Electricity Homework Solutions

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 ×

10−19 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× 10−19 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 µC isreleased from rest when it is 50 cm from asecond particle of charge −20 µC.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 µC = 5× 10−5 C ,

d = 50 cm = 0.5 m ,

Q = −20 µC = −2× 10−5 C , and

ke = 8.9875× 109 .


The force exerted on the particle is

F = ke|q1| |q2|

r2= ma

‖~a‖ = ke‖~q‖ ‖~Q‖md2

= ke

∣

∣5× 10−5 C∣

∣

∣

∣−2× 10−5 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 Coulomb’s 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× 10−8 C.Explanation:

Let : L = 0.15 m ,

m = 0.03 kg , and

θ = 5◦ .

L

aθ

m m

q q

From the right triangle in the figure above,we see that

sin θ =a

L.

Therefore

a = L sin θ

= (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.

θ

θ

mg

F

T

eT sin θ

T cos θ

Because the sphere is in equilibrium, theresultant of the forces in the horizontal andvertical directions must separately add up tozero:

∑

Fx = T sin θ − Fe = 0∑

Fy = T cos θ −mg = 0 .

From the second equation in the system

above, we see that T =mg

cos θ, so T can be

eliminated from the first equation if we makethis substitution. This gives a value

Fe = mg tan θ

= (0.03 kg)(

9.8 m/s2)

tan(5◦)

= 0.0257217 N ,

for the electric force.From Coulomb’s law, the electric force be-

tween the charges has magnitude

|Fe| = ke|q|2r2

,


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| r2

ke

=

√

(0.0257217 N) (0.0261467 m)2

(8.98755× 109 Nm2/C2)

= 4.4233× 10−8 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× 10−5 N.Explanation:

Let : q1 = 5 nC = 5× 10−9 C ,

q2 = 6 nC = 6× 10−9 C ,

q3 = −3 nC = −3× 10−9 C ,

r1,2 = 0.3 m ,

r1,3 = 0.1 m , and

kC = 8.98755× 109 N ·m2/C2 .

F1,2

F1,3F

θ

The repulsive force

F1,2 = kCq1 q2r21,2

= 8.98755× 109 N ·m2/C2

×(

5× 10−9 C) (

6× 10−9 C)

(0.3 m)2

= 2.99585× 10−6 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× 10−9 C) (

−3× 10−9 C)

(0.1 m)2

= −1.34813× 10−5 N

acts along the negative y-axis.Thus

Fnet =[

(2.99585× 10−6 N)2

+(−1.34813× 10−5 N)2]1/2

= 1.38102× 10−5 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:


tan θ =F1,3

F1,2

θ = tan−1

(

F1,3

F1,2

)

= tan−1

(

1.34813× 10−5 N

2.99585× 10−6 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 ∝ 1

r2, 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


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.

QA

QB≈ −17

9= −1.88889

∣

∣

∣

∣

QA

QB

∣

∣

∣

∣

≈ 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 =a√2correct

6. d =q

Qa

7. d =q

Q

√2 a

8. d =q

Q

a

2

9. d =q

Q

a√2

Explanation:We have the force on charge q on the x axis

due to charge Q on the y axis

~F =1

4π ε0

q Q

r2r ,

where r =√

a2 + d2. So the x component ofthe force on q is

Fx =1

4π ε0

q Q

r2cos θ

=1

4π ε0

q Q

a2 + d2

d√a2 + d2

=1

4π ε0

q Qd

(a2 + d2)3/2.

For maximum x component of the force,∂ Fx∂d

= 0 is required. Therefore

∂Fx∂d

=q Q

4π ε0

a2 − 2 d2

(a2 + d2)5/2= 0

a2 − 2 d2 = 0

d =a√2

.

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 ∆θ islabeled ∆q.


x

y

−−−−−−−−−−−−−−−−−−

∆θ

θ rx

y

III

III IV

B

A

O

∆q is given by

1. None of these

2. ∆q = Q

3. ∆q =Q∆θ

2π

4. ∆q =2Q∆θ

π

5. ∆q =Q∆θ

πcorrect

6. ∆q =Q

2π

7. ∆q =2Q

π

8. ∆q =Q

π

9. ∆q = 2πQ

10. ∆q = πQ

Explanation:The angle of a semicircle is π, thus the

charge on a small segment with angle ∆θ is

∆q =Q∆θ

π.

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 θ

r2

3. ∆Ex =k |∆q| cos θ

r2correct

4. ∆Ex =k |∆q| cos θ

r

5. ∆Ex =k |∆q| sin θ

r

6. ∆Ex = k |∆q| r2

7. ∆Ex = k |∆q| (sin θ) r2

8. ∆Ex = k |∆q| (cos θ) r2

9. ∆Ex = k |∆q| (cos θ) r

10. ∆Ex = k |∆q| (sin θ) 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 θ =k |∆q|r2

cos θ .

014 (part 3 of 3) 10 pointsDetermine the magnitude of the electric fieldat O . The total charge is −7.5 µC, 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 µC ,

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 θ

r2

=k |Q|π r2

cos θ∆θ


Therefore, the magnitude of the electric fieldat the center is given by

E = Ex =

∫ π/2

−π/2

k |Q|π r2

cos θ dθ

=2 k |Q|π r2

.

For the above values, the magnitude is givenby

E =2(

8.98755× 109 N ·m2/C2)

|(−7.5 µC)|π (14 cm)2

= 2.18941× 106 N/C .

The direction is along negative x axis.

x

y

−−−−−−−−−−−−−−−−−−

∆θ

θ 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 pyramid’s 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,

Φ = ~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

Φ = 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 µC, −9 µC, 27 µC, and −84 µC.Calculate the net electric flux through the

submarine.Correct answer: −6.88954× 106 N ·m2/C.Explanation:

Let : q1 = 5 µC = 5× 10−6 C ,

q2 = −9 µC = −9× 10−6 C ,

q3 = 27 µC = 2.7× 10−5 C , and

q4 = −84 µC = −8.4× 10−5 C .

From Gauss’s Law:

Φ =q1 + q2 + q3 + q4

ε0

=(5× 10−6 C) + (−9× 10−6 C)

8.854× 10−12 C2/N ·m2

+(2.7× 10−5 C) + (−8.4× 10−5 C)

8.854× 10−12 C2/N ·m2

= −6.88954× 106 N ·m2/C .


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 µC/m3. (The volume of a cylinder withradius r and length ` is V = π r2 `.)The value of the Permittivity of free space

is 8.85419× 10−12 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 ,

ρ = 1 µC/m3

= 1× 10−6 C/m3 , and

ε0 = 8.85419× 10−12 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

Φs = 2π r `E ,

and the charge enclosed by the surface is

Qenc = π r2 ` ρ .

Using Gauss’ law,

Φs =Qenc

ε0

2π r `E =π r2 ` ρ

ε0.

Thus

E =ρ

2 ε0r

=

(

1× 10−6 C/m3)

(0.01 m)

2 (8.85419× 10−12 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 =4π R3

3, and

Asphere = 4π 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.


The total amount of flux flowing throughthe Gaussian surface is given by

1. Φ =Q

ε0.

2. Φ =Q

4 ε0.

3. Φ =Q

2 ε0.

4. Φ =Q

8 ε0. correct

5. Φ =2Q

ε0.

6. Φ =4Q

ε0.

Explanation:Basic Concept: Gauss’ Law.

Solution: For spherical symmetric case,

Φ = 4π r2 E

=Qencl

ε0.

Φ =Qencl

ε0

=Q

ε0

4π

3

(

R

2

)3

4π

3R3

=Q

8 ε0.


The magnitude of the electric field ‖~E‖ at R

2is given by

1. ‖~E‖ = k Q

2R2. correct

2. ‖~E‖ = k Q

R2.

3. ‖~E‖ = 2 k Q

R2.

4. ‖~E‖ = 2 k Q2

R2.

5. ‖~E‖ = k Q2

R2.

6. ‖~E‖ = k Q2

2R2.

Explanation:Gauss’s Law gives us

4π r2 E =Qencl

ε0

=Q

ε0

4

3π

(

R

2

)3

4

3π R3

=Q

8 ε0,

E =Q

4π

(

R

2

)2

8 ε0

=Q

4π ε0 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, q′2

R3, q′′2


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

q1√2 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 + q2√2 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,Φinside = 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

q′2 = −q1 .

This verifies that the charge on the innersurface 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:


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 10−6 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‖ = Q

ε0 A

2. ‖~E‖ = Q

2 ε0 Acorrect

3. ‖~E‖ = Q

4 ε0 A

4. ‖~E‖ = Q

4π ε0 a2

5. ‖~E‖ = Q

4π ε0 a

6. ‖~E‖ = 2 ε0 QA

7. ‖~E‖ = ε0 QA


8. ‖~E‖ = 4π ε0 a2 Q

9. ‖~E‖ = 4π ε0 aQ

10. ‖~E‖ = ε0 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 Gauss’Law states that

ΦTOTAL

= 2E S

=1

ε0

Q

AS , so

E =Q

2 ε0 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 λ and thecorresponding charge on the outer conduc-tor per unit length is −λ (same in magni-

tudes but with opposite signs) and λ > 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 =λ

2π ε0 r1correct

3. E =λ√

2π ε0 r1

4. E =λ√

3π ε0 r1

5. E =2λ√

3π ε0 r1

6. E =λR1

4π ε0 r12

7. E =λR1

3π ε0 r12

8. E =λ2 R1

4π ε0 r12

9. E =λ

2π ε0 R1

10. None of these.

Explanation:Pick a cylindrical Gaussian surface with the

radius r1 and apply the Gauss’s law; we obtain

E · ` · 2π r1 =Q

ε0

E =λ

2π ε0 r1.

026 (part 2 of 4) 10 pointsThe electric field vector points

1. in the negative r direction


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 =λ

2π ε0 r2

3. E =λ√

2π ε0 r2

4. E =λ√

3π ε0 r2

5. E =2λ√

3π ε0 r2

6. E =λR1

4π ε0 r22

7. E =λR1

3π ε0 r22

8. E =λ2 R1

4π ε0 r22

9. E =λ

2π ε0 R1

10. None of these.

Explanation:Pick a cylindrical Gaussian surface with the

radius r2 and apply the Gauss’s 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 = − 1

2π ε0

q

l

∫ R1

R2

dr

r

= − 1

2π ε0

q

lln r

∣

∣

∣

∣

R1

R2

=q

2π ε0 lln

R2

R1.

Since C =q

V, we obtain the capacitance

C =2π ε0 l

ln

(

R2

R1

)

=2π (8.85419× 10−12 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 = (α + β x2) ıwhere x is in meters, α = 3 N/C, andβ = 2 N/(Cm2).


Ey

x

z

a

c

b

a

What is the magnitude of the net chargeenclosed by the surface?Correct answer: 1.1832× 10−12 C.Explanation:

Let : a = b = 0.4 m ,

c = 0.36 m ,

α = 3 N/C , and

β = 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

∆Φ =

∫

right side

Ex dA⊥ −∫

left side

Ex dA⊥

= a b[

α+ β(a+ c)2 − α− β a2]

= a b β (2 a c+ c2)

= a b c β (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 = ε0 ∆Φ

= [8.85419× 10−12 C2/(Nm2)]

× (0.133632 Nm2/C)

= 1.1832× 10−12 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

Φ = 520000 N ·m2/C .

By Gauss’ law,

Φ =

∮

~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,

Φ = E A = Eπ r2

E =Φ

π r2

=520000 N ·m2/C

π (0.2 m)2

= 4.13803× 106 N/C .


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 µC

3 µC −5 µCP

ı

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 µC = 1.5× 10−6 C ,

q2 = 3 µC = 3× 10−6 C ,

q3 = −5 µC = −5× 10−6 C ,

a = 0.2 m , and

ke = 8.98755× 109 N ·m2/C2 .

The potential at P is given by

V = ke∑

i

qiri.

From the sketch below, the height h is givenby

h =

√

a2 −(a

2

)2

=

√3

2a .

Notice that q1 > 0,q2 > 0, and q3 < 0.

VP = ke

(

q1h+

q2a/2

+q3a/2

)

=2 kea

(

q1√3+ q2 + q3

)

=2(

8.98755× 109 N ·m2/C2)

0.2 m

×(

1.5× 10−6 C√3

+ 3× 10−6 C

−5× 10−6 C)

= −101917 V .

033 (part 2 of 3) 10 pointsWhat is the vertical component of the electricforce on the 1.5 µC charge due to the 3 µCcharge?

1. F =ke (1.5 µC) (3 µC)

(0.2 m)2cot 30◦

2. F =ke (1.5 µC) (3 µC)

(0.2 m)2cot 60◦

3. F =ke (1.5 µC) (3 µC)

(0.2 m)2cos 30◦ correct

4. F =ke (1.5 µC) (3 µC)

(0.2 m)2cos 60◦

5. F =ke (1.5 µC) (3 µC)

(0.2 m)2tan 30◦

6. F =ke (1.5 µC) (3 µC)

(0.2 m)2tan 60◦

7. F =ke (1.5 µC) (3 µC)

(0.2 m)2

8. F =ke (1.5 µC) (3 µC)

(0.2 m)2sin 45◦

9. F =ke (1.5 µC) (3 µC)

(0.2 m)2tan 45◦

10. F =ke (1.5 µC) (3 µC)

(0.2 m)2cot 45◦

Explanation:

Fv = F cosα

=ke q1 q2

r2cosα

=ke (1.5 µC) (3 µC)

(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.


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

4π ε0 r

where r is the distance between the charges,

so, since ke =1

4π ε0,

U = ke a[

q1 q2 + q2 q3 + q3 q1

]

=(

8.98755× 109 N ·m2/C2)

× (0.2 m)[

(1.5× 10−6 C) (3× 10−6 C)

+ (3× 10−6 C) (−5× 10−6 C)

+ (−5× 10−6 C) (1.5× 10−6 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


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


supplied by the ignition circuit?Correct answer: 1800 V.Explanation:

Let : E = 3× 106 V/m and

d = 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× 10−31 kg , and

qe = 1.60218× 10−19 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 energy1

2me v

2 = −(−qe)∆V

∆V = mev2

2 qe

=(

9.10939× 10−31 kg)

×(

1.19917× 107 m/s)2

2 (1.60218× 10−19 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× 10−7 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× 10−7 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.,

r2

r1=

3R4R =

3

4.


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 potentialsV2

V1at the “centers” of the

two solid conducting spheres?

1.V2

V1=

4

3correct

2.V2

V1=

3

4

3.V2

V1=

3

2

4.V2

V1=

3

8

5.V2

V1=

16

9

6.V2

V1=

9

16

7.V2

V1=

9

8

8.V2

V1=

9

32

9.V2

V1= 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

V2

V1=

kq2r2

kq1r1

=r1

r2

=4R3R

=4

3.


What is the ratio of the electric fieldsE2

E1at

the “surfaces” of the two spheres?

1.E2

E1=

16

9correct

2.E2

E1=

9

16

3.E2

E1=

9

8

4.E2

E1=

9

32

5.E2

E1=

4

3

6.E2

E1=

3

4

7.E2

E1=

3

2

8.E2

E1=

3

8

9.E2

E1= 1

Explanation:


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,

E2

E1=

kq2r22

kq1r21

=

(

r1

r2

)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 fieldsE2

E1at

the “surfaces” of the two spheres?

1.E2

E1=

4

3correct

2.E2

E1=

3

4

3.E2

E1=

3

2

4.E2

E1=

3

8

5.E2

E1=

16

9

6.E2

E1=

9

16

7.E2

E1=

9

8

8.E2

E1=

9

32

9.E2

E1= 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,V2

V1= 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,

E2

E1=

kq2r22

kq1r21

(4)

=kq2r2

1

r2

kq1r1

1

r1

=V2

1

r2

V11

r1

, since V1 = V2

=r1

r2


=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 =r2

r1q1

=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,

E2

E1=

(

q2q1

)(

r1

r2

)2

=

6

7Q

8

7Q

(

4R3R

)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 µC 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 .


The potential difference from O to A isdefined as

∆V = VA − VO = −∫ A

O

~E · d~s .

We know that ~E = (250 V/m) ı . We needto choose a path to integrate along. Becausethe electric force is conservative, it doesn’tmatter 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 = −∫ B

O

~E · d~s−∫ A

B

~E · d~s

= −∫ x

0

E dx−∫ y

0

0 dy

= −E∫ x

Odx = −E∆x

= −(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 θ ds .

where cos θ =x

l⇒ x = l cos θ .

VA − VO = −E cos θ

∫ l

Ods

= −E l cos θ

= −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

∝ 1

r

G.

rR0

∝ 1

r

X .

rR0

∝ 1

r2


P.

rR0

∝ 1

r

Z.

rR0

∝ 1

r

M.

rR0

∝ 1

r2

∝ 1

r

Y.

rR0

∝ 1

r2

S.

rR0

∝ 1

r2

L.

rR0

∝ 1

r2

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 Gauss’s law, where Eis constant everywhere on the surface, gives

ΦE =

∮

~E · d~A

=

∮

E dA

= E

∮

dA

= E(

4π r2)

=qinε0

,

where we have used the fact that the surfacearea of a sphere A = 4π r2 . Now, we solve forthe electric field

E =qin

4π ε0 r2

=Q

4π ε0 r2, where R < r . (1)

This is the familiar electric field due to a pointcharge that was used to develop Coulomb’slaw.


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 Gauss’s 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 ∝ 1

r2

E

048 (part 2 of 4) 10 pointsWhich diagram describes the electric field vs

radial distance [E(r) function] for a uniformly

charged 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 Gauss’s law, so

E =Q

4π ε0 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 Gauss’s 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 ρ ≡ Q

V, we calculate qin :

qin = ρ V ′

= ρ

(

4

3π r3

)

.

By symmetry, the magnitude of the electricfield is constant everywhere on the sphericalgaussian surface and is normal to the surfaceat each point. Therefore, Gauss’s law in theregion r < R gives

∮

E dA = E

∮

dA

= E(

4π r2)

=qinε0

.

Solving for E gives

E =qin

4π ε0 r2

=ρ4

3π r3

4π ε0 r2

=ρ

3 ε0r .

Because ρ =Q

4

3π R3

(by definition) and since

k =1

4π ε0, this expression for E can be writ-

ten as

E =Qr

4π ε0 R3

=k Q

R3r , where R < r . (3)


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 ∝ 1

r2for 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 ∝ 1

r2

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 ∝ 1

r

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


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 = Vr−VR (where VR = kQ

Ras

shown in Eq. 4) at some interior point of thesphere, so

Vr = VR +∆V

= kQ

R−∫ r

RE dr

= kQ

R− k

Q

R3

∫ r

Rr 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 ∝ 1

rV

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


Explanation:We know that the potential due to a collec-

tion of N point charges is given by

V =1

4π ε0

N∑

i=1

qiri

=1

4π ε0

(

q

a+−qa

)

= 0



−q +q

a a

+q

2a

(b)

What is the electric potential E at the ori-gin?

1. zero

2. positive correct

3. negative


Explanation:

V =1

4π ε0

(−qa

+q

a+

q

2 a

)

> 0


−q

−q

2a

2a

+q

a

(c)


1. zero correct

2. positive

3. negative


Explanation:

V =1

4π ε0

(−q2 a

+−q2 a

+q

a

)

= 0


+q

+q

2a

2a

−q

a

(d)


1. zero correct

2. positive

3. negative


Explanation:

V =1

4π ε0

(−qa

+q

2 a+

q

2 a

)

= 0 .

keywords:

Charge on a Capacitor26:01, trigonometry, numeric, > 1 min, nor-mal.



A 15 pF capacitor is connected across a 75 Vsource.What charge is stored on it?

Correct answer: 1.125× 10−9 C.Explanation:

Let : C = 15 pF = 1.5× 10−11 F and

V = 75 V .

The capacitance is

C =q

Vq = C V

= (1.5× 10−11 F) (75 V)

= 1.125× 10−9 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 = ε0A

d.

Hence doubling d halves the capacitance,and Q = C V is also halved

(

C ′ = ε0A

2 d=

1

2ε0

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 ×

10−12 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× 10−12 F , and

ε0 = 8.85419× 10−12 C2/N ·m2 .

C =ε0 A

d

d =ε0 A

C

=

(

8.85419× 10−12 C2/N ·m2)

7× 10−12 F

×(

0.0012 m2)

= 0.00151786 m .

keywords:

AP B 1993 MC 15 16


26:03, trigonometry, multiple choice, > 1 min,fixed.

058 (part 1 of 2) 10 pointsConsider the circuit

100 V

2 µF

4 µF

3 µF

5 µF

a b

c

What is the equivalent capacitance for thisnetwork?

1. Cequivalent =10

7µF

2. Cequivalent =3

2µF

3. Cequivalent =7

3µF

4. Cequivalent = 7 µF correct

5. Cequivalent = 14 µF

Explanation:

EB

C1

C2

C3

C4

a b

c

Let : C1 = 2 µF ,

C2 = 4 µF ,

C3 = 3 µF ,

C4 = 5 µF , and

EB = 100 V .

The equivalent capacitance of capacitors C1

and C2 (parallel) is C12 = C1 + C2 = 6 µF .C12 and C3 are in series, so

1

C123=

1

C12+

1

C3=

C3 + C12

C12 C3

C123 =C12 C3

C3 + C12

=(6 µF) (3 µF)

6 µF + 3 µF= 2 µF .

C123 and C4 are parallel, so

C = C4 + C123

= 7 µF .

059 (part 2 of 2) 10 pointsWhat is the charge stored in the 5-µF lower-right capacitor?

1. Q1 = 360 µC

2. Q1 = 500 µC correct

3. Q1 = 710 µC

4. Q1 = 1, 100 µC

5. Q1 = 1, 800 µC

Explanation:

Let : C4 = 5 µF and

EB = 100 V .

The charge stored in a capacitor is given byQ = C V , so,

Q4 = C4 V

= (5 µF) (100 V)

= 500 µC .

keywords:

Capacitor Circuit 0226:03, trigonometry, numeric, > 1 min, nor-mal.

060 (part 1 of 2) 10 pointsA capacitor network is shown below.


100V

15 µF

15 µF

15 µF

15 µF

9µF

15µF

15µF

y

z

What is the equivalent capacitance betweenpoints y and z of the entire capacitor net-work?Correct answer: 14.4545 µF.Explanation:

Let : Ca = C = 15 µF ,

Cb = C = 15 µF ,

Cc = C = 15 µF ,

Cd = C = 15 µF ,

Ce = C = 15 µF ,

Cf = C = 15 µF ,

Cx = 9 µF = 9× 10−6 F and

EB = 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 .


E R

Ca

Cf

Cx

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 .


E R Cx

Cabcdef

y

zThese capacitors are parallel, so

Cyz = Cx + Cabcdef

= Cx +4

11C

= 9 µF +4

11(15 µF)

= 14.4545 µF .


061 (part 2 of 2) 10 pointsWhat is the charge on the 9 µF capacitorcentered on the left directly between points yand z?Correct answer: 0.0009 C.Explanation:

C ≡ q

Vq = Cx V

= (9× 10−6 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 κ 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 C′2

κ

The ratioC ′

12

C12of the resultant capacitances is

1. None of these.

2.C ′

12

C12=

2

1 + κ.

3.C ′

12

C12= κ .

4.C ′

12

C12=

1 + κ

2κ.

5.C ′

12

C12=

1 + κ

2.

6.C ′

12

C12=

2κ

1 + κ. correct

Explanation:

Let : C1 = C2 = C and

C ′2 = κC2 = κC ,

where κ is dielectric constant.V = constant. C1 and C2 are in series, so

1

C12=

1

C1+

1

C2=

C2 + C1

C1 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

=κC2

(1 + κ)C=

κC

1 + κ.


ThereforeC ′

12

C12=

2κ

1 + κ.


The ratioV ′

2

V2of potential differences across

capacitor C2 for the two cases is

1.V ′

2

V2=

2κ

1 + κ.

2.V ′

2

V2= κ .

3.V ′

2

V2=

2

1 + κ. correct

4.V ′

2

V2=

1 + κ

2κ.

5.V ′

2

V2=

1 + κ

2.

6. None of these.

Explanation:For Case One,

V2 =Q2

C2=

V C12

C2=

V

2.

For Case Two,

V ′2 =

Q′2

C ′2

=V C ′

12

C ′2

=V κC

1+κ

κC=

V

1 + κ.

ThereforeV ′

2

V2=

2

1 + κ.


The ratioU ′

Uof total energy stored in the

capacitors for the two cases is

1. None of these

2.U ′

U=

2

1 + κ.

3.U ′

U= κ .

4.U ′

U=

1 + κ

2κ.

5.U ′

U=

1 + κ

2.

6.U ′

U=

2κ

1 + κ. correct

Explanation:For Case One,

U =1

2C12 V

2 .

For Case Two,

U ′ =1

2C ′

12 V2 .

Therefore

U ′

U=

C ′12

C12=

2κ

1 + κ.

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 κ

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.

κ κ

The dielectric will

1. be pulled back into the capacitor. cor-rect

2. remain in place.


3. be pushed out of the capacitor.

Explanation:The capacitance of a capacitor with a di-

electric slab is

Cin = κCout , where κ > 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

κCout

=1

κUout , so

Uin < Uout ,

where Uout is with an air-filled gap and Uin

is 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 =k∆q

r2r

~E =∑

∆~Ei .

Symmetry of the configuration will causesome component of the electric field to bezero.


Gauss’ law states

ΦS =

∮

~E · d~A =Q

ε0.

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:

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