Version 001 Exam 1 shih (57480) 1

Size: px

Start display at page:

Download "Version 001 Exam 1 shih (57480) 1"

Katrina Harmon
6 years ago
Views:

1 Version 001 Exm 1 shih 57480) 1 This print-out should hve 6 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Holt SF 17Rev prt 1 of ) 10.0 points Three positive point chrges re rrnged in tringulr pttern in plne, s shown below. 3 nc 9 m 9 m 3 nc 5 nc 9 m 7 nc µn µn.6446 µn m 9 m 5 nc 9 m 7 nc Find the mgnitude of the net electric force on the 7 nc chrge. The Coulomb constnt is N m /C e e e e e e e e e e-09 Correct nswer: N. Let : q 1 = 3 nc = C, q = 7 nc = C, q 3 = 5 nc = C, x 1,y 1 ) = 0 m,9 m), x,y ) = 9 m,0 m), x 3,y 3 ) = 0 m,9 m), nd k C = N m /C. q 1 q F electric = k C r F net = F 1 F 3 F net = Fnet,x F net,y The distnces between the 7 nc chrge nd q 1 nd q re r1 = x x 1 ) y y 1 ) = x 0) 0y 1 ) = x y 1 = 9 m) 9 m) = 16 m, nd r3 = x 3 x ) y 3 y ) = 0x ) y 3 0) = x y 3 = 9 m) 9 m) = 16 m. Consider the mgnitudes of the forces. The repulsive force q q 1 F 1 = k C r1 = N m /C C ) C ) 16 m = N cts downwrd nd to the right long the line connecting q nd q 1, nd the repulsive force q q 3 F 3 = k C r3 = N m /C C ) C ) 16 m = N

2 Version 001 Exm 1 shih 57480) cts upwrd nd to the right long the line connecting q nd q 3. Now, lets consider the directions of the forces. Since x = y 1 = y 3, θ 1 = θ 3 = 45, nd F x = F 1 cos45 F 3 cos45 = N)cos N)cos45 = N F y = F 1 sin45 F 3 sin45 = N)sin N)sin45 = N F net = Fx F y [ = 9 N ) N ) ] 1 = N. 00 prt of ) 10.0 points Wht is the direction of this force mesured from the positive x-xis s n ngle between 180 nd 180, with counterclockwise positive)? Correct nswer: tnθ = F y F x θ = rctn Fy F x = ) = tn 1 10 ) N N The net force on the 7 nc chrge is N cting bove the positive x-xis. Mgnitude of Force points There re two identicl smll metl spheres with chrges 64.3 µc nd 41.6 µc. The distnce between them is 6 cm. The spheres replced incontct then set ttheir originl distnce. Clculte the mgnitude of the force between the two spheres t the finl position. The Coulomb constnt is N m /C Correct nswer: N. Let : q A = 64.3 µc = C, q B = 41.6 µc = C, nd d = 6 cm = 0.06 m. When the spheres re in contct, the chrges will rerrnge themselves until equilibrium is reched. Ech sphere will then hve hlf of the originl totl chrge: q = q A q B

3 Version 001 Exm 1 shih 57480) 3 = C C = C. The force between the two spheres is F = k q d = N m /C ) C) 0.06 m) = N. Uniformly Chrged Bent Wire points A uniformly chrged insulting rod of length 18.8 cm is bent into the shpe of semicircle s in the figure. The vlue of the Coulomb constnt is Nm /C. Let : L = 18.8 cm = m nd q = 8.63 µc = C. Cll the length of the rod L nd its chrge q. Due to symmetry E y = de y = 0 nd E x = desinθ = k e dqsinθ r, where dq = λdx = λrdθ, so tht y θ x 18.8cm 8.63 µc If the rod hs totl chrge of 8.63 µc, find the horizontl component of the electric field t O, the center of the semicircle. Define right s positive Correct nswer: N/C. O 3π/ E x = k eλ cosθdθ r π/ = k eλ r sinθ) 3π/ = k eλ r, π/ where λ = q L nd r = L π. Therefore, E x = k eqπ L = Nm /C ) m) C)π = N/C. Since the rod hs negtive chrge, the field is pointing to the left towrds the chrge distribution). A positive test chrge t O would feel n ttrctive force from the semicircle, pointing to the left.

4 Version 001 Exm 1 shih 57480) 4 An Oblique Field 005 prt 1 of ) 10.0 points A chrged cork bll is suspended on light string in the presence of uniform electric field s in the figure. The bll is in equilibrium. E = N/C)î N/C)ĵ. The ccelertion of grvity is 9.8 m/s. 0.5 m ĵ î T cosθ = mg qe y tnθ = T sinθ T cosθ = mg qe y )tnθ = qe x mg qe y = qe x tnθ qe x mg qe y qe y = mg qe x tnθ q = mg qe x E y E y tnθ mg tnθ q = E y tnθ E x E 34 Find the chrge on the bll Correct nswer: nc. 1.3 g Let : E x = N/C, E y = N/C, θ = 34, nd m b = 1.3 g = kg. In the î nd ĵ directions, force equilibrium tells us qe x T sinθ = 0 1) qe y T cosθ mg = 0, ) q = kg)9.8 m/s ) tn N/C) tn N/C) = nc. 006 prt of ) 10.0 points Find the tension in the string Correct nswer: N. The first eqution for force equilibrium gives T = qe x sinθ = nc) N/C) sin34 = N. T sinθ = qe x Chrge Inside Box 0

5 Version 001 Exm 1 shih 57480) points A cubic box of side, oriented s shown, contins n unknown chrge. The verticlly directed electric field hs uniform mgnitude E t the top surfce nd E t the bottom surfce. E E How much chrge Q is inside the box? 1. Q encl = 0. Q encl = ǫ 0 E correct 3. Q encl = E ǫ 0 4. Q encl = ǫ 0 E 5. insufficient informtion 6. Q encl = E ǫ 0 7. Q encl = 3 E ǫ 0 8. Q encl = 3 ǫ 0 E 9. Q encl = 6 ǫ 0 E 10. Q encl = 1 ǫ 0E Electric flux through surfce S is, by convention, positive for electric field lines going out of the surfce S nd negtive for lines going in. Here the surfce is cube nd no flux goes through the verticl sides. The top receives Φ top = E inwrd is negtive) nd the bottom Φ bottom = E. The totl electric flux is Φ E = E E = E. Using Guss s Lw, the chrge inside the box is Q encl = ǫ 0 Φ E = ǫ 0 E. Energy gined from A to B points A proton is relesed from rest in uniform electric field of mgnitude V/m directed long the positive x xis. The proton undergoes displcement of 0. m in the direction of the electric field s shown in the figure. v A = 0 A V/m v 0 0. m B Apply the principle of energy conservtion to find the mount of the kinetic energy gined fter it hs moved 0. m e e e e e e e e e e-15 Correct nswer: J.

6 Version 001 Exm 1 shih 57480) 6 The chnge in the potentil energy of the proton is U = q p V = C)38000 V) = J. Conservtion of energy in this cse is So K U = K f K i U = 0. K f K i = U = J. Uniformly Chrged Sphere 05 p 009 prt 1 of ) 10.0 points Consider the uniformly chrged sphere with rdius R, shown in the figure below. Q is the totl chrge inside the sphere p Find the totl flux pssing through the Gussin surfce sphericl shell) with rdius r. R r 9. Φ = Q r ǫ 0 R ) 10. Φ = Q ǫ 0 R r ) Guss lw sttes tht E d A = q in ǫ 0 = Φ. The net chrge inside the sphere of rdius r is q in = ρ 4 3 πr3 = Q 4 3 πr3 4 3 πr3 = Q r3 R 3, so the totl flux through the surfce is Φ = Q ǫ 0 r 3 R prt of ) 10.0 points Find the electric field t rdius r. 1. E = k QR r 3. E = k QR r 3 3. E = k Q r 1. Φ = Q ǫ 0. Φ = Q r ) ǫ 0 R 3 ) 3 3. Φ = Q R ǫ 0 r 4. Φ = Q r ) ǫ 0 R 5. Φ = Q R ) ǫ 0 r 3 6. Φ = Q ǫ 0 r R) 3 correct 7. Φ = Q R ǫ 0 r 8. Φ = Q ǫ 0 r ) 4. E = k Q r 5. E = k Qr R 3 6. E = k Q R 7. E = k Q R 8. E = k QR r 4 9. E = k Qr R 3 correct 10. E = k Qr R 4

7 Version 001 Exm 1 shih 57480) 7 From Guss lw, E 4πr = Φ, so the electric field is E = Φ 4πr = Q ǫ 0 r 3 R 3 1 4πr = Qr 4πǫ 0 R 3 = kqr R 3. Tipler PSE points The distnce between the K nd Cl ions in KCl is m. Find the energy required to seprte the two ions to n infinite distnce prt, ssuming them to be point chrges initilly t rest. Theelementlchrgeis Cndthe Coulomb constnt is N m /C ev ev ev ev ev ev correct Let : q = C, k = N m /C, nd d = m. The energy is W = K U = 0U i = ke)e = kq d d = N m /C ) m ) C) 1 ev J = 5.14 ev. Holt SF 18Rev points The three chrges shown in the figure re locted t the vertices of n isosceles tringle. The Coulomb constnt is N m /C nd the ccelertion of grvity is 9.8 m/s C 5.5 cm 5.5 cm C C 1. cm Clculte the electric potentil t the midpoint of the bse if the mgnitude of the positivechrge is C nd themgnitude of the negtive chrges re C Correct nswer: V. Let : q 1 = C, q = q 3 = C, = 5.5 cm, b = 1. cm, nd k e = N m /C. Let r 1, r nd r 3 be the distnces of the respective chrges from the midpoint of the bse of the tringle.

8 Version 001 Exm 1 shih 57480) 8 r 1 = r r 3 r = r 3 = b r 1 ) b = 1 4 b V = k e q r Then the totl electric potentil is V tot = V 1 V V 3 q 1 q q 3 = k e k e k e r 1 r r [ 3 q 1 = k e 4 b q q ] 3 b b [ q 1 = k e 4 b q ] q 3 b [ q 1 = k e 4 b q ] b = N m /C ) [ C m) 0.01 m) ] C) 0.01 m = V. Connected Spheres prt 1 of ) 10.0 points Two chrged sphericl conductors re connected by long conducting wire, nd chrge of 15 µc is plced on the combintion. One sphere hs rdius of 4.05 cm nd the other rdius of 6.05 cm. Wht istheelectricfieldner thesurfce of the sphere with the smller rdius? Correct nswer: V/m. Let : r 1 = 4.05 cm = m, r = 6.05 cm = m, Q = 15 µc = C, nd k e = N m /C. Since the two spheres re connected by long conducting wire, they re t the sme potentil, nd q 1 = q 1) r 1 r We re lso given the totl chrge of the combintion Q = q 1 q ) With two equtions nd two unknowns, we cn solve for the chrge q 1 on the smller sphere. From 1), q 1 = r 1q r = r 1Qq 1 ) r q 1 r = r 1 Qr 1 q 1 q 1 r 1 r ) = r 1 Q q 1 = r 1Q r 1 r. Hence the electric field ner the surfce of the sphere with rdius r 1 is E = k e q 1 r 1 = k e Q r 1 r )r 1 = N m /C ) C m m) m) = V/m.

9 Version 001 Exm 1 shih 57480) prt of ) 10.0 points Wht is the potentil of the lrger sphere? Correct nswer: V. Since the two spheres re t the sme potentil, it suffices to consider the potentil of either sphere. Since q 1 ws obtined in the first prt of the question, we my simply determine the potentil of the smller sphere. V = k e q 1 r 1 = k e Q r 1 r = N m /C ) C m m = V. Tipler PSE points Wht is the electrosttic potentil energy of n isolted sphericl conductor of 10 cm rdius tht ischrged to 1000V? The Coulomb constnt is N m /C Correct nswer: µj. Let : r = 10 cm = 0.1 m, V = 1 kv = 1000 V, nd k = N m /C. The potentil on sphericl conductor is V = kq r Q = rv k, so the electrosttic potentil energy is U = 1 QV = 1 ) rv V k = rv k 0.1 m)1000 V) = N m /C ) 106 µj 1 J = µj. Sphere Within Shell A points Consider solid conducting sphere with n inner rdius R 1 surrounded by concentric thick conducting sphericl shell which hs n inner rdius R nd outer rdius R 3. There is chrge Q on the sphere nd no net chrge on the shell. For this problem, we dopt the stndrd convention of setting the electric potentil t infinity to zero. q = 0 q 1 = Q O R 1 R B A R 3

10 Version 001 Exm 1 shih 57480) 10 Find the potentil t O, where R O < R V O = 0. V O = kq R 3 kq R kq R 1 correct 3. V O = 4. V O = kq kq R 3 R 1 5. V O = kq R 1 6. V O = kq kq R R 1 kq 7. V O = R 1 8. V O = kq R 1 9. V O = kq R 1 R 10. V O = kq R 1 We re still inside the sphericl shell. The potentil due to the shell requires knowing something bout where chrge exists on the shell. To this end, consider Gussin sphericl surfce inside the shell. This surfce cn hve no flux through it since no electric field cn be upheld inside conductor. By Guss s Lw, there cn therefore be no chrge enclosed. But we re lredy enclosing Q on thesphere, so chrgeq must reside onthe inner surfce of the shell: q inner = Q. Sincethenet chrgeontheshell iszero, the chrge on the outer surfce must be Q for the inner nd outer chrges to dd up to zero: for thin shell of rdius. Thus the potentil due to the inner shell rdius R ) is V = k q inner R = k Q R nd tht due to the outer shell rdius R 3 ) is V 3 = k q outer R 3 = k Q R 3, so the contribution from the two surfces of the shell is V 1 = k Q R 3 k Q R. However, we re now looking t point inside the sphere. The chrge is ll on the surfce of the sphere, so similrly to the sitution for the shell we hve V = k Q R 1 everywhere inside the sphere. Thus the totl potentil t O is V O = k Q R 3 k Q R k Q R 1. This potentil is the sme t O s it is everywhere inside the sphere. A conductor is n equipotentil object. Add Chrge to Four prt 1 of ) 10.0 points Four chrges re plced t the corners of squreofside,withq 1 = q = q, q 3 = q 4 = q, where q is positive. Initilly there is no chrge t the center of the squre. q = q q 3 = q q outer = Q. Now tht we know the exct distribution of chrge, we cn utilize the expression for potentil inside of sphericl chrge distribution: V = k q, q 1 = q q q 4 = q

11 Version 001 Exm 1 shih 57480) 11 Find the work required to bring the chrge q from infinity nd plce it t the center of the squre. 1. W = kq. W = kq 3. W = 4kq 4. W = kq 5. W = 8kq 6. W = 4kq 7. W = 0 correct 8. W = 4kq 9. W = 4kq 10. W = kq Bsed on the superposition principle, the potentil t the center due to the chrges t the corners is V = V 1 V V 3 V 4 = kq 1111) = 0, r where r is the common distnce from the center to the corners. The work required to bring the chrge q from infinity to the center is then W = qv = prt of ) 10.0 points Wht is the mgnitude of the totl electrosttic energy of the finl 5 chrge system? It my be useful to consider the symmetry property of the chrge distribution which leds to cncelltions mong severl terms. 1. U = 4 kq.. U = kq. correct 3. U = 4 kq. 4. U = kq. 5. U = kq. 6. U = 4 kq. 7. U = kq. 8. U = 4 kq. 9. U = kq. 10. U = 8 kq. No work is done in bringing in the fifth chrge nd plcing it t the center, so the totl electrosttic energy is contributed only by 4 corner chrges: U = U 1 U 13 U 14 U 3 U 4 U 34 = kq ) 1 = kq. Cylindricl Cpcitor 03 v points A 75 m length of coxil cble hs solid cylindricl wire inner conductor with dimeter of mm nd crries chrge of 5.1 µc. The surrounding conductor is cylindricl shell nd hs n inner dimeter of 6.55 mm nd chrge of 5.1 µc. Assume the region between the conductors is ir. The Coulomb constnt is N m /C. Wht is the cpcitnce of this cble?

12 Version 001 Exm 1 shih 57480) Correct nswer: nf. Let : k e = N m /C, Q = 5.1 µc, l = 75 m, = mm, nd b = 6.55 mm. The chrge per unit length is λ Q l. b V = E d s b dr = k e λ = k e Q l ln r b ). The cpcitnce of cylindricl cpcitor is given by C Q V = l 1 ) k e b ln 75 m = N m /C ) ) 9 ) nf 6.55 mm 1 F ln mm = nf Cpcitor Circuit 04 shortest 00 prt 1 of ) 10.0 points Consider the group of cpcitors shown in the figure V 4.43 µf 8.41 µf 1.84 µf b c d 8.66 µf Find the equivlent cpcitnce between points nd d Correct nswer: µf. Let : C 1 = 8.41 µf, C = 4.43 µf, C 3 = 8.66 µf, C 4 = 1.84 µf, nd E B = 11.8 V. E B C 1 C b c For cpcitors in series, C 3 1 = 1 C series C i V series = V i, C 4 d nd the individul chrges re the sme.

13 Version 001 Exm 1 shih 57480) 13 For prllel cpcitors, C prllel = C i Q prllel = Q i, nd the individul voltges re the sme. InthegivencircuitC ndc 3 reconnected prllel with equivlent cpcitnce C 3 = C C 3 = 4.43 µf8.66 µf = µf. C 1, C 3, nd C 4 re connected in series with equivlent cpcitnce 1 C b = 1 1 ) 1 C 1 C 3 C 4 1 = 8.41 µf µf µf = µf. ) 1 01 prt of ) 10.0 points Determine the chrge on the 4.43 µf cpcitor t the top center prt of the circuit Correct nswer: µc. The voltge drops cross C 1 nd C 4 re then nd V 1 = Q 1 C 1 = V 4 = Q 4 C 4 = µc 8.41 µf = V µc 1.84 µf = V. The remining voltge is V remin = V totl V 1 V 4 = 11.8 V1.899 V V = V. This remining voltge is the sme cross the prllel cpcitors, so Q = C V remin = 4.43 µf)1.019 V) = µc. Two Dielectric Slbs v1 0 prt 1 of 3) 10.0 points A prllelpltecpcitor hs cpcitncec 0, plte seprtion d nd plte re A. Two dielectric slbs of dielectric constnts κ 1 nd κ, ech of thickness d, re inserted between the pltes. Chrges Q nd Q re put on the left nd right pltes, respectively. dielectric constnt left dielectric constnt right κ κ 1 l A = l d Withoutthedielectrics,whtisthefieldE 0 between the pltes? 1. E 0 = Q ǫ 0 A d. E 0 = Q 4ǫ 0 A 3. E 0 = Q ǫ 0 A d 4. E 0 = Q ǫ 0 A correct 5. E 0 = Q 4ǫ 0 A d 6. E 0 = Q ǫ 0 A

14 Version 001 Exm 1 shih 57480) E 0 = Q ǫ 0 A Solution: From Guss Lw, we know tht Φ = Q encl ǫ 0. For flt surfce, like plte, the flux is the norml component of the electric field times the re. Also, for prllel plte cpcitor, thefieldisonlyinthegpneglecting edge effects), so Φ = E 0 A E 0 = Φ A = Q ǫ 0 A. 03 prt of 3) 10.0 points With the dielectrics inserted, the fields in κ 1 nd κ re respectively 1. E 1 = E 0 κ nd E = E 0 κ 1. E 1 = E 0 κ nd E = E 0 κ 1 3. E 1 = E 0 κ 1 nd E = E 0 κ correct 4. E 1 = E 0κ κ 1 κ nd E = E 0κ 1 κ 1 κ 5. E 1 = E 0 κ 1 nd E = E 0 κ 6. E 1 = E 0κ 1 κ κ 1 κ nd E = E 0κ 1 κ κ 1 κ 7. E 1 = E 0κ 1 κ 1 κ nd E = E 0κ κ 1 κ In ny region with dielectric constnt κ, the permittivity ǫ 0 becomes κǫ 0, so the fields in κ 1 nd κ re E 1 = Q κ 1 ǫ 0 A = E 0 κ 1 E = Q κ ǫ 0 A = E 0 κ. 04 prt 3 of 3) 10.0 points Let κ 1 = 3, κ = 4.41, A =.74 m, nd d = 4.7 mm. Wht is the resultnt cpcitnce? e e e e e e e e e e-08 Correct nswer: F. The potentil difference between the two pltes is the sum of the potentil differences cross the two dielectrics. The potentil difference cross region with constnt electric field is the product of the field nd the distnce, so V = V 1 V = E 0 d κ 1 E 0 d κ = Q d 1 1 ). ǫ 0 A κ 1 κ The cpcitnce is given by C = Q V = ǫ 0A d 1 1 κ 1 1 κ ) = κ 1κ ǫ0a κ 1 κ d [ ] 3)4.41) = F/m ).74 m ) m = F. Chrge in Wire points The current in wire decreses with time ccording to the reltionship I =.3 ma) e t

15 Version 001 Exm 1 shih 57480) 15 where = s 1. Determine the totl chrge tht psses through the wire from t = 0 to the time the current hs diminished to zero Correct nswer: C. q = t t=0 Idt I = dq dt = A)e s1t dt t=0 = A) e s1 t s 1 = C. 0 Correct nswer: A. Let : V = 0.59 V, l = 1.3 m, A = 0.7 mm = m, nd ρ = Ω m. The resistnce is so the current is I = V A ρl R = V I = ρl A, = 0.59 V) m ) Ω m)1.3 m) = A. Current in Tungsten Wire points A 0.59 V potentil difference is mintined cross 1.3 m length of tungsten wire tht hscross-sectionlreof0.7mm ndthe resistivity of the tungsten is Ω m. Wht is the current in the wire?

ragsdale (zdr82) HW2 ditmire (58335) 1

ragsdale (zdr82) HW2 ditmire (58335) 1 rgsdle (zdr82) HW2 ditmire (58335) This print-out should hve 22 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. 00 0.0 points A chrge of 8. µc