This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 17.44 “The worlzdone by the total force on the electron is equal to the change in its kinetic energy. The electric force is the only force acting and, in this particular problem, it is constant. Hence, using the OWE
theorem, I“ll/rte”; a1 = =:» F 0 AF: KEﬁnal — KEinitiai Let i be in the direction of the initial motion of the electron, and ChOose the origin at the place where the
electron enters the ﬁeld. The electric force is opposite in direction to the change in electron’s position vector.
When the electron has zero speed, let the distance into the ﬁeld it has traveled 'be :0. With these choices,
we have (—eEl) e (mi) = 0 J — émvg _ meg _ (9.11 x 1031 kg)(4.00 x 106 m/S)2 2” x — 2e}; ‘ were x 10—19 C){3OU N/C) =>z=0.152rn. 17.60 a) The electric ﬁeld is directed from regions of high potential to regions of lower potential, so the direction
of the ﬁeld is from the sheet with potential 100 V to the sheet with potential ~1OU V. b) From Strategic Example 17.1 on pages 770—771 of the text, we have AVi=Ed =; 192%
200V 3 c) The magnitude of the torque is 1': f5x]_il =pEsin9
=> 7 = (6.0 x 10—30 Cm)(5.00 X 103 we J sin 120° => 1— : 2.6 x 10—25 Nrn. The vector product right—hand rule indicates that the direction of f" is perpendicular to and out of the page. d) The potential energy of the dipole is PE: —f)']*j= —pEc036 => PE 2 46.0 x 10—30 Crn)(5.00 x 103 we J cos 120° = 1.5 x 10‘26 .1. We can express this answer in units of electron—volts. by dividing it by the appropriate conversion factor, —26
PE: ﬁlm—“J— : 9.4 x 10—3eV. 1.602 x 10—19 J/eV e) The minimum value of the potential energy of the dipole occurs when the dipole is parallel to the ﬁeld,
so PEmin = —pEcosO° = 45.0 x 1030 cmxaoe x 103 we) = —3.0 x 10—23 J. One important applicatiOn of the results of this problem is the microwave oven. Most food has a high
percentage of water, and these water molecules experience torques and undergo various rotations as electric
ﬁelds (microwaves) go by them in different directions. Since the thermodynamic internal energy is the
macrosc0pic manifestation of the motion of molecules, the terques exerted on the molecules cause the food
to warm up and “cook.” This heating is more efﬁcient and quicker because the food heats up uniformly
rather than from the outside in. So it is the tOrque microwaves exert on water molecules that is the basis
for microwave cooking. (A book, called The Zapping of America, was published in the 1960’s claiming that microwaves used for
transmitting telephone signals over long distances “cook” peOple just as food gets cooked in a microwave
oven. This claim, however, never was substantiated.) 18.8
a) The potential diﬁerence between points A and B is
VABZVA—VB=10V —(—15V)=25V.
b) The potential difference between points B and C is
VgngB—Vc=(—15V)—20V = —35V. c) The potential difference between points C and A is
VOA =Vc—VA :20V —10V =10V. d) The potential difference between points A and C is
VACZVA#VC= 10V —20V =—10V. 18.24 The three capacitOrs are connectad in parallel, so Case1 +02 +03 = 2.0 #1? +30 as +4.0 as = 9.0 as. 18.25 The three capacitOrs are connected in series, so 1=i+i+i_1+1 1 13
0., 01 02 03—min 3.0pF+4.0,uF_12pF 12 .._.—.—__ — — ... _ _ ._  “a _ __18.35_ . __,. __. a) Note that the two capacitors are connected in parallel with the independent voltage source, and so they
both have the same potential difference across them, 15 V. b) The magnitude of the charge on each capacitor is found from the potential difference and its capacitance, _ El 3
C‘— IVI =:> 1Ql 01V}. For the 4.0 psF capacitor,
Q = (4.0 x 10c6 F)(15V) = 6.0 x 10—5 C = 60 no.
Fer the 5.0 nF capacitor,
Q: (6.0 x 10—5F)(15V) = 9.0 x 10—5 C = 90st}.
c) The two capacitors are connected in parallel, so the equivalent capacitance is their sum 06., a 4.0 as + 6.0 as = 10.0 F. a) The magnitude of the electric ﬁeld at the surface of a uniform spherical charge distribution of radius R
is 1 WI 2
= — = R E
E 41mg R2 2} 471'60
. 5 2 1 “\I 0
2:5 = wzﬁljl X 105 C_ 9.00 x 109 Nm2/02 Since the direction of the ﬁeld is toward the surface, i.e., toward the charge, the charge must be negative,
which means that Q = —4.51 x 105 C. b) The energy density is ' 1 N C 2
Energy density = £50192 : l4w€0 2 ( 00 / ) 1
=—~————————=4.42 105J 3.
2 2 4w 2 4719.00 x 109 Mug/02) X h“ 18.50 The capacitance is r: timesthat of an airﬁlled parallel plate canaciter, so 59A _ sonar? _ 47rear2
C _ ET _ K’ d _ 5’ 4d (0'60 X 10—2 m)? — 2 0 x 1011 F = 20 pF.
(9.00 x 109 Nm2/02 )(4)(0.25 X 10—3 m) . ' = (50) ...
View
Full
Document
This note was uploaded on 10/12/2008 for the course PHYSICS EN.171.102 taught by Professor Barnett during the Fall '07 term at Johns Hopkins.
 Fall '07
 Barnett
 Physics

Click to edit the document details