hw 1 solutions

lui (el22284) – HW 1 – li – (58545) 1

This print-out should have 27 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 (part 1 of 3) 10.0 pointsThe dot or scalar product of two (3d) vec-

tors ~a = 〈a1, a2, a3〉 and ~b = 〈b1, b2, b3〉 isdefined as

~a •~b =3∑

i=1

aibi = a1b1 + a2b2 + a3b3 .

In 3d Euclidean space, the scalar producthas a geometric interpretation, given by

~a •~b =‖~a‖‖~b‖ cos θ ,

where θ is the angle between ~a and ~b and thenorm(length) of a vector ‖~a‖ is defined by

‖~a‖=√~a • ~a .

In words, the scalar product of two vectors canbe thought of as the product of the magnitude

of ~a with the magnitude of the projection of ~bonto the direction of ~a. It is used to calculatethe product of vector quantities when only theparallel components of each vector contribute(e.g., Work = Force • Displacement).

Let ~a = 〈11, 8.25, 0〉 and ~b =〈3.63, 7.425, 0〉.Calculate ~a •~b.

Your answer must be within ± 1.0%Correct answer: 101.186.

Explanation:

From the definition, this is simply calcu-

lated by

~a •~b = a1b1 + a2b2 + a3b3

= 11× 3.63 + 8.25× 7.425 + 0× 0

= 101.186 .

002 (part 2 of 3) 10.0 points

Determine θ, the angle between ~a and ~b. Ex-

press your answer in degrees.

Your answer must be within ± 1.0%

Correct answer: 27.0766.

Explanation:

From the geometric interpretation of the

scalar product,

θ = cos−1

(

~a •~b‖~a‖‖~b‖

)

= cos−1

~a •~b√

a21+ a2

2+ a2

3

√

b21+ b2

2+ b2

3

= cos−1

(

101.186

13.75× 8.26484

)

= 27.0766 .


If two vectors are perpendicular, what is their

scalar product?

Your answer must be within ± 1.0%

Correct answer: 0.

Explanation:From the geometric interpretation, one can

see that when θ = 90◦, the scalar product isequal to zero.

004 (part 1 of 3) 10.0 pointsRecall that, in a 2d plane polar coordinate

system, a point in space is described by its

lui (el22284) – HW 1 – li – (58545) 2

radial distance r from the origin and the angleφ between the +x-axis and a line drawn fromthe origin to the point. For example, the point(3, π/4) looks like:

The relations between cartesian (x, y) coor-dinates and plane polar (r, φ) are given by

x = r cosφ , y = r sinφ , x2 + y2 = r2 .

Question: A circle of radius R centered atthe origin is described in plane polar coordi-nates by r = R. What are the Cartesian (i.e.,(x,y)) coordinates of a point on the circle atsome arbitrary angle φ, expressed in terms ofpolar quantities? (Note: you will frequentlyparametrize a point on a circle in this way.)

1. (R cosφ,R sinφ) correct

2. (R sinφ,R cosφ)

3. (R, φ)

4. (x, y)

5. (x cosφ, y sinφ)

6. (x2 + y2, tan−1

(

y

x

)

)

Explanation:From the Cartesian/plane polar rela-

tions given above, the x-coordinate isR cosφ and the y-coordinate is R sinφ, so(R cosφ,R sinφ) is the correct choice.

005 (part 2 of 3) 10.0 pointsIn 3d, we may extend this idea to cylindricalcoordinates (r, φ, z), which you may think ofas a plane polar coordinate system with thez-dimension tacked on.

This coordinate system is useful when deal-ing with problems that have radial symmetryabout some central axis.Question: You are given a hollow cylinder

of radius R whose central axis is the z-axisand whose base rests on the xy-plane. Whatare the Cartesian coordinates of an arbitrarypoint on the surface of the cylinder in termsof cylindrical quantities?

1. (R, φ, z)

2. (R cosφ,R sinφ, z) correct

3. (x cosφ, y sinφ, z)

4. (x2 + y2, tan−1

(

y

x

)

, z)

5. (x, y, z)

6. (R sinφ,R cosφ, z)

Explanation:Since the cylindrical coordinate system is

just a plane polar system with the z-axis ap-pended to it, the correct answer is the same asin 1 above, but with the z coordinate added:(R cosφ,R sinφ, z)

006 (part 3 of 3) 10.0 pointsIn cases where we have radial symmetry aboutthe origin (usually called spherical symme-try), it is most useful to use spherical coor-dinates (r, θ, φ). In the spherical coordinatesystem, a point is defined by: its radial dis-tance from the origin r; the azimuthal angleφ, defined as the angle between the +x-axis

lui (el22284) – HW 1 – li – (58545) 3

and the projection on the xy-plane of the linedrawn from the origin to the point; the po-

lar angle θ, defined as the angle between the+z axis and the line drawn from the originto the point. Note that for the polar angle,0 ≤ θ ≤ π.

Question: You are given a solid sphere ofradius R centered at the origin. What are theCartesian coordinates of an arbitrary pointwithin the sphere in terms of spherical quan-tities. You will need to use the figure todetermine the x,y, and z coordinates in termsof their spherical counterparts.

1. (R sin θ cosφ,R sin θ sinφ,R cos θ)

2. (r sin θ cosφ, r sin θ sinφ, r cos θ) correct

3. (r cos θ cosφ, r cos θ sinφ, r sin θ)

4. (r sinφ cos θ, r sinφ sin θ, r cosφ)

5. (x, y, z)

6. (r sin θ sinφ, r sin θ cosφ, r cos θ)

Explanation:Examining the figure, the projection onto

the xy plane of the line drawn from the originto an arbitrary point is seen to be r sin θ. Thecomponent of this projection along the x-axisis r sin θ cosφ, and the projection along they-axis is r sin θ sinφ. Finally, the projectiononto the z-axis of the line drawn from the ori-gin to the arbirary point is seen to be r cos θ.Combining everything, the correct answer is(r sin θ cosφ, r sin θ sinφ, r cos θ).

007 (part 1 of 5) 10.0 pointsA planet is located at

~P = 〈−1× 1010, 8× 1010,−2× 1010〉.

A star is located at

~S = 〈6× 1010,−5× 1010, 3× 1010〉.

What is ~R, the vector pointing from the planetto the star?

1. ~R =〈60000000000,−130000000000, 40000000000〉

2. ~R =〈80000000000,−120000000000, 60000000000〉

3. ~R =〈100000000000,−130000000000, 30000000000〉

4. ~R = 〈7× 1010,−1.3× 1011, 5× 1010〉correct

5. ~R =〈110000000000,−110000000000, 50000000000〉

Explanation:This is a vector subtraction problem. To

find ~R, we subtract ~S − ~P by respective com-ponents:

Sx − Px = 6× 1010 − (−1× 1010) = 7× 1010

Sy − Py = −5× 1010 − (8× 1010) = −1.3× 1011

Sz − Pz = 3× 1010 − (−2× 1010) = 5× 1010

So

~R = ~S − ~P

= 〈7× 1010,−1.3× 1011, 5× 1010〉 .


What is∣

∣

∣

~R∣

∣

∣?

lui (el22284) – HW 1 – li – (58545) 4

Your answer must be within ± 1.0%Correct answer: 1.55885× 1011.

Explanation:

To find∣

∣

∣

~R∣

∣

∣, we use the Pythagorean theo-

rem.

∣

∣

∣

~R∣

∣

∣=√

(7× 1010)2 + (−1.3× 1011)2 + (5× 1010)2

=√

2.43× 1022

= 1.55885× 1011.

009 (part 3 of 5) 10.0 pointsFor the remaining three parts of this problem,you will find the components of R, the unitvector in the direction of ~R. Begin by findingRx.Your answer must be within ± 1.0%


Explanation:We simply divide ~Rx by the magnitude:

Rx =~rx∣

∣

∣

~R∣

∣

∣

=7× 1010

1.55885× 1011= 0.44905.


Find Ry.Your answer must be within ± 1.0%

Correct answer: −0.83395.

Explanation:We divide ~Ry by the magnitude:

Ry =~Ry∣

∣

∣

~R∣

∣

∣

=−1.3× 1011

1.55885× 1011= −0.83395.


Find Rz.Your answer must be within ± 1.0%


Explanation:We divide ~Rz by the magnitude:

Rz =~Rz∣

∣

∣

~R∣

∣

∣

=5× 1010

1.55885× 1011= 0.32075.

012 (part 1 of 2) 10.0 pointsA particle undergoes two displacements.

The first has a magnitude of 174 cm andmakes an angle of 114◦ with the positive xaxis. The resultant displacement has a mag-nitude of 150 cm and is directed at an angleof 37.5◦ to the positive x axis.Find the magnitude of the second displace-

ment.Your answer must be within ± 1.0%

Correct answer: 201.47 cm.

Explanation:

Let : R1 = 174 cm at θ1 = 114◦ and

R = 150 cm at θ = 37.5◦ .

r1

r2

R

R = R1 +R2

R2 = R−R1 , so

R2x = (150 cm) cos 37.5◦ − (174 cm) cos 114◦

= 189.775 cm , and

R2y = (150 cm) sin 37.5◦ − (174 cm) sin 114◦

= −67.6427 cm ,

and the magnitude is

|R2| =√

R22x +R2

2y

=√

(189.775 cm)2 + (−67.6427 cm)2

= 201.47 cm .


lui (el22284) – HW 1 – li – (58545) 5

What is the direction of the second displace-ment (with positive measured counterclock-wise from the x axis, and between the limitsof −180◦ and +180◦)?Your answer must be within ± 1.0%

Correct answer: −19.6179◦.

Explanation:

tan θ =

(

s2ys2x

)

θ = arctan

(

s2ys2x

)

= arctan

(−67.6427 cm

189.775 cm

)

= −19.6179◦ .

014 10.0 pointsVectors ~A, ~B, ~C, ~D, and ~E are shown inthe figure. For convenience, the tails of eachvector are arbitrarily located at (0,0).

y

−5 −3 −1 0 1 2 3 4 5

x

−5

−4

−3−2

−10

12

34

5

A BC

D

E

Identify

~R = −~A + ~B − ~C − ~D − ~E ,

1. None of these figures is correct.

2.

y

−5 −3 −1 0 1 2 3 4 5

x

−5−4

−3−2

−101

23

45 R

3.y

−5 −3 −1 0 1 2 3 4 5

x

−5

−4−3

−2

−10

12

34

5

R

4.

y

−5 −3 −1 0 1 2 3 4 5

x

−5

−4

−3−2

−10

12

34

5

R

correct

5.

lui (el22284) – HW 1 – li – (58545) 6

y

−5 −3 −1 0 1 2 3 4 5

x

−5−4

−3−2

−101

23

45

R

Explanation:

Let : (xa, ya) = (−2,−4) ,

(xb, yb) = (2,−4) ,

(xc, yc) = (1,−4) ,

(xd, yd) = (−3, 5) , and

(xe, ye) = (1,−3) .

xr = −(−2) + (2)− (1)− (−3)− (1) = 5 and

yr = −(−4) + (−4)− (−4)− (5)− (−3) = 2 ,

so (xr, yr) = (5, 2) .

y

−5 −3 −1 0 1 2 3 4 5

x

−5

−4

−3−2

−10

12

34

5 −A

B

−C

−D

−E

R

015 10.0 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

What is the ratio of the electric fieldstrength at point A to the electric fieldstrength at point B?

1.E

A

EB

=2

1

2.E

A

EB

=4

1correct

3.E

A

EB

=1

1

4.E

A

EB

=1

2

5.E

A

EB

=8

1

Explanation:

Let : rB= 2 r

A.

The electric field strength E ∝ 1

r2, so

EA

EB

=

1

r2A

1

r2B

=r2B

r2A

=(2 r)2

r2= 4 .

016 10.0 pointsAn electron in a region where there is anelectric field experiences a force of magnitude2.4× 10−16 N.What is the magnitude of the electric field

at the location of the electron? The charge onan electron is −1.602× 10−19 CYour answer must be within ± 1.0%

Correct answer: 1498.13 N/C.

Explanation:The field is defined to be the force per

unit charge experienced by a particle (so longas the particle has a charge small enoughthat it does not change the background fieldsignificantly).Thus, we have:

∣

∣

∣

~E∣

∣

∣=

∣

∣

∣

~F∣

∣

∣

q

=2.4× 10−16 N

1.602× 10−19 C= 1498.13 N/C

lui (el22284) – HW 1 – li – (58545) 7

017 (part 1 of 2) 10.0 pointsYou want to create an electric field

~E = 〈0, 3904 N/C, 0〉

at the origin. Find the y coordinate whereyou would need to place a proton, in order tocreate this field at the origin.The value of k is 8.98755× 109 N ·m2/C2

and the value of the proton charge is1.6× 10−19 C.Your answer must be within ± 1.0%

Correct answer: −6.06912× 10−7 m.

Explanation:The equation for the electric field, in gen-

eral, is

~E =1

4πǫ0

q

|~r|2r.

In this case, q = 1.6× 10−19 C for a proton,and

r = 〈0, 1, 0〉,since the field points in the positive y direc-tion. |~r| is the unknown we want to solve for.We can write

Ey = 3904 N/C =1

4πǫ0

1.6× 10−19 C

|~r|2

⇒ |~r|2 = 1

4πǫ0

1.6× 10−19 C

3904 N/C

⇒ y = ±√

1

4πǫ0

1.6× 10−19 C

3904 N/C

= ±6.06912× 10−7 m .

We choose the minus sign to place the pro-ton below the origin; this way the field willpoint upward like we want it to.

018 (part 2 of 2) 10.0 pointsInstead of a proton, where would you placean electron to produce the same field at theorigin?Your answer must be within ± 1.0%

Correct answer: 6.06912× 10−7 m.

Explanation:This is simple. The proton and electron

carry the same magnitude of charge, but theelectron’s charge is negative. The calculationfrom above will be exactly the same, but wechoose a positive sign at the end to place theelectron above the origin. This way the fieldwill still point upward as we want it to.

019 (part 1 of 3) 10.0 pointsThe following figure shows a particle with apositive charge of 2e at location A creatingan electric field ~E at point B. Any chargedparticle placed at location B experiences aforce ~F .

+A

B

~E

~F

Now, imagine a positive charge of 3e isplaced at point B. What is the value ofthe electric field at location B due to thecharge at A? (Remember that e representsthe fundamental unit of charge, 1.6 × 10−19

C. Also, rBA is the position vector pointingfrom point A to point B.)

1. ~E =1

4πǫ0

2e2

|~rBA| rBA

2. ~E =1

4πǫ0

6e2

|~rBA|2rBA

3. ~E =1

4πǫ0

2e

|~rBA|2rBA correct

4. ~E =1

4πǫ0

6e

|~rBA| rBA

5. ~E =1

4πǫ0

4e

|~rBA|2rBA

Explanation:The electric field at point B is not related

to the charge at point B, only the one atpoint A, so the answers with e2 are incorrect.Recalling the form of the electric field,

lui (el22284) – HW 1 – li – (58545) 8

~E =1

4πǫ0

q

|~r|2r,

we see that it has the distance squared in thedenominator. Therefore the correct choice is

~E =1

4πǫ0

2e

|~rBA|2rBA.

020 (part 2 of 3) 10.0 pointsWhat is the force on the charged particle atB?

1. ~F =1

4πǫ0

3e2

|~rBA| rBA

2. ~F =1

4πǫ0

3e

|~rBA|2rBA

3. ~F =1

4πǫ0

6e2

|~rBA|2rBA correct

4. ~F =1

4πǫ0

9e2

|~rBA|2rBA

5. ~F =1

4πǫ0

6e

|~rBA|2rBA

Explanation:Force involves both particles, and has the

general form

~F =1

4πǫ0

Q1Q2

|~r|2r,

where~r is the vector pointing between the twocharges. In this case, Q1 = 2e and Q2 = 3e,so we end up with

~F =1

4πǫ0

6e2

|~rBA|2rBA.

021 (part 3 of 3) 10.0 pointsThe positive particle at B is removed and anelectron is placed at location B.What is the magnitude of the force on the

electron?

1. F =1

4πǫ0

e

|~rBA|2

2. F =1

4πǫ0

2e2

|~rBA|2correct

3. F =1

4πǫ0

2e

|~rBA|

4. F =1

4πǫ0

3e2

|~rBA|2

5. F =1

4πǫ0

6e2

|~rBA|Explanation:

The magnitude of the force is just∣

∣

∣

~F∣

∣

∣. So

we just take the length of the force vector,which (following the explanation from part 2)is given by

~F =1

4πǫ0

−2e2

|~rBA|2rBA.

(Remember the electron has a negativecharge.) Taking the length, we get

∣

∣

∣

~F∣

∣

∣=

∣

∣

∣

∣

∣

1

4πǫ0

−2e2

|~rBA|2rBA

∣

∣

∣

∣

∣

F =1

4πǫ0

2e2

|~rBA|2∣

∣

∣rBA

∣

∣

∣

=1

4πǫ0

2e2

|~rBA|2(1)

=1

4πǫ0

2e2

|~rBA|2

022 10.0 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.

0.16m

3◦

0.04 kg 0.04 kg

Find the magnitude of the charge oneach sphere. The acceleration of gravity is9.8 m/s2 and the value of Coulomb’s constantis 8.98755× 109 N ·m2/C2 .

lui (el22284) – HW 1 – li – (58545) 9

Your answer must be within ± 1.0%Correct answer: 2.53204× 10−8 C.

Explanation:

Let : L = 0.16 m ,

m = 0.04 kg , and

θ = 3◦ .

L

aθ

m m

q q

From the right triangle,

sin θ =a

La = L sin θ = (0.16 m) sin 3◦

= 0.00837375 m .

The separation of the spheres is r = 2 a =0.0167475 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 .

Dividing,

F sin θ

F cos θ=

Fe

mgFe = mg tan θ

= (0.04 kg)(

9.8 m/s2)

tan 3◦

= 0.0205439 N .

From Coulomb’s law, the electric force be-tween the charges has magnitude

|Fe| = ke|q|2r2

|q| =√

|Fe| r2ke

=

√

(0.0205439 N) (0.0167475 m)2

(8.98755× 109 N ·m2/C2)

= 2.53204× 10−8 C .


In the region shown in the diagram belowthere is an electric field due to a point chargelocated at the center of the circle. The arrowsindicate the magnitude and direction of theelectric field at the locations shown.

A

B

D

C

What is the sign of the source charge? Youranswer should either be a plus sign or a minussign.Your answer must be within ± 1.0%

Correct answer: -.

Explanation:Electric fields point away from positive

charges and toward negative charges.The fields are pointing toward the circle, so

there should be a net negative charge insidethe circle.

lui (el22284) – HW 1 – li – (58545) 10

024 (part 2 of 5) 10.0 pointsNow, a particle whose charge is −6× 10−9 Cis placed at location A.Which arrow best indicates the direction of

the electric force on this charge? Enter ’j’ ifthe force is the zero vector.

ab

c

de

f

g

h

Your answer must be within ± 1.0%Correct answer: b.

Explanation:Like charges repel, so the direction of the

force on a negatively charged particle at pointA should be away from the circle, and thus, indirection b.

025 (part 3 of 5) 10.0 pointsThe electric field at location A has the value〈−6000 N/C,−6000 N/C, 0〉.The unit vector in the direction of ~E can be

written in the form:

E = 〈Ex, Ey, Ez〉

where we understand Ez = 0.What is Ex?Your answer must be within ± 1.0%


Explanation:For any vector, the unit normal can be

found by taking the vector and dividing it byits magnitude, since

(

~v

|~v|

)

·(

~v

|~v|

)

=~v · ~v|~v|2

=|~v|2

|~v|2= 1

We can easily find the magnitude of ~E:

∣

∣

∣

~E∣

∣

∣=√

E2x +E2

y

=√

(−6000 N/C)2 + (−6000 N/C)2

= 8485.28 N/C

So, finally, we find

Ex =Ex∣

∣

∣

~E∣

∣

∣

=−6000 N/C

8485.28 N/C= −0.707107


What is Ey?Your answer must be within ± 1.0%


Explanation:For any vector, the unit normal can be

found by taking the vector and dividing it byits magnitude, since

(

~v

|~v|

)

·(

~v

|~v|

)

=~v · ~v|~v|2

=|~v|2

|~v|2= 1

We can easily find the magnitude of ~E:

∣

∣

∣

~E∣

∣

∣=√

E2x +E2

y

=√

(−6000 N/C)2 + (−6000 N/C)2

= 8485.28 N/C

So, finally, we find

Ey =Ey∣

∣

∣

~E∣

∣

∣

=−6000 N/C

8485.28 N/C= −0.707107

lui (el22284) – HW 1 – li – (58545) 11

027 (part 5 of 5) 10.0 pointsThe electric force on the −6× 10−9 C chargecan be written in the form

〈Fx, Fy, Fz〉where it is understood that Fz = 0.What is the value of Fx?Your answer must be within ± 1.0%

Correct answer: 3.6× 10−5 N.

Explanation:The force on a charged particle is merely

the charge of that particle times the field atits location.Therefore, we have:

~F = q~E

Fx = qEx

= (−6× 10−9 C)(−6000 N/C)

= 3.6× 10−5 N

hw 1 solutions

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