (1) L kxλ (x 2 +D 2 ) 3/2dx (2) L kdλ (x 2 +D 2 ) 3/2dx (3) L kxλ x 2 +D 2dx (4) L kdλ x 2 +D 2dx (5) L kλ x 2 +D 2dx The mgnitude of the electric field due to n infinitesiml line element dx locted t x is de = k(λdx)/(x 2 +D 2 ). Use the similrity of two norml tringles, s ws done in the solution to Problem 3, to find tht the x component of this force is given by de x = dex/(x 2 +D 2 ) 1/2. Integrte this from x = to x = L. 6. glss rod forms semi-circle of rdius r = 5 cm with chrge of +q distributed uniformly in the upper qudrnt nd q distributed uniformly in the lower qudrnt. Wht is the direction of the electric field t the center P of the semi-circle? (1) ĵ (2) ĵ (3) î (4) î (5) The electric field is zero. +q q r y P x The electric field due to the positively chrged prt of the rod points in the î ĵ direction, wheres the negtively chrged prt produces n electric field pointing in the î ĵ direction. Since the mgnitudes of the two fields re exctly the sme, their x components will cncel out, leving only the component in the ĵ direction. 7. Two thin concentric sphericl shells mde of copper hve net chrges of q 1 = 46 nc, q 2 = 69 nc nd rdii of r 1 = 15 cm, r 2 = 34 cm s shown in the figure. Wht is the surfce chrge density on the inner surfce of the smller shell, with rdius r 1, in units of C/m 2? q 2 q 1 r 2 r 1 (1) (2) (3) (4) (5) t equilibrium, no chrge will pper on the inner surfce of conductor unless the cvity contins chrged body. lterntively, in the present geometry, you my drw sphericl, concentric Gussin surfce whose rdius is slightly lrger thn the inner rdius of the inner shell. Since the electric field in conductor is zero t equilibrium, the field on the Gussin surfce will be zero. Guss theorem tells tht the enclosed chrge, nmely the chrge on the inner surfce, is zero. 8. Three very lrge prllel plnes re seprted by distnces d nd 2d. They crry chrges, whose densities re +σ, 2σ, nd +3σ, s shown in the figure. Wht is the mgnitude of the electric field t point in the middle between the second nd third pltes? (Hint: Electric fields re vectors tht dd s such.) +σ 2σ +3σ (1) 2σ ǫ (2) σ ǫ (3) σ 2ǫ (4) 3σ 2ǫ d 2d (5) 3σ ǫ The mgnitudes of the fields produced by the three pltes re σ/(2ǫ ), 2σ/(2ǫ ), nd 3σ/(2ǫ ). The first one points to the right, wheres the other two point to the left. So the mgnitude of the net field is ( 1+2+3)σ/(2ǫ ),

3 9. The figure shows four solid spheres,, B, C, D, ech with chrge Q uniformly distributed throughout its volume. For ll spheres, point P is t the sme distnce from the center of the sphere, s shown. Rnk the spheres ccording to the mgnitude of the electric field they produce t point P, gretest first. (1) nd B, C, D (2) The field is eqully strong in ll of the cses. (3) B,, C nd D (with C nd D equl) (4) B,, C, D (5) D, C, nd B (with nd B equl) P P P P B C D round the center of ech sphere, drw concentric sphericl Gussin surfce tht goes through point P. Since the res of these surfces re identicl, the electric field t P depends only on the mount of the enclosed chrge, not on the re of the Gussin surfce. The Gussin surfces for nd B completely enclose Q, so these spheres produce the strongest field. The Gussin surfce for C encloses lrger frction of Q thn does the Gussin surfce for D. So sphere C comes second, nd sphere D the lst. 1. smll bll, with chrge of 4 µc on it, is plced in sphericl cvity inside sphericl conductor, s illustrted in the figure. The net chrge on the conductor is 3 µc. Wht is the mount of chrge on the outer surfce of the conductor? + (1) 1 µc (2) 7 µc (3) (4) 3 µc (5) 4 µc The 4 µc chrge on the bll ttrcts chrge of 4 µc to the inner surfce of the conductor. s result, 3 ( 4) µc will pper on the outer surfce so tht the net chrge on the conductor will remin 3 µc. 11. solid sphericl conductor is given net positive chrge. The electrosttic potentil of the conductor is (1) constnt throughout the volume. (2) lrgest t the center. (3) lrgest on the surfce. (4) lrgest somewhere between the center nd surfce. (5) zero t the center. t equilibrium, the electric field inside conductor is lwys zero nd, consequently, the electrosttic potentil is constnt throughout. 12. Two conducting spheres re plced fr prt from ech other. The smller of the two crries totl chrge Q. The lrger one, whose rdius is three times tht of the smller one, crries no net chrge. fter the two spheres re connected by thin conducting wire, the mounts of chrge on the smller nd lrger spheres re, respectively: (1) Q/4 nd 3Q/4 (2) Q/1 nd 9Q/1 (3) Q/3 nd 2Q/3 (4) Q/2 nd Q/2 (5) Q nd 2Q Some of the Q will move to the lrger sphere so tht the electrosttic potentils V of the two sphere will be equl. Since V of given sphere is proportionl to the chrge on it nd inversely proportionl to its rdius, the chrges on the two spheres must be in the rtio of 1 to 3, the rtio of the two rdii. The smller sphere gets Q/4 nd the lrger sphere gets 3Q/4.

4 13. Eight prticles re plced on squre of side length, s shown in the figure. Ech prticle is chrged with n mount indicted. Wht is the electric potentil t the center of the squre, if V = t infinity? +q +2q 3q +q q 3q 2q +q (1) 4 2kq/ (2) 5kq/ (3) 3kq/ (4) kq/( 5) (5) kq/( 2) The electrosttic potentils produced by the four prticles on the edges will cncel out t the center. dding the potentils produced by the four prticles t the corners gives kq/(/ 2) k(3q)/(/ 2)+kq/(/ 2) k(3q)/(/ 2). 14. n electron is moving rightwrd between two prllel chrged pltes seprted by distnce d = 1. cm, s shown in the figure. The plte potentils re V 1 = 3 V nd V 2 = 2 V. If the initil speed of the electron t the left plte is 1. km/s, wht is its speed just s it reches the right plte? d V 1 V 2 (1) 1,9 km/s (2) 23 km/s (3) 64 km/s (4) 7.6 km/s (5) The electron will never rech the right plte. Since the electron is negtively chrged wheres the potentil of the right plte is higher thn tht of the left plte, the electron will rech the right plte. Let V be the potentil difference, 1 V. ccording to the conservtion of energy, 1 2 mv2 f 1 2 mv2 i = ev, where v f nd v i re the finl nd initil speeds of the electron. fter plugging in numbers, you will immeditely find tht the second term on the left side is much smller thn the right side of the eqution. So just solve the simplified, pproximte eqution 1 2 mv2 f = ev for v f. slower, error-prone wy of solving this problem is to first find the electric field E = V/d, then the ccelertion = ee/m. Next find the trvelling time t of the electron by solving the kinetic eqution, which is qudrtic (yikes), nd compute v f = v i +t. 15. prllel-plte cpcitor hs plte seprtion d nd cpcitnce C. If slb of metl of thickness d/3 is inserted in the middle of the gp between the two pltes, s shown, wht will be the cpcitnce between the pltes? d d/3 (1) 3C/2 (2) 2C (3) C (4) C/2 (5) The originl cpcitnce is given by C = ǫ /d, where is the re of ech plte. fter the insertion of the slb, let us chrge the cpcitor with chrge q +q on the top plte nd q on the bottom plte. Then q will pper on the top surfce of the slb nd +q on the bottom surfce, ttrcted to the chrges on the pltes. This mens tht the rrngement is equivlent to two cpcitors, ech with plte seprtion of d/3, in series. Ech of these cpcitors hs cpcitnce of ǫ /(d/3), which is 3C. Two such cpcitnces in series mke 3C/2. lterntively, compre the originl cpcitor with the finl cpcitor, both chrged with q, nd notice tht the electric field in the gp(s) is the sme for the two. Since the totl gp size for the second cpcitor is 2/3 of the first one, the potentil difference cross it is lso 2/3 of tht of the first one. (Note tht the potentil difference cross the slb is zero.) The sme q but 2/3 of V this mens tht its cpcitnce, C = q/v, is 3/2 of the originl cpcitnce.

5 16. In the figure, potentil difference of V = 24 V is pplied cross the rrngement of cpcitors with cpcitnces C 1 = C 2 = 4. µf nd C 3 = 1. µf. Wht is the chrge on one of the pltes of cpcitor C 1? C 1 (1) 48 µc V (2) 96 µc (3) 72 µc (4) 12 µc (5) 6. µc C 2 C 3 The equivlent cpcitnce of C 1 nd C 2 in series is 2. µf. The equivlent cpcitor will therefore hve chrge of 24 V 2. µf = 48 µc. Peek inside the equivlent cpcitor nd find tht this mount of chrge ppers on C 1, s well s C 2 for tht mtter. 17. Wht is the equivlent cpcitnce, in µf, of the circuit shown if ech cpcitor hs cpcitnce of 1 µf? C 1 C 2 C3 C 4 (1) 6/11 (2) 11/6 (3) 5/6 (4) 6 (5) 1/6 V C 5 C 6 The circuit consists of three groups of cpcitors, in series. The first group comprises C 1 lone, the second group three cpcitors in prllel, nd the third group two cpcitors in prllel. Therefore, the equivlent cpcitnce is tht of 1 µf, 3 µf, nd 2 µf in series 6/11 µf becuse 1+1/3+1/2 = 11/ Wires C nd D re mde of different mterils nd hve length L C = L D = 1. m. The resistivity nd rdius of wire C re Ωm nd 1. mm, nd those of wire D re Ωm nd.3 mm. The wires re joined s shown, nd current of 2. is pssed through the combo. Wht is the rte t which energy is dissipted in wire C? C D L C L D (1) 5 W (2) 1 W (3) 3 W (4) 14 W (5) 21 W The resistnce of wire C is R = ( )(1.)/[π( ) 2 ] = 4.π Ω. The rte t which energy is dissipted is the power, P = i 2 R, which is bout 5 W V bttery is connected to resistive strip consisting of three sections with the sme cross-sectionl re but different conductivities. The figure gives the electric potentil V versus position x long the strip. The conductivity of section 3 is (Ωm) 1. Wht is the conductivity of section 1, in 1 7 (Ωm) 1? (1) 6. (2) 3. (3) 1.5 (4) 4.5 (5) V 1 x 2 3 Recll E = ρj, where the three symbols re the electric field, conductivity, nd current density. Since the cross-sectionl res of the three sections re the sme, J is the sme throughout. Therefore, the E of ech section is proportionl to ρ, inversely proportionl to its conductivity. Noting tht E is the slope of the curve of V versus x, find tht the E in section 1 is one hlf the E in section 3. This mens tht the conductivity of section 1 is twice the conductivity of section 3. lterntively, solve V = ir = i(ρl/) = il(σ) for conductivity σ to find σ = il/(v). Since i nd re constnt throughout, σ is proportionl to L/V, which is twice s lrge in section 1 s in section 3 ccording to the grph.

6 2. resistor dissiptes.5 W when potentil difference of 3. V is pplied. When the potentil difference is chnged to 1. V, wht is the power dissipted in this resistor? (1).56 W (2) 4.5 W (3) 1.5 W (4).5 W (5).17 W P = V 2 /R. s V is decresed by fctor of 3, P will decrese by fctor of 9.

200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes

200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes PHYSICS 132 Smple Finl 200 points 5 Problems on 4 Pges nd 20 Multiple Choice/Short Answer Questions on 5 pges 1 hour, 48 minutes Student Nme: Recittion Instructor (circle one): nme1 nme2 nme3 nme4 Write