# 1051test2answersW10 - Circle instructor Yethiraj or...

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Unformatted text preview: Circle instructor: Yethiraj or Morrow ,1 Name: Lab period: Student Number: MEMORIAL UNIVERSITY OF NEWFOUNDLAND DEPARTMENT OF PHYSICS‘AND PHYSICAL OCEANOGRAPHY Physics 1051 Winter 2010 Term Test 2 March 12, 2010 INSTRUCTIONS: 1. Do all questions. Marks are indicated in the left margin. Budget time accordingly. 2. Write your name and student number on each page. 3. You may use a calculator. All other aids are prohibited. 4. Write answers neatly in space provided. If necessary, continue onto the back of the page. 5. Do not erase or use “whiteout” to correct answers. Draw a line neatly through material to be replaced and continue with correction. 6. Assume all information given is accurate to 3 signiﬁcant ﬁgures. 7. Don’t panic. If something isn’t clear, ASK! SEE LAST PAGE FOR SOME POTENTIALLY USEFUL FORMULAE AND CONSTANTS For ofﬁce use onl : Circle instructor: Yethiraj or Morrow 2 Name: Student Number: M Lab period: [10] 1. Four charges are positioned on the x-y plane as shown: ql =+4.5><10'9 C is at x=0cm,y=+2cm. q2 =~4.le0'°€ isat x=-2cm,y=00m. (a) What is the total electric potential at the origin? (b) What is the electric ﬁeld at the origin? Give your answer in unit vector notation. (c) There is an equipotential line that passes through the origin. Draw it on the ﬁgure and brieﬂy justify the line you have drawn. (d) What is the electric force F3! on charge ql due to charge q3 ? Give your answer in unit vector notation. (1) (a) Vwo al‘ (0105 (B) The Fidel: JAM- i-u (Large- 1 w 3 Few dowuwaroil 2 4 fake lef-l-uarci- —. S "A (-3) i? 1‘ ("3) (‘8.ﬁ‘1xxoﬁ) (Q'SX'Oﬁ) g‘l-ot x\0 E; " :3 (2):.“ K3,“? C ® E3: El A S A N “‘ -l. XIO l / E4 = E? :- o‘ l C 4- '5 " 2 02x10; J“) N/ _. *4 _, 2.02x10 l ‘t ‘ c E 2. 2 EC '1 ( ‘7‘ (a) As Show“. ‘ ‘ alum .4 @ Even, O‘n lin€ {S eﬁuiJiSi‘Gni’ +6 05 Keﬁqm /?05 _.‘: f‘ﬁr Ly Ea [q|’%p) LM ‘WM MA °?PGJTL( ‘5 ‘6‘. {Kl-mkd. i? ‘4 ~‘1 2 3' ' mad '1) lW-Hx:o)(4-5H° :4 <9 1! Lot-m3; :- ~ x/zf" / A L» W ﬂag“ Circle instructor: Yethiraj or Morrow 3 Name: Lab period: Student Number: [10] 2. (a) A conducting spherical shell with an inner radius of a = 1.5 cm and an outer radius of b = 2.0 cm carries a net charge of q = —5.0x10'” C. The cavity at the centre of the shell is empty. (i) What is the magnitude of the electric ﬁeld at a distance r = 2.5 cm from the centre of the sphere? (ii) What is the surface charge density on the inner surface of the shell (i.e. at radius a)? Brieﬂy justify your answer. (iii) What is the potential diﬁerence AV = Vb — Va between the outer and the inner surface of the sphere? Brieﬂy justify your answer. (3) GM; LN For r>l> .4}. ﬂ ‘ ® CI>E= E1. attr‘ r ‘L :9 E 31 Go .. W i”) A fPlj Gauss! ion.) (2.5): x )o_4. 7’ EV . r E 3 :Ix'z 7‘ ‘0‘2' @ Here E20 .'. W (#- 4-ﬁr} 9 QcWAC/60 => Gin-w!- =0 0 . ch 7 ®fnner O iuner ‘- 4m; in") Cauduchw is eﬁdiPoRMl-ial '—'3> AV “ VL“V.‘ '1‘ D 5 _. —‘ ® (er) Ekwuw=o AV: gE.AS 2 o (b) What is the magnitude of the electric ﬁeld 10.0 cm away from a very f‘ long line of charge withalinear charge density of 6.3><10'9 C/m? You i l g 1;: may ﬁnd the drawing helpful. g r’ “KP GKJ Lon 9“"F‘3C' "r O berm E= E; ’t (‘73 SJWJ“) 9 “‘9‘” “a; 094L162) Mace So 3% “14.5.3le +e‘v‘ r344“! “a” ‘ if » Er dA A ® 4%: Scam a» E 59“ ‘ E“ (2”!) «CD Also C135: Game. :— 1 N ® .5; es RM Q) “a C23 ham" )A a E, CZWFIQ.) E2. 3 ) Q0 2Wr€° Circle instructor: Yethiraj or Morrow 4 Name: Lab period: Student Number: [10] 3. A charged particle starts from rest and is accelerated through a potential difference AV 2 Vh — Va as shown. It then enters a region of space containing a uniform magnetic ﬁeld of 6.0 T directed into the page as shown. The charge travels along a semi-circular path with a diameter of 15.0 mm as shown. (a) Is the charged particle positive or negative? (b) If the mass of the particle is m = 6.64 x10"27 kg and the magnitude of its charge is I q I: 1.6 x 10‘19 C , what is the speed of the charged particle? (c) What is the potential difference, AV = Vb — Va , through which the charged particle was initially accelerated? }’ B=6.0T (aBF:°M rcak’r-kmd rule—2 x x x .4 .s UPVJ RDS “ﬁrmx x \r x B roiui's I x x A Va Vb 15mmx X @ F; =- 1 V X B |-—:————J—>x--": Tw—‘X Upward; a}, > o x x x (5) av B = 01‘ Y. V— ‘18“ U-GMO"H)( G'°)(°‘°‘5>/2 .. a M 1+ TM 4-64! x1647 _ 8 ~ o.o|\ KID = Ll x lo6 "/5 (Cause/(Vﬂkﬂk d— W (Parhoh SEA; «6ku HS‘i‘) a. ‘1' r ‘27 {Loam 2- _\ .cuxlo " AV =~ “‘/2W“’ = iii/I” 6L )‘6 XlO-)1 4 V .. \ =7 2'5) y D CLed: ) V 4V. ..2-Sl"l0lt V (b t/ Circle instructor: Yethiraj or Morrow Lab [10] (3% shown bk gaurﬁ (C) (7) Name: Student Number: period: 4. A rod of length L = 15.0 cm is oriented along the x-axis as shown. A charge Q = +2 x10’6 C is uniformly spread along the rod. Point P is located a distance [1 = 10.0 cm above the left end of the rod as shown. (a) What is the linear charge density, 2. , on the rod? (b) On the diagram, show the direction of the contribution, d1? , to the electric ﬁeld at P ﬁorn the charge on segment dx located a distance x from the left end of the rod as shown. (c) Write an expression for the y—component dEy of the contribution to the electric ﬁeld at P from the charge on segment dx. (d) Calculate the y-component of the electric ﬁeld, E y , at a point P due to the entire rod? ' —L ) = 6) - zwo L. ans -5 {.33 A to CAM -— ' 5 Q=2xlO c dgy ‘2 SHE —- . ———->X t:"‘*' b" “‘9 ’ k L=15.0cm —> x513 New dE E = J“ ““L 2 use!» hztxz V's-x" dEy = 6‘2? Sine - text olx (A2; xe)3/z XaL fo E), - f‘dE‘y _. JV 3/ X‘o “la-XL) " x=0 2 (kg: [ Y __ D V X‘rk‘ xat. ‘ liekk L. ALJt’LH“ ’ .55); : (ammo‘ﬂtsmo‘sﬂous) hJLt—kt (ow) (o 0 +(O"5)1. Circle instructor: Yethiraj or Morrow Lab period: Name: Student Number: Some Potentially Useful Formulae and Constants: " 41% A Flzzk “'2'” 12 ‘ q r U12=ke q; 2 12 E=k,—q?f B r _. AU=—qu-d§ w A [5:]:¢ .1.“ 3* tr? AV=VB—VA=—IE-d§ A ~ dq. “Mp-r AU=qAV ¢E=j1§ dA E: final/hi5; dx afv dz q’aide ¢‘= III V=k,g r Fﬂzqixg’ V=k¢ZgL v2 i r; 0,:— r V=k‘I-4-q- V _4 3 r mm—g‘ﬂ'r Cmcle = err (circumference) 2 Amm=47rr A :7"2 circle Physical constants: lg: 1 =8.99x109N-m2/C2 47:60 £0 = 8.85x10‘” C2 /N-m2 Mathematical formulae: e =1.602x10'19 C me =9.11x1o'31 kg ﬂul I_x_dx_-__1__ ,2 " r (x2+y2)3/2 — x2+y2 I£=mx I.___3’E___ =__i__ any)“ Isinéldﬂ = -cos€ Icosﬁd6=sin0 .1? = AB cosa = 11,13, + AyBy + AB, A . X g z (‘4sz —Asz)f+(Asz -AIBZ)J +(A‘By ‘AYBIV his in: ...
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## This note was uploaded on 03/08/2012 for the course PHYSICS 1051 taught by Professor Michaelmorrow during the Winter '12 term at Memorial University.

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1051test2answersW10 - Circle instructor Yethiraj or...

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