Electric Potential 4
-
Upload
fatima-tuzarah -
Category
Documents
-
view
173 -
download
0
description
Transcript of Electric Potential 4
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 1
This print-out should have 66 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering. The due time is Centraltime.
Charge in Lightning 0323:01, trigonometry, numeric, > 1 min, nor-mal.
001 (part 1 of 1) 10 pointsA strong lightning bolt transfers about 25 C
to Earth.The charge on an electron is 1.60218
1019 C.How many electrons are transferred?
Correct answer: 1.56038 1020 .Explanation:
Let : q = 25 C .
The charge is proportional to the numberof electrons, so
q = n qe
n =q
qe
=25 C
1.60218 1019 C= 1.56038 1020 .
keywords:
AP EM 1993 MC 5523:04, trigonometry, multiple choice, < 1 min,fixed.
002 (part 1 of 1) 10 pointsTwo metal spheres that are initially un-charged are mounted on insulating stands,as shown.
X Y
A negatively charged rubber rod is broughtclose to but does not make contact with sphereX. SphereY is then brought close toX on the
side opposite to the rubber rod. Y is allowedto touchX and then is removed some distanceaway. The rubber rod is then moved far awayfrom X and Y.What are the final charges on the spheres?
Sphere X Sphere Y
1. Zero Zero
2. Negative Negative
3. Negative Positive
4. Positive Negative correct
5. Positive Positive
Explanation:The force is repulsive if the charges are of
the same sign, so when the negatively chargedrod moves close to the sphere X, the neg-atively charged electrons will be pushed tosphere Y. If X and Y are separated beforethe rod moves away, those charges will re-main on X and Y. Therefore, X is positivelycharged and Y is negatively charged.
keywords:
Acceleration of a Particle23:05, trigonometry, numeric, > 1 min, nor-mal.
003 (part 1 of 1) 10 pointsA particle of mass 50 g and charge 50 C isreleased from rest when it is 50 cm from asecond particle of charge 20 C.Determine the magnitude of the initial ac-
celeration of the 50 g particle.Correct answer: 719 m/s2.Explanation:
Let : m = 50 g ,
q = 50 C = 5 105 C ,d = 50 cm = 0.5 m ,
Q = 20 C = 2 105 C , andke = 8.9875 109 .
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 2
The force exerted on the particle is
F = ke|q1| |q2|
r2= ma
~a = ke ~q ~Q
md2
= ke
5 105 C 2 105 C(0.05 kg) (0.5 m2)
= 719 m/s2 .
keywords:
Hanging Charges23:05, trigonometry, numeric, > 1 min, nor-mal.
004 (part 1 of 1) 10 pointsTwo identical small charged spheres hang inequilibrium with equal masses as shown inthe figure. The length of the strings are equaland the angle (shown in the figure) with thevertical is identical.The acceleration of gravity is 9.8 m/s2
and the value of Coulombs constant is8.98755 109 Nm2/C2 .
0.15m
5
0.03 kg 0.03 kg
Find the magnitude of the charge on eachsphere.Correct answer: 4.4233 108 C.Explanation:
Let : L = 0.15 m ,
m = 0.03 kg , and
= 5 .
L
a
m m
q q
From the right triangle in the figure above,we see that
sin =a
L.
Therefore
a = L sin
= (0.15 m) sin(5)
= 0.0130734 m .
The separation of the spheres is r = 2 a =0.0261467 m . The forces acting on one of thespheres are shown in the figure below.
mg
F
T
e
T sin
T cos
Because the sphere is in equilibrium, theresultant of the forces in the horizontal andvertical directions must separately add up tozero:
Fx = T sin Fe = 0Fy = T cos mg = 0 .
From the second equation in the system
above, we see that T =mg
cos , so T can be
eliminated from the first equation if we makethis substitution. This gives a value
Fe = mg tan
= (0.03 kg)(9.8 m/s2
)tan(5)
= 0.0257217 N ,
for the electric force.From Coulombs law, the electric force be-
tween the charges has magnitude
|Fe| = ke |q|2
r2,
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 3
where |q| is the magnitude of the charge oneach sphere.
Note: The term |q|2 arises here because thecharge is the same on both spheres.This equation can be solved for |q| to give
|q| =|Fe| r2ke
=
(0.0257217 N) (0.0261467 m)2
(8.98755 109 Nm2/C2)= 4.4233 108 C .
keywords:
Serway CP 15 1123:05, trigonometry, numeric, > 1 min, nor-mal.
005 (part 1 of 2) 10 pointsThree charges are arranged in a triangle asshown.The Coulomb constant is 8.98755
109 N m2/C2.
+
+
0.1 m
0.3 m5 nC
3 nC
6 nC
y
x
What is the net electrostatic force on thecharge at the origin?Correct answer: 1.38102 105 N.Explanation:
Let : q1 = 5 nC = 5 109 C ,q2 = 6 nC = 6 109 C ,q3 = 3 nC = 3 109 C ,
r1,2 = 0.3 m ,
r1,3 = 0.1 m , and
kC = 8.98755 109 N m2/C2 .
F1,2
F1,3F
The repulsive force
F1,2 = kCq1 q2r21,2
= 8.98755 109 N m2/C2
(5 109 C) (6 109 C)
(0.3 m)2
= 2.99585 106 N
acts along the negative x-axis, and the attrac-tive force
F1,3 = kCq1 |q3|r21,3
= 8.98755 109 N m2/C2
(5 109 C) (3 109 C)
(0.1 m)2
= 1.34813 105 N
acts along the negative y-axis.Thus
Fnet =[(2.99585 106 N)2
+(1.34813 105 N)2]1/2= 1.38102 105 N .
006 (part 2 of 2) 10 pointsWhat is the direction of this force (as an anglebetween 180 and 180 measured from thepositive x-axis, with counterclockwise posi-tive)?Correct answer: 102.529 .Explanation:
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 4
tan =F1,3F1,2
= tan1(F1,3F1,2
)
= tan1(1.34813 105 N2.99585 106 N
)= 77.4712
below the negative x-axis. From the positivex-axis, the angle is
180 + 77.4712 = 102.529 .
keywords:
AP B 1993 MC 6823:07, trigonometry, multiple choice, < 1 min,fixed.
007 (part 1 of 1) 10 pointsThe diagram shows an isolated, positivecharge Q, where point B is twice as far awayfrom Q as point A.
+Q A B
0 10 cm 20 cm
The ratio of the electric field strength atpoint A to the electric field strength at pointB is
1.E
A
EB
=8
1.
2.E
A
EB
=4
1. correct
3.E
A
EB
=2
1.
4.E
A
EB
=1
1.
5.E
A
EB
=1
2.
Explanation:
Let : rB= 2 r
A.
The electric field strength E 1r2, so
EA
EB
=
1
r2A
1
r2B
=r2
B
r2A
=(2 r)2
r2= 4 .
keywords:
Two Charge Field23:13, trigonometry, multiple choice, > 1 min,wording-variable.
008 (part 1 of 3) 10 pointsTwo point-charges at fixed locations pro-
duce an electric field as shown below.
A B
X
Y
A negative charge placed at point X wouldmove
1. toward charge B. correct
2. toward charge A.
3. along an equipotential plane.
Explanation:The electric field runs from a positive po-
tential to a negative potential, so it pointsfrom a positive charge to a negative charge.Therefore the charge B is positive. A negativecharge will move toward a positive potential,which creates lower potential energy and ahigher kinetic energy.
009 (part 2 of 3) 10 pointsThe electric field at point X is
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 5
1. stronger than the field at point Y . cor-rect
2. weaker than the field at point Y .
3. the same as that the field at point Y .
Explanation:The field at X is stronger than the field
at Y , since the number of field lines per unitvolume at X is greater than the number offield lines per unit volume at Y .
010 (part 3 of 3) 10 pointsEstimate the ratio of the magnitude of
charge A to the magnitude of charge B. Youranswer must be within 5%.Correct answer: 1.88889 .Explanation:The number of field lines is proportional to
the magnitude of the charge.
QAQB
179
= 1.88889QAQB 1.88889 .
keywords:
Maximum force on one charge23:05, calculus, multiple choice, > 1 min,fixed.
011 (part 1 of 1) 10 pointsCharge Q is on the y axis a distance a fromthe origin and charge q is on the x axis adistance d from the origin.What is the value of d for which the x
component of the force on q is the greatest?
1. d = 0
2. d = a
3. d =2 a
4. d =a
2
5. d =a2correct
6. d =q
Qa
7. d =q
Q
2 a
8. d =q
Q
a
2
9. d =q
Q
a2
Explanation:We have the force on charge q on the x axis
due to charge Q on the y axis
~F =1
4pi 0
q Q
r2r ,
where r =a2 + d2. So the x component of
the force on q is
Fx =1
4pi 0
q Q
r2cos
=1
4pi 0
q Q
a2 + d2d
a2 + d2
=1
4pi 0
q Qd
(a2 + d2)3/2.
For maximum x component of the force, Fxd
= 0 is required. Therefore
Fxd
=q Q
4pi 0
a2 2 d2(a2 + d2)5/2
= 0
a2 2 d2 = 0d =
a2
.
keywords:
Charged Semicircle23:10, calculus, numeric, > 1 min, normal.
012 (part 1 of 3) 10 pointsConsider the setup shown in the figure be-low, where the arc is a semicircle with radiusr. The total charge Q is negative, and dis-tributed uniformly on the semicircle. Thecharge on a small segment with angle islabeled q.
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 6
x
y
rx
y
III
III IV
B
A
O
q is given by
1. None of these
2. q = Q
3. q =Q
2pi
4. q =2Q
pi
5. q =Q
picorrect
6. q =Q
2pi
7. q =2Q
pi
8. q =Q
pi
9. q = 2piQ
10. q = piQ
Explanation:The angle of a semicircle is pi, thus the
charge on a small segment with angle is
q =Q
pi.
013 (part 2 of 3) 10 pointsThe magnitude of the x-component of theelectric field at the center, due to q, is givenby
1. Ex =k |q|r2
2. Ex =k |q| sin
r2
3. Ex =k |q| cos
r2correct
4. Ex =k |q| cos
r
5. Ex =k |q| sin
r
6. Ex = k |q| r2
7. Ex = k |q| (sin ) r2
8. Ex = k |q| (cos ) r2
9. Ex = k |q| (cos ) r
10. Ex = k |q| (sin ) rExplanation:Negative charge attracts a positive test
charge. At O, E points toward q . Accord-ing to the sketch, the vector Ex is pointingalong the negative x axis. The magnitude ofthe Ex is given by
Ex = E cos =k |q|r2
cos .
014 (part 3 of 3) 10 pointsDetermine the magnitude of the electric fieldat O . The total charge is 7.5 C, the radiusof the semicircle is 14 cm, and the Coulombconstant is 8.98755 109 N m2/C2.Correct answer: 2.18941 106 N/C.Explanation:
Let : Q = 7.5 C ,r = 14 cm , and
k = 8.98755 109 N m2/C2 .
By symmetry of the semicircle, the y-component of the electric field at the centeris
Ey = 0 .
Combining part 1 and part 2,
Ex =k |q| cos
r2
=k |Q|pi r2
cos
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 7
Therefore, the magnitude of the electric fieldat the center is given by
E = Ex =
pi/2pi/2
k |Q|pi r2
cos d
=2 k |Q|pi r2
.
For the above values, the magnitude is givenby
E =2(8.98755 109 N m2/C2) |(7.5 C)|
pi (14 cm)2
= 2.18941 106 N/C .
The direction is along negative x axis.
x
y
rx
y
III
III IV
B
A
O
E
keywords:
Flux Through a Pyramid24:01, trigonometry, numeric, > 1 min, nor-mal.
015 (part 1 of 1) 10 pointsA (6 m by 6 m) square base pyramid withheight of 4 m is placed in a vertical electricfield of 52 N/C.
6 m
4 m
52 N/C
Calculate the total electric flux which goesout through the pyramids four slanted sur-faces.Correct answer: 1872 Nm2/C.Explanation:
Let : s = 6 m ,
h = 4 m , and
E = 52 N/C .
By Gauss law,
= ~E ~ASince there is no charge contained in the pyra-mid, the net flux through the pyramid mustbe 0 N/C. Since the field is vertical, the fluxthrough the base of the pyramid is equal andopposite to the flux through the four sides.Thus we calculate the flux through the baseof the pyramid, which is
= E A = E s2
= (52 N/C) (6 m)2
= 1872 Nm2/C .
keywords:
Flux Through a Submarine24:02, trigonometry, numeric, > 1 min, nor-mal.
016 (part 1 of 1) 10 pointsThe following charges are located inside a sub-marine: 5 C, 9 C, 27 C, and 84 C.Calculate the net electric flux through the
submarine.Correct answer: 6.88954 106 N m2/C.Explanation:
Let : q1 = 5 C = 5 106 C ,q2 = 9 C = 9 106 C ,q3 = 27 C = 2.7 105 C , andq4 = 84 C = 8.4 105 C .
From Gausss Law:
=q1 + q2 + q3 + q4
0
=(5 106 C) + (9 106 C)
8.854 1012 C2/N m2
+(2.7 105 C) + (8.4 105 C)
8.854 1012 C2/N m2= 6.88954 106 N m2/C .
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 8
keywords:
Long Cylindrical Insulator 0324:03, trigonometry, numeric, > 1 min, nor-mal.
017 (part 1 of 1) 10 pointsConsider a long, uniformly charged, cylindri-cal insulator of radius R with charge density1 C/m3. (The volume of a cylinder withradius r and length ` is V = pi r2 `.)The value of the Permittivity of free space
is 8.85419 1012 C2/N m2
R
1 cm
What is the magnitude of the electric fieldinside the insulator at a distance 1 cm fromthe axis (1 cm < R)?Correct answer: 564.705 N/C.Explanation:
Let : r = 1 cm = 0.01 m ,
= 1 C/m3
= 1 106 C/m3 , and0 = 8.85419 1012 C2/N m2 .
Consider a cylindrical Gaussian surface ofradius r and length ` much less than thelength of the insulator so that the compo-nent of the electric field parallel to the axis isnegligible.
`
rR
The flux leaving the ends of the Gaussiancylinder is negligible, and the only contribu-tion to the flux is from the side of the cylinder.Since the field is perpendicular to this surface,the flux is
s = 2pi r `E ,
and the charge enclosed by the surface is
Qenc = pi r2 ` .
Using Gauss law,
s =Qenc0
2pi r `E =pi r2 `
0.
Thus
E =
2 0r
=
(1 106 C/m3) (0.01 m)
2 (8.85419 1012 C2/N m2)= 564.705 N/C .
keywords:
Uniformly Charged Sphere 0424:03, trigonometry, multiple choice, < 1 min,fixed.
018 (part 1 of 2) 10 points
Given : Vsphere =4pi R3
3, and
Asphere = 4pi R2 .
Consider a sphere, which is an insula-tor, where charge is uniformly distributedthroughout.Consider a spherical Gaussian surface with
radiusR
2, which is concentric to the sphere
with a radius R.
R
R
2
p
Q is the totalcharge insidethe sphere.
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 9
The total amount of flux flowing throughthe Gaussian surface is given by
1. =Q
0.
2. =Q
4 0.
3. =Q
2 0.
4. =Q
8 0. correct
5. =2Q
0.
6. =4Q
0.
Explanation:Basic Concept: Gauss Law.
Solution: For spherical symmetric case,
= 4pi r2 E
=Qencl0
.
=Qencl0
=Q
0
4pi
3
(R
2
)34pi
3R3
=Q
8 0.
019 (part 2 of 2) 10 points
The magnitude of the electric field ~E at R2
is given by
1. ~E = k Q2R2
. correct
2. ~E = k QR2
.
3. ~E = 2 k QR2
.
4. ~E = 2 k Q2
R2.
5. ~E = k Q2
R2.
6. ~E = k Q2
2R2.
Explanation:Gausss Law gives us
4pi r2 E =Qencl0
=Q
0
4
3pi
(R
2
)34
3pi R3
=Q
8 0,
E =Q
4pi
(R
2
)28 0
=Q
4pi 0 2R2
=k Q
2R2.
keywords:
Shell Game 01 v224:07, trigonometry, multiple choice, < 1 min,fixed.
020 (part 1 of 3) 10 pointsConsider the following spherically symmetricsituation: We have a charge q1 on a metallicball at the center, inside of a conducting shellof inner radius R2 and outer radius R3. Thereis a total charge of q2 on the shell.
O
q2
q1
A
B
Ca
bc
R1, q1
R2, q2
R3, q2
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 10
Find E at A where OA = a.
1. EA= k
q1a2
correct
2. EA= k
q12 a2
3. EA= k
q1b2
4. EA= k
q1c2
5. EA= 0
6. EA= k
q13 a2
7. EA= k
q12 a2
8. EA= k
2 q1a2
9. EA= k
3 q1a2
10. EA= k
4 q1a2
Explanation:Pick a Gaussian surface (sphere since we
are in spherical symmetry) center at the pointcharge and of radius a. This surface containsonly the point charge, so qencl = q1. Theformula for E gives
EA=
k q1a2
.
021 (part 2 of 3) 10 pointsFind E at B, where OB = b.
1. EB= 0 correct
2. EB= k
q1a2
3. EB= k
q1b2
4. EB= k
q12 b2
5. EB= k
q1c2
6. EB= k
q22 b2
7. EB= k
q1 + q22 b2
8. EB= k
q1 q2b2
9. EB= k
3 q1b2
10. EB= k
4 q1b2
Explanation:For an electrostatic situation, inside of a
conductor, there is no charge; i.e., qinside = 0.Also, ~Einside = 0 and there is no flux inside,inside = 0.Thus
EB= 0 .
Notice also that since the electric field at Bis zero, the total enclosed charge is zero, orq1 + q
2 = 0. Therefore
q2 = q1 .This verifies that the charge on the inner
surface of a conducting shell is q1, whereq1 is the charge is the charge enclosed by theshell.
022 (part 3 of 3) 10 pointsFind E at C, where OC = c.
1. EC= 0
2. EC= k
q1a2
3. EC= k
q1 + q2b2
4. EC= k
q1 q22 a2
5. EC= k
q1c2
6. EC= k
q12 b2
7. EC= k
q1 + q2c2
correct
8. EC= k
q1 q2c2
9. EC= k
3 q1c2
10. EC= k
4 q1c2
Explanation:
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 11
Here the Gaussian surface is a sphere cen-tered at the point charge q1 and of radius c.The enclosed charge in this sphere is all thecharge, or q1 + q2. The electric field at C is
EC= k
q1 + q2c2
.
keywords:
Solid Conducting Sphere24:08, trigonometry, multiple choice, < 1 min,fixed.
023 (part 1 of 1) 10 pointsA positive charge of 106 coulomb is placedon an insulated solid conducting sphere.Which of the following is true?
1. When a second conducting sphere isconnected by a conducting wire to the firstsphere, charge is transferred until the elec-tric potentials of the two spheres are equal.correct
2. The electric field inside the sphere is con-stant in magnitude, but not zero.
3. The electric field in the region surround-ing the sphere increases with increasing dis-tance from the sphere.
4. An insulated metal object acquires a netpositive charge when brought near to, but notin contact with, the sphere.
5. The charge resides uniformly throughoutthe sphere.
Explanation:Every point in the conductor becomes equi-
potential, and the electric field is defined asthe gradient of the electric potential, so insidethe conducting sphere, all points are equi-potential and there is no electric field.Outside the conducting sphere, the electric
field is the same when there are net charges atthe center of the sphere, so the electric fielddecreases with increasing distance from thesphere.
If there is no net charge on the insulatedmetal object when brought near to, but notin contact with the sphere, there is also nonet charge on it. Only the charge distributionchanges.Since there is repulsion among like charges,
charges reside uniformly on the surface of thesphere.
keywords:
Field From a Charged Plate JM24:06, trigonometry, multiple choice, < 1 min,fixed.
024 (part 1 of 1) 10 pointsA uniformly charged conducting plate witharea A has a total charge Q which is positive.The figure below shows a cross-sectional viewof the plane and the electric field lines due tothe charge on the plane. The figure is notdrawn to scale.
E E
+Q
+++++++++++
P
Find the magnitude of the field at point P ,which is a distance a from the plate. Assumethat a is very small when compared to thedimensions of the plate, such that edge effectscan be ignored.
1. ~E = Q0 A
2. ~E = Q2 0 A
correct
3. ~E = Q4 0 A
4. ~E = Q4pi 0 a2
5. ~E = Q4pi 0 a
6. ~E = 2 0 QA
7. ~E = 0 QA
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 12
8. ~E = 4pi 0 a2 Q
9. ~E = 4pi 0 aQ
10. ~E = 0 Qa2
Explanation:Basic Concepts Gauss Law, electrostatic
properties of conductors.Solution: Let us consider the Gaussian
surface shown in the figure.
E
+Q
+++++++++++
E
S
Due to the symmetry of the problem, thereis an electric flux only through the right andleft surfaces and these two are equal. If thecross section of the surface is S, then GaussLaw states that
TOTAL
= 2E S
=1
0
Q
AS , so
E =Q
2 0 A.
keywords:
Coaxial Cable 0124:05, calculus, multiple choice, < 1 min, nor-mal.
025 (part 1 of 4) 10 pointsA long coaxial cable consists of an inner cylin-drical conductor with radius R1 and an outercylindrical conductor shell with inner radiusR2 and outer radius R3 as shown. The ca-ble extends out perpendicular to the planeshown. The charge on the inner conductorper unit length along the cable is and thecorresponding charge on the outer conduc-tor per unit length is (same in magni-
tudes but with opposite signs) and > 0.
QR1
R2
R3
Q
Find the magnitude of the electric field atthe point a distance r1 from the axis of theinner conductor, where R1 < r1 < R2.
1. E = 0
2. E =
2pi 0 r1correct
3. E =
2pi 0 r1
4. E =
3pi 0 r1
5. E =2
3pi 0 r1
6. E =R1
4pi 0 r12
7. E =R1
3pi 0 r12
8. E =2 R1
4pi 0 r12
9. E =
2pi 0 R1
10. None of these.
Explanation:Pick a cylindrical Gaussian surface with the
radius r1 and apply the Gausss law; we obtain
E ` 2pi r1 = Q0
E =
2pi 0 r1.
026 (part 2 of 4) 10 pointsThe electric field vector points
1. in the negative r direction
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 13
2. in the positive r direction correct
Explanation:The field points from positive charge to
negative change.Since the center conductor is negatively
charged, the electric field vector points in thenegative r direction.
027 (part 3 of 4) 10 pointsFind the magnitude of the electric field at thepoint a distance r2 from the axis of the innerconductor, where R3 < r2.
1. E = 0 correct
2. E =
2pi 0 r2
3. E =
2pi 0 r2
4. E =
3pi 0 r2
5. E =2
3pi 0 r2
6. E =R1
4pi 0 r22
7. E =R1
3pi 0 r22
8. E =2 R1
4pi 0 r22
9. E =
2pi 0 R1
10. None of these.
Explanation:Pick a cylindrical Gaussian surface with the
radius r2 and apply the Gausss law. Becausethere is no net charge inside the Gaussiansurface, the electric field E = 0 .
028 (part 4 of 4) 10 pointsFor a 100 m length of coaxial cable with innerradius 1 mm and outer radius 1.5 mm.Find the capacitance C of the cable.
Correct answer: 13.7207 nF.Explanation:
Let : ` = 100 m ,
R1 = 1 mm , and
R2 = 1.5 mm .
We calculate the potential across the capaci-tor by integrating E ds. We may choosea path of integration along a radius; i.e.,E ds = Edr.
V = 12pi 0
q
l
R1R2
dr
r
= 12pi 0
q
lln r
R1
R2
=q
2pi 0 lln
R2R1
.
Since C =q
V, we obtain the capacitance
C =2pi 0 l
ln
(R2R1
)
=2pi (8.85419 1012 c2/N m2)
ln
(1.5 mm
1 mm
) (100 m)
= 13.7207 nF .
keywords:
Charge in a Closed Surface24:02, calculus, numeric, > 1 min, normal.
029 (part 1 of 2) 10 pointsA closed surface with dimensions a = b =0.4 m and c = 0.36 m is located as in the fig-ure. The electric field throughout the regionis nonuniform and given by ~E = ( + x2) where x is in meters, = 3 N/C, and = 2 N/(Cm2).
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 14
Ey
x
z
a
c
b
a
What is the magnitude of the net chargeenclosed by the surface?Correct answer: 1.1832 1012 C.Explanation:
Let : a = b = 0.4 m ,
c = 0.36 m ,
= 3 N/C , and
= 2 N/(Cm2) .
The electric field throughout the region isdirected along the x-axis and the direction ofd ~A is perpendicular to its surface. Therefore,~E is parallel to d ~A over the four faces ofthe surface which are perpendicular to theyz plane, and ~E is perpendicular to d ~A overthe two faces which are parallel to the yzplane. That is, only the left and right sidesof the right rectangular parallel piped whichencloses the charge will contribute to the flux.The net electric flux through the cube is
=
right side
Ex dA
left side
Ex dA
= a b[+ (a+ c)2 a2]
= a b (2 a c+ c2)
= a b c (2 a+ c)
= (0.4 m) (0.4 m) (0.36 m)
[2 N/(Cm2)] [2 (0.4 m) + 0.36 m]= 0.133632 Nm2/C ,
so the enclosed charge is
q = 0
= [8.85419 1012 C2/(Nm2)]
(0.133632 Nm2/C)= 1.1832 1012 C .
030 (part 2 of 2) 10 pointsWhat is the sign of the charge enclosed in thesurface?
1. positive correct
2. negative
3. Cannot be determined
Explanation:Since there is more flux coming out of the
surface than going into the surface, the signof the enclosed charge must be positive.
Flux Through a Loop 0124:01, calculus, numeric, > 1 min, normal.
031 (part 1 of 1) 10 pointsA 40 cm diameter loop is rotated in a uniformelectric field until the position of maximumelectric flux is found. The flux in this positionis measured to be 520000 N m2/C.What is the electric field strength?
Correct answer: 4.13803 106 N/C.Explanation:
Let : r = 20 cm = 0.2 m and
= 520000 N m2/C .By Gauss law,
=
~E d~A
The position of maximum electric flux will bethat position in which the plane of the loop isperpendicular to the electric field; i.e., when~E d~A = E dA. Since the field is constant,
= E A = Epi r2
E =
pi r2
=520000 N m2/C
pi (0.2 m)2
= 4.13803 106 N/C .
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 15
keywords:
Three Point Charges 1725:01, trigonometry, multiple choice, > 1 min,normal.
032 (part 1 of 3) 10 pointsConsider three point charges at the vertices ofan equilateral triangle. Let the potential bezero at infinity.The value of the Coulomb constant is
8.98755 109 N m2/C2.
0.2m
60
1.5 C
3 C 5 CP
What is the electrostatic potential at thepoint P at the center of the base of the equi-lateral triangle given in the diagram?Correct answer: 101917 V.Explanation:
Let : q1 = 1.5 C = 1.5 106 C ,q2 = 3 C = 3 106 C ,q3 = 5 C = 5 106 C ,a = 0.2 m , and
ke = 8.98755 109 N m2/C2 .The potential at P is given by
V = kei
qiri.
From the sketch below, the height h is givenby
h =
a2
(a2
)2=
3
2a .
Notice that q1 > 0,q2 > 0, and q3 < 0.
VP = ke
(q1h+
q2a/2
+q3a/2
)
=2 kea
(q13+ q2 + q3
)
=2(8.98755 109 N m2/C2)
0.2 m
(1.5 106 C
3+ 3 106 C
5 106 C)= 101917 V .
033 (part 2 of 3) 10 pointsWhat is the vertical component of the electricforce on the 1.5 C charge due to the 3 Ccharge?
1. F =ke (1.5 C) (3 C)
(0.2 m)2cot 30
2. F =ke (1.5 C) (3 C)
(0.2 m)2cot 60
3. F =ke (1.5 C) (3 C)
(0.2 m)2cos 30 correct
4. F =ke (1.5 C) (3 C)
(0.2 m)2cos 60
5. F =ke (1.5 C) (3 C)
(0.2 m)2tan 30
6. F =ke (1.5 C) (3 C)
(0.2 m)2tan 60
7. F =ke (1.5 C) (3 C)
(0.2 m)2
8. F =ke (1.5 C) (3 C)
(0.2 m)2sin 45
9. F =ke (1.5 C) (3 C)
(0.2 m)2tan 45
10. F =ke (1.5 C) (3 C)
(0.2 m)2cot 45
Explanation:
Fv = F cos
=ke q1 q2
r2cos
=ke (1.5 C) (3 C)
(0.2 m)2cos 30
034 (part 3 of 3) 10 pointsFind the total electrostatic energy of the sys-tem, again with the zero reference at infinity.Correct answer: 0.80888 J.
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 16
Explanation:The total electrostatic energy of the sys-
tem is the sum of the electrostatic energiesbetween each pair of charges:
U = U12 + U23 + U31
The electrostatic energy between the chargesqi and qj is given by
Uij =qi qj
4pi 0 r
where r is the distance between the charges,
so, since ke =1
4pi 0,
U = ke a[q1 q2 + q2 q3 + q3 q1
]=(8.98755 109 N m2/C2) (0.2 m)
[(1.5 106 C) (3 106 C)
+ (3 106 C) (5 106 C)+ (5 106 C) (1.5 106 C)
]= 0.80888 J .
keywords:
Moving a Charge25:02, trigonometry, numeric, > 1 min, nor-mal.
035 (part 1 of 1) 10 pointsIt takes 120 J of work to move 1 C of chargefrom a positive plate to a negative plate.What voltage difference exists between the
plates?Correct answer: 120 V.Explanation:
Let : W = 120 J and
q = 1 C .
The voltage difference is
V =W
q=
120 J
1 C= 120 V .
keywords:
AP B 1993 MC 7025:03, trigonometry, multiple choice, < 1 min,fixed.
036 (part 1 of 1) 10 pointsTwo negatively charged spheres with differentradii are shown in the figure below.
Q Q
The two conductors are now conneted by awire.
Which of the following occurs when the twospheres are connected with a conducting wire?
1. No charge flows.
2. Negative charge flows from the largersphere to the smaller sphere until the elec-tric field at the surface of each sphere is thesame.
3. Negative charge flows from the largersphere to the smaller sphere until the elec-tric potential of each sphere is the same.
4. Negative charge flows from the smallersphere to the larger sphere until the elec-tric field at the surface of each sphere is thesame.
5. Negative charge flows from the smallersphere to the larger sphere until the electricpotential of each sphere is the same. correct
Explanation:When the wire is connected, charge will flow
until each surface is at the same potential.When disconnected the potential of each
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 17
sphere is given by
V =ke q
r.
The smaller sphere is at a more negative po-tential than the larger sphere, so negativecharge will flow from the smaller sphere tothe large one until they are at the same po-tential.
keywords:
Equipotential Surfaces 0225:03, trigonometry, multiple choice, > 1 min,fixed.
037 (part 1 of 2) 10 pointsConsider the figure
+Q
#1+++++++++++
Q
#2
A
B
C Dx
y
Of the following elements, identify all thatcorrespond to an equipotential line or surface.
1. line AB only correct
2. line CD only
3. both AB and CD
4. neither AB nor CD
Explanation:Consider the electric field
+Q
#1
+++++++++++
Q
#2
A
B
C D x
y
An equipotential line or surface (AB) isnormal to the electric field lines.
038 (part 2 of 2) 10 pointsConsider the figure
A
q +B
+q
C
DOf the following elements, identify all that
correspond to an equipotential line or surface.
1. line AB only
2. line CD only correct
3. both AB and CD
4. neither AB nor CD
Explanation:Consider the electric field:
A
+
B
C
D
An equipotential line or surface (CD) isnormal to the electric field lines.
keywords:
Starting a Car 0325:04, trigonometry, numeric, > 1 min, nor-mal.
039 (part 1 of 1) 10 pointsThe gap between electrodes in a spark plugis 0.06 cm. To produce an electric sparkin a gasoline-air mixture, an electric field of3 106 V/m must be achieved.On starting a car, what is the magnitude of
the minimum voltage difference that must be
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 18
supplied by the ignition circuit?Correct answer: 1800 V.Explanation:
Let : E = 3 106 V/m andd = 0.06 cm = 0.0006 m .
Assuming the electric field between the twoelectrodes is constant, then the potential dif-ference between the electrodes is
V = E d
=(3 106 V/m) (0.0006 m)
= 1800 V .
keywords:
Accelerating an Electron25:05, trigonometry, numeric, > 1 min, nor-mal.
040 (part 1 of 1) 10 pointsThrough what potential difference would anelectron need to be accelerated for it toachieve a speed of 4 % of the speed of light(2.99792 108 m/s), starting from rest?Correct answer: 408.799 V.Explanation:
Let : s = 4% = 0.04 ,
c = 2.99792 108 m/s ,me = 9.10939 1031 kg , andqe = 1.60218 1019 C .
The speed of the electron is
v = 0.04 c
= 0.04(2.99792 108 m/s)
= 1.19917 107 m/s ,By conservation of energy
1
2me v
2 = (qe)V
V = mev2
2 qe
=(9.10939 1031 kg)(1.19917 107 m/s)22 (1.60218 1019 C)
= 408.799 V .
keywords:
Point Charge25:05, trigonometry, numeric, > 1 min, nor-mal.
041 (part 1 of 1) 10 pointsAt distance r from a point charge q, the elec-tric potential is 600 V and the magnitude ofthe electric field is 200 N/C.Determine the value of q.
Correct answer: 2.00277 107 C.Explanation:
Let : ke = 8.98755 109 N m2/C2 ,V = 600 V , and
e = 200 N/C .
E =ke q
r2and V =
ke q
r, so that
V
E= r.
The potential is
V =ke q
r=
ke qVE
=ke q E
V
q =V 2
keE
=(600 V)2
(8.98755 109 N m2/C2) (200 N/C)= 2.00277 107 C .
keywords:
Conducting Spheres 0225:09, trigonometry, multiple choice, > 1 min,wording-variable.
042 (part 1 of 4) 10 pointsConsider two solid conducting spheres withradii r1 = 4R and r2 = 3R ; i.e.,
r2r1
=3R4R =
3
4.
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 19
The two spheres are separated by a largedistance so that the field and the potential atthe surface of sphere #1 only depends on thecharge on #1 and the corresponding quan-tities on #2 only depend on the charge on#2.Place an equal amount of charge on both
spheres, q1 = q2 = Q .
r1
q1#1
r2
q2 #2
After the electrostatic equilibrium on eachsphere has been established, what is the ratio
of the potentialsV2V1
at the centers of the
two solid conducting spheres?
1.V2V1
=4
3correct
2.V2V1
=3
4
3.V2V1
=3
2
4.V2V1
=3
8
5.V2V1
=16
9
6.V2V1
=9
16
7.V2V1
=9
8
8.V2V1
=9
32
9.V2V1
= 1
Explanation:For a solid conducting sphere, the charge is
uniformly distributed at the surface. FromGauss Law, the electric field outside the
sphere is given by E(r) = kQ
r2, where Q
is the total charge on the sphere and r is thedistance from the center of the sphere. By in-tegration with respect to r, the potential can
be expressed as V (r) = kQ
r, so the potential
at the surface of the sphere is
V (r) = kQ
r, (1)
where R is radius of the sphere and r R .For the electrostatic case, the potential is
constant throughout a conducting body, sothe potential at the center is the same asanywhere on the conductor.Thus at two centers
V2V1
=kq2r2
kq1r1
=r1r2
=4R3R
=4
3.
043 (part 2 of 4) 10 points
What is the ratio of the electric fieldsE2E1
at
the surfaces of the two spheres?
1.E2E1
=16
9correct
2.E2E1
=9
16
3.E2E1
=9
8
4.E2E1
=9
32
5.E2E1
=4
3
6.E2E1
=3
4
7.E2E1
=3
2
8.E2E1
=3
8
9.E2E1
= 1
Explanation:
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 20
For a conducting sphere, the charge is uni-formly distributed at the surface. Based onGauss law, the electric field on the surface ofa conducting sphere of radius R with chargeQ is
E(r) = keQ
r2, where r R . (2)
Thus on the surface r = R of the twospheres,
E2E1
=
kq2r22
kq1r21
=
(r1r2
)2
=
(4R3R
)2
=
(4
3
)2
=16
9.
044 (part 3 of 4) 10 pointsNow connect the two spheres with a wire.
r1
q1#1
r2
q2 #2
There will be a flow of charge through thewire until equilibrium is established.
What is the ratio of the electric fieldsE2E1
at
the surfaces of the two spheres?
1.E2E1
=4
3correct
2.E2E1
=3
4
3.E2E1
=3
2
4.E2E1
=3
8
5.E2E1
=16
9
6.E2E1
=9
16
7.E2E1
=9
8
8.E2E1
=9
32
9.E2E1
= 1
Explanation:When the spheres are connected by a wire,
charge will flow from one to the other untilthe potential on both spheres is the same.
As noted,V2V1
= 1, defines equilibrium.
The spheres are connected by a wire and nocurrent is flowing (at equilibrium), thereforethe ends of the wire are at the same potential
V2 = V1 . (3)
For a conducting sphere, the charge is uni-formly distributed at the surface. Based onGauss law, on the surface of a conductingsphere of radius R with charge Q is
E(r) = keQ
r2, where r R , and
V (r) = kQ
r, where r R .
Thus on the surface r = R of the twospheres,
E2E1
=
kq2r22
kq1r21
(4)
=kq2r2
1
r2
kq1r1
1
r1
=V2
1
r2
V11
r1
, since V1 = V2
=r1r2
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 21
=4R3R
=4
3.
045 (part 4 of 4) 10 pointsNow, what is the charge q1 on sphere #1?
1. q1 =8
7Q correct
2. q1 =6
7Q
3. q1 =7
8Q
4. q1 =7
6Q
5. q1 =4
7Q
6. q1 =3
7Q
7. q1 =7
4Q
8. q1 =7
3Q
9. q1 = Q
Explanation:When the spheres are connected by a wire,
charge will flow from one to the other untilthe potential on both spheres is the same.In this case, this implies that
keq1r1
= keq2r2
, or
q2 =r2r1
q1
=3R4R q1
=3
4q1 . (5)
The total charge of the system remains con-stant; i.e., from the initial condition q1 =q2 = Q, the total change on both spheres isq1 + q2 = 2Q. Using q2 from Eq. 5, we have
q1 + q2 = 2Q
q1 +3
4q1 = 2Q
7
4q1 = 2Q
q1 =8
7Q .
And the charge on sphere # 2 is q2 =6
7Q ,
since q1 + q2 =8
7Q+
6
7Q = 2Q .
Check Eq. 4: On the surfaces of the twospheres,
E2E1
=
(q2q1
)(r1r2
)2
=
6
7Q
8
7Q
(4R
3R)2
=
(3
4
)(4
3
)2
=4
3.
Third of eighteen versions.
keywords:
Change in Potential e225:04, calculus, multiple choice, < 1 min, nor-mal.
046 (part 1 of 1) 10 pointsA uniform electric field of magnitude 250 V/mis directed in the positive x-direction. Sup-pose a 12 C charge moves from the origin topoint A at the coordinates, (20 cm, 50 cm).
x
y 250 V/m
O
A
(20 cm, 50 cm)
What is the absolute value of the change inpotential from the origin to point A?Correct answer: 50 V.Explanation:
Let : x = 20 cm ,
y = 50 cm , and
~E = 250 V/m .
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 22
The potential difference from O to A isdefined as
V = VA VO = AO
~E d~s .
We know that ~E = (250 V/m) . We needto choose a path to integrate along. Becausethe electric force is conservative, it doesntmatter which path we take; they all give thesame answer. There are two choices of pathfor which the math is simple (see the figurebelow.)
x
yE
O
A(x, y)
I
I
B
II
Path I:
VA VO = (VA VB) + (VB VO),
From O to B, ~E and d~s are both along thex-axis, so ~E d~s = E dx. From B to A, ~E andd~s are perpendicular, so ~E d~s = 0.
VA VO = BO
~E d~s AB
~E d~s
= x
0
E dx y
0
0 dy
= E xO
dx = Ex= (250 V/m) (0.2 m)= 50 V .
The absolute value is
|V | = 50 V .
Path II: In this case, ~E d~s = E cos ds .where cos =
x
l x = l cos .
VA VO = E cos lOds
= E l cos = E x .
which is the same as the result for the otherpath.
keywords:
Potential Diagrams 0225:04, calculus, multiple choice, > 1 min,wording-variable.
047 (part 1 of 4) 10 pointsConsider a sphere with radius R and charge
Q
Q
and the following graphs:
Q.
rR0
1r
G.
rR0
1r
X .
rR0
1r2
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 23
P.
rR0
1r
Z.
rR0
1r
M.
rR0
1r2
1r
Y.
rR0
1r2
S.
rR0
1r2
L.
rR0
1r2
Which diagram describes the electric fieldvs radial distance [E(r) function] for a con-ducting sphere?
1. Y correct
2. S
3. L
4. X
5. Z
6. G
7. Q
8. P
9.MExplanation:The electric field for R < r with
the sphere conducting and/or uniformlynon-conducting: Because the charge distri-bution is spherically symmetric, we select aspherical gaussian surface of radius R < r,concentric with the conducting sphere. Theelectric field due to the conducting sphere isdirected radially outward by symmetry and istherefore normal to the surface at every point.Thus, ~E is parallel to d~A at each point. There-fore ~E d~A = E dA and Gausss law, where Eis constant everywhere on the surface, gives
E =
~E d~A
=
E dA
= E
dA
= E(4pi r2
)=
qin0
,
where we have used the fact that the surfacearea of a sphere A = 4pi r2 . Now, we solve forthe electric field
E =qin
4pi 0 r2
=Q
4pi 0 r2, where R < r . (1)
This is the familiar electric field due to a pointcharge that was used to develop Coulombslaw.
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 24
The electric field for r < R with thesphere conducting: In the region insidethe conducting sphere, we select a sphericalgaussian surface r < R, concentric with theconducting sphere. To apply Gausss lawin this situation, we realize that there is nocharge within the gaussian surface (qin = 0),which implies that
E = 0 , where r < R . (2)
Y.
rR0
E 1r2
E
048 (part 2 of 4) 10 pointsWhich diagram describes the electric field vsradial distance [E(r) function] for a uniformlycharged non-conducting sphere?
1. S correct
2. L
3. X
4. Z
5. G
6. Q
7. P
8. Y
9.MExplanation:The electric field for R < r with
the sphere conducting and/or uniformlynon-conducting: In the region outside theuniformly charged non-conducting sphere, wehave the same conditions as for the conduct-ing sphere when applying Gausss law, so
E =Q
4pi 0 r2, where R < r , (1)
as in Part 1.The electric field for r < R with the
sphere uniformly non-conducting: Inthis case we select a spherical gaussian sur-face at a radius r where r < R, concentricwith the uniformly charged non-conductingsphere. Let us denote the volume of thissphere by V . To apply Gausss law in thissituation, it is important to recognize that thecharge qin within the gaussian surface of thevolume V is less than Q. Using the volume
charge density QV
, we calculate qin :
qin = V
=
(4
3pi r3
).
By symmetry, the magnitude of the electricfield is constant everywhere on the sphericalgaussian surface and is normal to the surfaceat each point. Therefore, Gausss law in theregion r < R gives
E dA = E
dA
= E(4pi r2
)=
qin0
.
Solving for E gives
E =qin
4pi 0 r2
=4
3pi r3
4pi 0 r2
=
3 0r .
Because =Q
4
3pi R3
(by definition) and since
k =1
4pi 0, this expression for E can be writ-
ten as
E =Qr
4pi 0 R3
=k Q
R3r , where R < r . (3)
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 25
Note: This result forE differs from the one weobtained in the Part 3. It shows that E 0as r 0. Therefore, the result eliminates theproblem that would exist at r = 0 if E varied
as1
r2inside the sphere as it does outside the
sphere. That is, if E 1r2
for r < R, the field
would be infinite at r = 0, which is physicallyimpossible. Note: Also the expressions forParts 1 and 2 match when r = R.
S.
rR0
E 1r2
E
049 (part 3 of 4) 10 pointsWhich diagram describes the electric poten-tial vs radial distance [V (r) function] for aconducting sphere?
1. Z correct
2. G
3. Q
4. P
5. Y
6. S
7. L
8. X
9.MExplanation:The electric potential for R < r with
the sphere conducting and/or uniformlynon-conducting: In the previous parts wefound that the magnitude of the electric fieldoutside a charged sphere of radius R is
E = kQ
r2, where R < r ,
where the field is directed radially outward
when Q is positive.In this case, to obtain the electric potential
at an exterior point, we use the definition forelectric potential:
V = r
E dr
= k Q r
dr
r2
= kQ
r, where R < r . (4)
Note: This result is identical to the expressionfor the electric potential due to a point charge.The electric potential for r < R with
the sphere conducting: In the region insidethe conducting sphere, the electric field E =0 . Therefore the electric potential everywhereinside the conducting sphere is constant; thatis
V = V (R) = constant , where R < r .
(5)
Z.
rR0
V 1r
V
050 (part 4 of 4) 10 pointsWhich diagram describes the electric poten-tial vs radial distance [V (r) function] for auniformly charged non-conducting sphere?
1. G correct
2. Q
3. P
4. Y
5. S
6. L
7. X
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 26
8. Z
9.MExplanation:The electric potential for R < r with
the sphere conducting and/or uniformlynon-conducting: In the region outside theuniformly charged non-conducting sphere, wehave the same conditions as for the conduct-ing sphere when applying the definition forthe electric potential; therefore,
V = r
E dr
= k Q r
dr
r2
= kQ
r, where R < r . (4)
The electric potential for r < R withthe sphere uniformly non-conducting:Because the potential must be continuous atr = R , we can use this expression to obtainthe potential at the surface of the sphere; i.e.,the potential at a point on the conducting
sphere is V = kQ
rFrom Part 2 we found that the electric field
inside an uniformly charged non-conductingsphere is
E =k Q
R3r , where r < R . (6)
We can use this result in the definition forthe electric potential to evaluate the potential
difference V = VrVR (where VR = k QR
as
shown in Eq. 4) at some interior point of thesphere, so
Vr = VR +V
= kQ
R rRE dr
= kQ
R k Q
R3
rRr dr , from Eq. 6
= k2Q
2R+ k
Q
2R3(R2 r2)
= k3Q
2R k Q
2R3r2
= kQ
2R
(3 r2) , where r < R .
G.
rR0
V 1rV
keywords:
Finding Zero Potential25:06, trigonometry, multiple choice, < 1 min,fixed.
051 (part 1 of 4) 10 pointsAll of the charges shown are of equal magni-tude.
q +q
a a
(a)
What is the electric potential E at the ori-gin? Assume zero potential at infinity.
1. zero correct
2. positive
3. negative
4. Cannot be determined
Explanation:We know that the potential due to a collec-
tion of N point charges is given by
V =1
4pi 0
Ni=1
qiri
=1
4pi 0
(q
a+qa
)= 0
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 27
052 (part 2 of 4) 10 points
q +q
a a
+q
2a
(b)
What is the electric potential E at the ori-gin?
1. zero
2. positive correct
3. negative
4. Cannot be determined
Explanation:
V =1
4pi 0
(qa
+q
a+
q
2 a
)> 0
053 (part 3 of 4) 10 points
q
q
2a
2a
+q
a
(c)
What is the electric potential E at the ori-gin?
1. zero correct
2. positive
3. negative
4. Cannot be determined
Explanation:
V =1
4pi 0
(q2 a
+q2 a
+q
a
)= 0
054 (part 4 of 4) 10 points
+q
+q
2a
2a
qa
(d)
What is the electric potential E at the ori-gin?
1. zero correct
2. positive
3. negative
4. Cannot be determined
Explanation:
V =1
4pi 0
(qa
+q
2 a+
q
2 a
)= 0 .
keywords:
Charge on a Capacitor26:01, trigonometry, numeric, > 1 min, nor-mal.
055 (part 1 of 1) 10 points
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 28
A 15 pF capacitor is connected across a 75 Vsource.What charge is stored on it?
Correct answer: 1.125 109 C.Explanation:
Let : C = 15 pF = 1.5 1011 F andV = 75 V .
The capacitance is
C =q
Vq = C V
= (1.5 1011 F) (75 V)= 1.125 109 C
keywords:
Capacitance Comparison 0226:02, trigonometry, multiple choice, > 1 min,fixed.
056 (part 1 of 1) 10 pointsA parallel plate capacitor is connected to abattery.
+Q Q
d
2 d
If we double the plate separation,
1. the capacitance is doubled.
2. the electric field is doubled.
3. the potential difference is halved.
4. the charge on each plate is halved. cor-rect
5. None of these.
Explanation:The capacitance of a parallel plate capaci-
tor is
C = 0A
d.
Hence doubling d halves the capacitance,and Q = C V is also halved(
C = 0A
2 d=
1
20
A
d=
1
2C
).
keywords:
Plate Separation26:02, trigonometry, numeric, > 1 min, nor-mal.
057 (part 1 of 1) 10 pointsA parallel-plate capacitor has a plate area of12 cm2 and a capacitance of 7 pF.The permittivity of a vacuum is 8.85419
1012 C2/N m2.What is the plate separation?
Correct answer: 0.00151786 m.Explanation:
Let : A = 12 cm2 = 0.0012 m2 ,
C = 7 pF = 7 1012 F , and0 = 8.85419 1012 C2/N m2 .
C =0 A
d
d =0 A
C
=
(8.85419 1012 C2/N m2)
7 1012 F (0.0012 m2)
= 0.00151786 m .
keywords:
AP B 1993 MC 15 16
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 29
26:03, trigonometry, multiple choice, > 1 min,fixed.
058 (part 1 of 2) 10 pointsConsider the circuit
100 V
2 F
4 F
3 F
5 F
a b
c
What is the equivalent capacitance for thisnetwork?
1. Cequivalent =10
7F
2. Cequivalent =3
2F
3. Cequivalent =7
3F
4. Cequivalent = 7 F correct
5. Cequivalent = 14 F
Explanation:
EB
C1
C2
C3
C4
a b
c
Let : C1 = 2 F ,
C2 = 4 F ,
C3 = 3 F ,
C4 = 5 F , and
EB = 100 V .The equivalent capacitance of capacitors C1and C2 (parallel) is C12 = C1 + C2 = 6 F .C12 and C3 are in series, so
1
C123=
1
C12+
1
C3=
C3 + C12C12 C3
C123 =C12 C3
C3 + C12
=(6 F) (3 F)
6 F + 3 F= 2 F .
C123 and C4 are parallel, so
C = C4 + C123
= 7 F .
059 (part 2 of 2) 10 pointsWhat is the charge stored in the 5-F lower-right capacitor?
1. Q1 = 360 C
2. Q1 = 500 C correct
3. Q1 = 710 C
4. Q1 = 1, 100 C
5. Q1 = 1, 800 C
Explanation:
Let : C4 = 5 F and
EB = 100 V .
The charge stored in a capacitor is given byQ = C V , so,
Q4 = C4 V
= (5 F) (100 V)
= 500 C .
keywords:
Capacitor Circuit 0226:03, trigonometry, numeric, > 1 min, nor-mal.
060 (part 1 of 2) 10 pointsA capacitor network is shown below.
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 30
100V
15 F
15 F
15 F
15 F
9F
15F
15F
y
z
What is the equivalent capacitance betweenpoints y and z of the entire capacitor net-work?Correct answer: 14.4545 F.Explanation:
Let : Ca = C = 15 F ,
Cb = C = 15 F ,
Cc = C = 15 F ,
Cd = C = 15 F ,
Ce = C = 15 F ,
Cf = C = 15 F ,
Cx = 9 F = 9 106 F andEB = V = 100 V .
E R
Ca
Cf
Cb
Cd
Cx
Ce
Cc
y
z
For capacitors in series,
1
Cseries= 1
Ci
Vseries =
Vi ,
and the individual charges are the same.For parallel capacitors,
Cparallel =
Ci
Qparallel =
Qi ,
and the individual voltages are the same.
The capacitors Cb, Cc, and Cd are in series,so
1
Cbcd=
1
C+
1
C+
1
C=
3
C
Cbcd =1
3C .
This reduces the circuit to
E R
Ca
Cf
Cx
Ce
Cbcd
y
z
The capacitors Ce and Cbcd are parallel, so
Cbcde = C + Cbcd = C +1
3C =
4
3C .
This reduces the circuit to
E R
Ca
CfCx
Cbcde
y
z
The capacitors Ca, Cbcde and Cf are in series,so
1
Cabcdef=
1
C+
3
4C+
1
C=
11
4C
Cabcdef =4
11C .
This reduces the circuit to
E R Cx
Cabcdef
y
zThese capacitors are parallel, so
Cyz = Cx + Cabcdef
= Cx +4
11C
= 9 F +4
11(15 F)
= 14.4545 F .
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 31
061 (part 2 of 2) 10 pointsWhat is the charge on the 9 F capacitorcentered on the left directly between points yand z?Correct answer: 0.0009 C.Explanation:
C qV
q = Cx V
= (9 106 F) (100 V)= 0.0009 C .
keywords:
Capacitors in Series26:05, trigonometry, multiple choice, > 1 min,fixed.
062 (part 1 of 3) 10 pointsConsider the two cases shown below. In CaseOne two identical capacitors are connected toa battery with emf V . In Case Two, a di-electric slab with dielectric constant fills thegap of capacitor C2. Let C be the resultantcapacitance for Case One and C the resul-tant capacitance for Case Two.
Case One
V
C1 C2
Case Two
V
C1 C2
The ratioC 12C12
of the resultant capacitances is
1. None of these.
2.C 12C12
=2
1 + .
3.C 12C12
= .
4.C 12C12
=1 +
2.
5.C 12C12
=1 +
2.
6.C 12C12
=2
1 + . correct
Explanation:
Let : C1 = C2 = C and
C 2 = C2 = C ,
where is dielectric constant.V = constant. C1 and C2 are in series, so
1
C12=
1
C1+
1
C2=
C2 + C1C1 C2
C12 =C1 C2
C1 + C2.
For Case One,
C12 =C1 C2
C1 + C2=
C2
2C=
C
2.
For Case Two,
C 12 =C1 C
2
C1 + C 2=
C2
(1 + )C=
C
1 + .
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 32
ThereforeC 12C12
=2
1 + .
063 (part 2 of 3) 10 points
The ratioV 2V2
of potential differences across
capacitor C2 for the two cases is
1.V 2V2
=2
1 + .
2.V 2V2
= .
3.V 2V2
=2
1 + . correct
4.V 2V2
=1 +
2.
5.V 2V2
=1 +
2.
6. None of these.
Explanation:For Case One,
V2 =Q2C2
=V C12C2
=V
2.
For Case Two,
V 2 =Q2C 2
=V C 12C 2
=V C1+C
=V
1 + .
ThereforeV 2V2
=2
1 + .
064 (part 3 of 3) 10 points
The ratioU
Uof total energy stored in the
capacitors for the two cases is
1. None of these
2.U
U=
2
1 + .
3.U
U= .
4.U
U=
1 +
2.
5.U
U=
1 +
2.
6.U
U=
2
1 + . correct
Explanation:For Case One,
U =1
2C12 V
2 .
For Case Two,
U =1
2C 12 V
2 .
Therefore
U
U=
C 12C12
=2
1 + .
keywords:
Dielectric in a Capacitor 0126:05, trigonometry, multiple choice, > 1 min,wording-variable.
065 (part 1 of 1) 10 pointsa) An isolated capacitor has a dielectric slab
between its plates.b) The capacitor is charged by a battery.c) After the capacitor is charged, the battery
is removed.d) The dielectric slab is then moved half way
out of the capacitor.e) Finally, the dielectric is released and is set
free to move on its own.
The dielectric will
1. be pulled back into the capacitor. cor-rect
2. remain in place.
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 33
3. be pushed out of the capacitor.
Explanation:The capacitance of a capacitor with a di-
electric slab is
Cin = Cout , where > 1 .
NOTE
When the battery is removed, the chargeon the plates of the capacitor will remainconstant. Charge is neither created nor de-stroyed.
Uout =1
2
Q2
Cout, and
Uin =1
2
Q2
Cin
=1
2
Q2
Cout
=1
Uout , so
Uin < Uout ,
where Uout is with an air-filled gap and Uinis with a dielectric-filled gap. A system willmove to a position of lower potential energy.After the dielectric is moved half way out
of the capacitor, the potential energy storedin the capacitor will be larger than it wouldhave been with the dielectric left in place.Therefore, the dielectric will be pulled back
into the capacitor.
keywords:
Dipole in an External Field 026:08, calculus, multiple choice, > 1 min,fixed.
066 (part 1 of 1) 10 pointsA dipole (electrically neutral) is placed in anexternal field.
+
(a)
+
(b)
+
(c)
+
(d)
For which situation(s) shown above is thenet force on the dipole equal to zero?
1. (a) only
2. (c) only
3. (c) and (d) correct
4. (a) and (c)
5. (b) and (d)
6. (a) and (d)
7. (a), (b), and (c)
8. (b), (c), and (d)
9. Another combination
10. None of these
Explanation:Basic Concepts: Field patterns of pointcharge and parallel plates of infinite extent.The force on a charge in the electric field is
given by
~F = q ~E
and the torque is defined as
~T = ~r ~F
~E =kq
r2r
~E =
~Ei .
Symmetry of the configuration will causesome component of the electric field to bezero.
-
Version One Homework 1 Juyang Huang 24018 Jan 16, 2008 34
Gauss law states
S =
~E d~A = Q
0.
Solutions: The electric dipole consists oftwo equal and opposite charges separated bya distance. In either situation (c) or (d), theelectric field is uniform and parallel every-where. Thus, the electric force on one chargeis equal but opposite to that on another sothat the net force on the whole dipole is zero.By contrast, electric fields are nonuniform forsituations both (a) and (b).
keywords: