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Unformatted text preview: PH 132 SPRING 2003
HOMEWORK # 4 Assigned: 02/03/03 50 SHEETS 22142 100 SHSETS
22144 200 SHEETS 2214] BP. The electric ﬁeld inside a nonconducting sphere o
with charge spread uniformly throughont its volume,
directed and has magnitude qr . W o): o tineoRl' 50') ‘= Here q (positive or negative] is the total charge within and r is the distance from the sphere’s center. (a) Takin
the center of the sphere, ﬁnd the electric potential V(r
sphere. (b) What is the difference in electric potential
point on the surface and the sphere’s center? (c) If q which of those two points is at the highet potential? EM: 6!) V (r4 R)
b)av= wen—VCR) 
one >0, is We wan,
or 1‘5 V(R)> We)? Elea‘n‘c Pa'mﬁhf due 1‘0 a. méhuous dt'ah'émﬁm Cafe/143.: 0) 714a MSWlo Pm b) e Posi‘ﬂvepo v(a)— vac) >0 5 O S HEETS 22142 100 SHEETS
22144 200 SHEETS 2214] a
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a... [ a
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ﬁnd P095”: anti, I Z Hid: 63)er W 0""2
b)6how Hie deem; polmb‘al.
an(1 a chaotic! shad 3.5 V: V0 " (fa); lﬂP. Figure 2530 shows. edgem, an inﬁnite nonconductin'
with positive surface charge density a on one side. (a) 25— l 8 and Eq. 24! 3 to show that the electric potential of an
sheet of charge can be written V = V0 — (tr/259); where
electric pmentia] a1 Ihc surface of die sheet and z is the  ' Work M £25 an Eleofn'c Field, Eka‘n’c warm! I») V“ 3655 "LgJ Comba'm'wj 1%{2 {we «1414‘;th gI'eIds: W=jff°d?
i'
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% V(z)= Vm 67‘? Use 441: ram {m a) V( = W — 1 £53: 5
a) 0) 23(26.» 22 I4] 50 SHEETS
22442 H30 SHEETS
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_._____..~_i____.__ @ﬂ’ﬂ‘V(o°)=o ' a”? jar'36 2’ We. V(x)=o _ and let the paniclcs have
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Xiaﬂdﬂw bdwn 7r andcgz 69!. 2526 P403) There are 3 case; 7‘» examine‘1'. x>d J 12'. mo, ZIT.()<x<d P . I xﬁx W y E if 1?: tha)=o,+§¢ G}de _ i
X 0’ x,=x—d g;;(fj:;f‘;j;’"w~’°%  E’ + g! 3’ k 5 .‘~. I
xd  x x—d
rk[ (K~d)3¢x]: k “Zex 1d = 556%) \ i '
1W xécd) X Xd [ x>d",sa (x00>O. Also "sz ~d ,so (ﬂex#0. 712:5 mam V5040: 2
Unﬁt): [i3 L‘X‘3Z]; 133 w]_o d—t/X:O => i
X 6"” X d‘X : 22 I41 50 SHEETS
22142 100 SHEETS
22444 200 SHEETS @ In Fig. 2533, point P is a: the center or Lhc rectangle. with wm‘
b: 0 at inﬁnity. what is the as: electric potential at P due to the “W charged particles? P vp 4% 1 Eledﬁ'p Poknhhl due pa‘nf charges
6
V 2 2 _._ d
k I"  q. a. an 2 .4 a :5 g '—
e., I; ) G‘G‘QJL‘W *4) 2 '3 '3 Z
.. _2.0Q __ 3Oz + 5. _2_0 +3
. R Vd'+l§)‘ ’ g (id’ﬂ‘y' {3% ,y" +2 ﬁ%1 (5.0 3, 0 +5.0 r3. 0) + 0—32620?! 0)] = _10_'.3:kq,'—1Q—g,é£.—l—O——
kcz dt+(_)1v d. "a l‘Iq’ 0’ V? 64.45%] = §£(2.94—9)= ANS 3’ 22 141 50 SHEETS
n 22!42 100 SHSE'I'S
22144 200 SHEETS % In Fig. 25”, what is the net potential at point P due to the '
fou: point charges, if V = 0 at inﬁnity? 22.141 so SHEETS
22142 IOO SHEETS
22.144 zoo SHEETS alan ’0) 5 155. A plastic rod has been formed into a cirble of radius R. 11 has . ‘ = 1r . I
a positive charge +Q uniformly distributed along owe—quarter of I . ﬂow] 6 o A, A! 6 haer  +Q .
. its cirwmfcrenoe and a negative 3.); wh—at 13 that “7
dmgcof—GQuaﬁfounlydiSlrib' mm: (a) at the we! C #31“ . from 9 " Y "‘3' 2” charge :  9Q
med along the rest of the circum circle and (b) 31 [301!“ a WP"; '5‘ '
fermcc (Fig. 25—37]. With V = 0 an the central axts of the are c a v (00) : O '    ' _ dim“: z from mmnwr? $5411 Eledn‘c. Pdmﬁal due. 1‘0 a cmh‘nm dzlsirxbuﬁ'm 04 charge 33 is (new 1’.) do pal:t b) ﬁnsi' in 3% «azasz exprwz'm 19>" We) and 441M (152. H 6r #1! Spark! Case; at” " [0) look “1‘ 0'? f'nﬁm'ksr'm" dmwm‘ 01‘ charjerﬂﬁ 6% a dz‘Jche r from
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22142 100 SHEETS
22144 200 SHEETS a
ma dfx: 28?. Figure. 2540 shows a plastic
rod of length L and unimel pos the axis, at distance d from (med
iLive charge Q lying on an x axis; and of the rod. : With V = 0 an iTIﬁlliiy, ﬁnd the elccu'ic potential at point P. on: Efedn'c PMELI due 1‘0 :2 Wm gamma deluge.
chaged rod [00k 42‘ ﬂ)? lh'ﬂhﬁéSIMd/JIJM afﬁx?! 4% , of/Mjfé d7, ﬁch
Tﬁe 1:5 mﬁﬁrm guyilk ("061,50 we ('0? we m. x. 4/3: 2/2: 4/50, 2 *f? =3§= £945. 507:: My, 12; We;
d+x L(d+x) a pm»?! my? 22 ‘41 50 SHEETS
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12144 200 SHEETS  ﬁctive mcxmiod {attire I 41. assumin the charges EreImi— ‘
work required to set up the four ' Lialiy inﬁmlely far 2pm. £1
chmgc conﬁguration of ﬁg. 25 whaf Eﬂgﬁ Work regalrz’d in 5d Up * 
+58 cauﬁﬁumﬁbd 8 0 9' FIGURE 2550
Exercise 56. yeahc W21! Mark 6/ paw 61:49:: We pod HM {15.2.4643 c/Mrgc (9} deargc, '19'00615 Me work. (3" 4:153 )‘o bn'nj ad 0976 I51. ' I++nbsnam¥obr.'rjzhz. f l
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22444 200 SHEETS 1214! 6
AMMD 59?. In the rectangle of Fig. 25
42. the sides have lengths 5.0 cm
I and 15 cm. q] = 5.0 ,uC. and
qu= +2.0.uc.wmv=0ar
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I
r (,wEZM. '
£=C7JSM
wzaosbw _9
it 12—30201?) C : Z..Ox /O—6C
2;. _ inﬁnity, what are the electric po
tentials (a) at corner A and (b) at
corner 8? (c) How much work is required to move a third : '
45 = +3.0 Jc.r.C from B to A along a diagonal of the rectang
(d) Does this work increase or decrease the electric energy the threecharge system? Is more, Iess, or the same work required if
43 is moved along paths that are
(e) inside the rectangle but not on
a diagonal and (1’) outside the  reclangle?,§ﬂf
E 3'
. Q———£————_,A NOTE: warez—56m
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22144 200 SHEETS 2244] CH53525 52f. Consider two wider separated conducting spheres, I and 2. '  R g 2 R +
the second having twice the diameter oflhe ﬁrst. The smaller sphere I V ° 1 ‘ initially has a positive charge 4!. and the larger one is initially un ‘ r .. _
charged. You now connect the spheres with a long thin wire. In'ﬁal Charje of 5we I — q,
(a) How are the ﬁnal potentials V, and V2 of Ihe spheres relaxed? ' t r _ (b) What are the ﬁnal charges :1, and g: on the spheres, in terms {MW chdﬁe 0(5PMC 2  0
of q? (o) What is the ratio of the ﬁnal surface charge density of sphereIwmalofsoheI'eZ? I . V1,“;
. b) q, dz r9676. ’ I
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6: qt 3; (we: 43R: 2 f z :5.) 2(2') HMS PH 132 _ Suggested Problems 2E. The electric potential difference between [he groundand a
cloud in a particular thunderstorm is 1.2 x 109 V. What is the
magnitude of the change in the electric potential energy (in mul
tiples of Lhe electronvolt} of an electron that moves between the
ground and the cloud? «we .... ..........._... .  .. .... . . 45. When an electron moves from A to 3 along an electric ﬁeld
line in Fig. 2529. the electric field does 3.94 X 10‘” J of work
on it. What are the electric po '
tential differences (a) V8 — VA, 131 m
(b) Vc — VA, and (c) Vc — v8? 5:];
line \
\ N — ”
\'\
B \
H. __ __ C, Equipozenljzls
th. 2529 Exercise 4. i
l
l
l
I
l
l
l
! 6E Two large, parallel, conduct ing plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of 3.9 X IO‘ '5 N acts on an electron
placed anywhere between the two plates. {Neglect fringing.)
(a) Find the electric ﬁeld at the position of the electron. (b) What
is the potential difference between the plates? on aa—
aqxtoﬂst’u M: #17:: héxlo “H...— O 50 SH£ET5 22142 100 SHEETS
22144 200 SHEETS 22141 ® H 559. {a} Use the result of Problem 28 to ﬁnd the 21mg ﬁ ' '
ponent E, a1 point P, in ﬁg. 2540. (Him: First substitute ' able .1: for the distance d in the result.) (b) Use symmetry:
mine. the elcctric ﬁeld component IEJ 31 P. . ssm POIM) WIT/7’ cwmpcwm. x19 qugm 3%
Q g xL.
X : ——_.— —— ‘* i
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3% = 2':
Ki? éﬂﬁg‘iﬁéiu ;
l
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— 9772”; 9x QM *3
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22144 200 SHEETS 2214] 6
mm I 55?. (a)learthhadanctsutface charge density of1.0 electron 6,1ng '
' per square mctcr (a very artiﬁcial assumption). what would its po ' D _ ’
mm be? (Set V = 0 at inﬁnity.) (1:) What would be the electric <52: _— __,¢‘3—‘ 3; “M
H ﬁeld due to E31111 just outside its surface? I ¢$w3>w¢7fh6
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This note was uploaded on 04/23/2008 for the course PH 132 taught by Professor Ramsdell/wick during the Fall '08 term at Clarkson University .
 Fall '08
 Ramsdell/Wick
 Physics

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