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Unformatted text preview: t) The apparent weight ai'a steel sphere immersed in various liquids is measured using a
spring seale. The greatest reading is ebtaiaed tier that liquid: V V
{35 lat“: ' “"3 ‘ (“73’ string tlte smallest density Eel'1
: having the largest density gr). C] subieet te the greatest atmesplterie pressure u 1]) having the greatest yelume U.
E} in which the sphere was submerged the deepest __
illCa LQJ‘ETZF Jeep:an
2) A certain blaelt af 1Lianed has a yelume ul' iﬁti eat3 and Heats urt water
{density = lﬂtltl kgtma] with 25% of its 1relun'te submerged. The downward three that
must be applied to held the bleek completely under water is abeut: rillLEE if ,__ Mutt N ___‘
)ﬂJTI'N (gilt? "a: Wild s a swift;
E)1.lxltl"hl =2 Patti: = {Pf“(ma F+ fleesslhlﬁﬂ Wvﬁh = ‘3 3} A 31] kg selid sphere, made et'a metal with a density ef' 23GB kgtmi‘. hangs suspended
by a eerd. 1When the sphere is immersed in a liquid ef‘ unknewa density, the tensieu in
the card is reduced la 23 N. The density at the liquid is elesest te‘. Gill titltt lustre3L T patch; a} use them3 _
a: use t:me : m: “T. _.
n) tsae ligme [3 up a) teen kgrmi '6 rs = _ _ ___
a {ennui { l We}
4] A U—tube is partially ﬁlled with water. One side ufthe tube is inside ul'a hex with still air and the ether side has a wind blewing aeress the apes end with a yeieeity at"
St} ltnu’lt. What is the difference in the height el' water between the twe sides at the tube? a Note: ptair] = 1.2 kgfml. ptwster] = ltlt'ltl hgtm].
l .
A) Uﬂlbem ﬁstdé—Egaqu; —" Yﬁ‘t guided: L 1.12 t . n} will: f’ﬁt?“ ' if’mfﬁ CD
E] Ilﬁﬁl‘ﬂ Ill) maﬁ‘ Eh:ch «is. Fe + Mr : ﬁg
ﬂips ; [A ,__ Ee—EJ "ISL {:'r:5t3%: la'ﬁmﬁl\ 3"] aims {it 5} Bernoulli "s equation can be derived from the conservation of: @nrrsr B] mass
C) angular momentum I
D} volume Lt Ll—Uﬂt“ [tob) CH E} pressure a} Kepler‘s second law, “A tine from the Sun to a given planet sweeps out equal areas in
equal times.“ can be derived from conservation of: A} energy mmass gee Leﬂorr pars} or“ lsooL I'll Elli HT I'l'tﬂlf'l'l ent LITI'I
volume E) pressure 7"} 'l‘wo uncharged metal spheres+ i. and M. are in contact. A negatively charged plastic
rod is brought close to 1., but not touching it._ as shown below. The two spheres are then
slightly separated an d, after this separation, the rod is removed. As a result of these
steps: insulating support: A) both spheres are neutral
l3) both spheres are positively charged
C) hotlt spheres are negatively charged D l. is negatively and M is positively charged
is positively and M is negatively charged The real Twenties “nut ﬁﬁw‘ @ nit/tan m Selma/a as: strpm’ﬂngl! 11% m c: ens3t. mm W {DA lE rimmcl) M LW in MLL... afar
ﬁdiilﬁ‘tlﬁﬁ'irei ’Tltfﬁ'UGDtmﬁ' "lli, agave. 3} What equal. positive chargee maid have ta he placed an the Earth and the Sun ed as
tﬂ balance their gravitational attraedcn‘? {Mass at" Sun ~ 2 a iﬂ‘w kg} at auxin“’11:: Leir'a LEEM ﬂ}: LLWQE 9r
a} same” t: L — .i r
C)5.?aiﬂ”C L i = (3; »%b dig95mm LEW} mﬁL
aanane ﬁgs: 5"“ m 5% mmb mt“
(fibula t: a J ‘ ie ii} ‘l‘we particles, X and "'1’, are 4.1] m apart. H has a charge of 20 and Y has a charge at"
Q. The ratie cf the magnitude ai'the electrostatic faree an X to that en ‘t’ is: IQ} The diagram below ahewa the electric ﬁeld iinea in a regian of space enniaining {Wu
small charged apheres, Y and 2. We can deduce that: A} ‘i’ is positive and E’. is negative (at,
H) the magnitude at‘the eleeLt‘it: ﬁeld is the aarnc everywhere
C the electric ﬁeld is strangest midway between Y and Z. a email negative charge piaecd at point X wnuid experience a three to the leFt
E] ‘t’ and Z must have the same Sign I". «ELM/ta exng at XI £0”: ll} In the figure below. a small EU g sphere is suspended from point P by an insulating
thread that is ﬁt) em long. The sphere bears an unknown oleego. Q. A positive point charge, q = +1.0 p0, is brought to a position direetliI below P and the sphere is repelled
to a new position. 3:] em to the right ol'q. as shown. The eharge Q is elosest to: ’F— Tﬁi‘E}: Cr“
engr‘i'esaﬁ :73 A) L3 til:
3} 2.13 ttC :EM pC 3.0 pC 3.5 ttC
12} The diagrams below show four possible orientations of an eleetrie dipole in a
uniform eleetrie ﬁeld [for 2 and 4 the angle with respeet lo the applied ﬁeld is 45" and
135°). Rtuilt them ueeonling to the magnitude of the torque exerted on the dipole by the ﬁeld. e_ast to greatest. a is as T” a s a
———+—} ——,4——> ——+—% ———k—+
1 2 A 3 4
mulls it" " iWEi
[$14.12.] 
{311.143 131. E fﬁ‘g
D 3,2and4tie,llten1 ‘3 I: 1:
EnttddtieJhenﬂ 1:2 1 “CL! : 13} A point eharge, Hg is plaeed 1t] em from an inﬁnite sheet of eharge that has
o = +ltl pC‘frnl. ll‘tlte magnitude oi" the eleetrie ﬁeld is here hall" way along the line from
the slteet to the point eharge, what is the size ol'Q'? Q1. Ellm let"
@aisae F r Leg; ’ 0 a; Lotti: mans: ﬁr: —" 31.. '
L1} lent: :—
ei iﬂﬂttC I Q i. H F Qua—ea. ' t} l
l: . — 14} A hollow eondueting sphere has an inner radius ofﬂﬁt} m and an outer radius of
LED m. The sphere eitrries u elmrge oi" Sﬂﬂ nC. A point charge of+ 30B of: is present
at the eenter of the sphere. The surface charge rlensit}.r on the outer sur face is closest to: = WHL 4 Qtﬁﬂﬂ '7 O CS}er : «Damn e. “TICtiff? : Twang ﬂ Ii nCr’m" C} .to "Hot: D)28nCJ'm _ _.1(§/ “L £1144 wormI 0" IS} The {'1 gore beiow shows l‘oor eggdshnped Gaussian surl'aees. labeled I, 2. 3. and 4.
enclosing IL}. I. 2. 3 [mini charges. respectively. The point ehsrges are +0. +20 and —50.
Rank the surfaces in terms of the magnitude of total ﬂux. lﬂlmlj. through each. item
lowest to highest. lﬁ] It snlid, insulating sphere of radius R contains a uniform volume distribution nf
pnsitive eharge. Which [if the graphs helth etirteeti},r gives the magnitude, E, of the eieetrie field as at funetien of r‘?
l s e E“ QJ‘ES‘eiL J 'E 3‘3 F,
.l ( mast} 31119.9
1" r r ﬁw’ﬁﬁﬂ r Di R R eﬁvj.
D E iqﬁée 93 1?} A leng eyi indrieai insulatnr (r = [l3 In] has a cross section as sh ewn. A smaller,
eylindrieel. heilew spaee (r = {12 In} is drilled parallel tti the axis at" the larger eylinder,
hut centered en a paint half tire},r between the eenter and edge ef the larger cylinder. Ifthe
volume charge density ed" the insulating material is p = id mCImJ. what it the magnitude
ef the eleetrie ﬁeld at the eenler til” the hollow space? .3 ii iii“ we [(77 i3] 4.5XIUHNEC 6 _ _\
C}e.ﬂslu*t~tt{f  J 394m
D} “singlets: r  Var1“ Nit? . . " _>' :
E} 181K“) 1 I d; I] r
K
1; ff
afh. ,r'" _ : t3?“J 0.01%‘fi3' “with
%{Q;WWD%V ﬁts#59361  E‘( W 1
‘9" 5‘ l ' efﬁxﬁtmwiﬁ‘fnd 13} Two charges Q are fixed at the yertiees of an equilateral triangle with sides of
length a. Using it = lt'ﬁl'ecu, the work reguired to move a third charge, q, from the third
vertex to the center ol'thc littcjoining the two lixed charges is: wmt': sweetie . + SD I‘LH]
Itthan s 9L SE;
C qufal C» qua.
E) (alsthan [9} It ring ol‘eharge has a radius of 1 m and a linear charge dettsity of 3 mem. If
Wm] = it. what is the electric potential 2 m from the center of the ring along the axis of
symmetry (along the line perpendicular to the ring, going through its center't‘:1 A}l.1xiﬂ:V 1? El QM“ sf .2; m Seeit'll one}: 0.9;“ Tia—r I. ' J ‘ j+l1 llﬁujtrm
(givinmiy .3.“ ME} W "J 'd’iie“LL U 31 ‘3
tssx m‘v 4m The Paar {moms (P‘):
x 1 .I'
P d ‘1 U“ a”? on 1d] ‘l'wo itletttieaL small, inttulatiagH at = H] g, spheres each have + ] rat: oi'eharge
uttiﬁtrtuly distributed throughout their 1walttmes. They are initiain at rest with their
centers separated by l tn. After they are released they are both free to move. Assuming
that the only force acting on the spheres is the electrostatic one, what is the magnitude of their linal velocities?
A} 3t] mtg denimi mum 1?. {mumd L B 94.? mt‘s L LN
94? mars L 43:” '.lt' C3 2 c} *1(3 \
D} ﬁﬂﬂﬂ rat’s " E} EQUUU 11135 b 'L
15—: \J i 2]} Suppose one has available: C 1 {£0 .3“— T'we sheets efcapper L m A at: m “1; a sheet afmica {thickness = 0.1 m, K = is) UL Pr A sheet of glass {thickness = 1i} mm. K = 'i'] ii eat all; Wag, emu:
A slab of parafﬁn {thickness =  cm, K = 2} cl Ta ehtain the largest capacitance. place the tth clapper sheets en either side cf: A a M15 mm air gap
he mica ‘ _
C} the glass (Zeus Patti; —a “Him
a} the parafﬁn mesaﬂwe I Y1
E.) the mica, glass, and parafﬁn {a campasite slab of thickness = [.21 em] J (.1. #1:": L c.—
email emf first“) mice. saler“ puma—He C ' (“’5‘ EU as 5"?"
Kid I‘m") go {3.0 '35.“: 5.1 : oﬁ Galas. 22} The diagram shows six, 6 pl" capaciters. The capacitance between paints a and b is: 23} A battery is used ta charge a parallelrplﬂtc capacitor, after which it is
Then the plates are pulled apart to twice their arigiaal separation. This p ccss will dauhlc the:
A} capacitance W
(JPhi
B. surface charge density (In each pittle C3 E"
[cred energy I If C
D} magnitude of the electric field between the two plates (J .: 2;; “I? C: 1 a
E.) charge an each plate 1
 Eda—d match =25ch d3 is caraw which of the following equivalem oapacitancos is not possible?
F
1. Aj4p]: —H—1 C=L+ '
EpF [1 [L H,
@2pl: 7% ‘— 3} i3 pF 
EJJﬁpF it C“1*Lﬂ+m‘%?F 25} Three capacitors an: arrangod as ShUWI'l in the sketch bolow. What is tho ratio of [he
chargo on the plates ol'Ct ~= 2 pF, C2 = 4 #13 and C3 = o 14!“? J' ‘2: L4.) LT; {Elk/Fl“ [— r——4 kinJ ——i i ‘1 E k. i. I !
I i
a 5—  h V —_—> ,5, (31:02:03 = A} I 4:9 1 '. I :  1*" Q m.
E; 3 : 2: I
F.) 9 ‘ 4: I all 5L3} £3: (gli ®,,::[email protected] EXTRA CREDIT 6’???) mini" is: we 396W 5mm
all ~me uni t1 feeuismii 25] Twer charges Q; = Li} C and Q: = 2.13 C are Ieeated cm the; axis at I] = i] and
x1 = 2J2} em. A I]1ird charge Q3 is in he placed between them a: position x; such that Ihe whole system is in equilibrium. Find Q; and .13. G Q3 a G:
I If 
NsiVminﬁl ice—ﬂ—Ic xg a?
G.EEem,D.34C u U U c521. lc. ﬂu? * [Q _s_1 1.:
} essum,+n_12c '9‘: r’ﬁgv‘h‘sﬁ l D] ﬂ.52em,+ﬂ.34C a; 5:1 _ is; E} ltisnetperssihlete achieve :hisequilihrium. { “iiL : Tc? 2) ‘ 1 2
g * (313 1.. I a F— (E—XB.‘
m": ' (Er*5} [ 3 “i Vx =V~r=Vz
I. Vx=FY}Va
(3} V5... supra; 'I D} Vs =l”r<1"z
E} V); { Vv‘iijz Wall! ...
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 Fall '05
 Ogilvie

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