midterm 1 - turner

porter (djp944) midterm 01 review turner (56545) 1

This print-out should have 22 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

001 10.0 pointsA particle of mass 2.88 1026 kg and charge9e is observed to have an instantaneous ac-celeration of 1.4 1012 m/s2. What is themagnitude of the electric field at the parti-cles location? (e = 1.6 1019C)Answer in units of N/C

002 10.0 pointsA hollow sphere of radius 2 cm has a chargeof Q = 4 nC spread uniformly over itssurface. The center of the ball is at ~rs =5, 0, 0 cm. A point charge of charge q =7.5 nC is located at ~rp = 6, 0, 0 cm. What isthe magnitude of the net electric field at thepoint A = (0, 8, 0),cm?Answer in units of N/C

003 10.0 pointsA dipole is located at the origin and iscomposed of charged particles with charge+2e and 2e, separated by a distance s =2 1010m along the y-axis. The +2e chargeis on the +y axis. Calculate the force on aproton at a location A = 0, 0, 3 108mdue to the dipole. (e = 1.6 1019C)

1. 0, 0,3.41 1015N

2. 0,6.82 1015, 0N

3. 0, 0, 3.41 1015N

4. 0, 0, 6.82 1015N

5. 0, 6.82 1015, 0N

6. 0,3.41 1015, 0N

7. 0, 3.41 1015, 0N

8. 0, 0,6.82 1015N

004 10.0 points

A very thin spherical plastic shell of radiusR = 13 cm carries a uniformly distributednegative charge of Q = 11 nC on its outersurface. An uncharged solid metal block isplaced nearby. The block is w = 11 cm thickand is 11 cm away from the surface of thesphere. Calculate the magnitude of the elec-tric field at the exact center of the conductorgenerated only by the polarization of the con-ductor. In other words, do not include theelectric field generated by the sphere.Answer in units of N/C

005 10.0 points

Two small, negatively charged plasticspheres are placed near a neutral iron blockas shown above. Which arrow best indicatesthe the direction of the net electric field atlocation A?

1. g

2. j

3. b


4. c

5. h

6. e

7. a

8. f

9. d

006 10.0 pointsThe electric field at a location C points northand has a magnitude of 6 105 N/C. Whereshould you place a proton and an electron(charge |q| = 1.61019C), at equal distancesfrom C, to produce this field?Answer in units of m

007 10.0 pointsConsider the algebraic analysis for the electricfield of a uniformly charged ring, at a locationon the axis of the ring. The ring has radiusR and total charge Q. We need to divide thering into pieces Q. Which of the followingis the correct expression for Q?

1.Q

2piR

2.Q

2pi

3.Q

piR2

4.Q

2piRR

5. none of these

6.Q

2piR

008 10.0 pointsConsider a thin plastic rod bent into a semi-circular arc of radius R = 51 cm with a uni-formly distributed charge of Q = 30 nC.

Determine the the magnitude of the electricfield at the origin generated by the rod.Answer in units of N/C

009 10.0 pointsConsider a square with side a. Four chargesq, +q, +q, and q are placed at the cornersA, B, C, and D, respectively

+

+

D

A

C

B

aO

What is the magnitude of the electric fieldat the center O?

1. EO =1

32

k q

a2

2. EO =12

k q

a2

3. EO = 32k q

a2

4. EO =k q

a2

5. EO =1

42

k q

a2

6. EO = 42k q

a2

7. EO = 3k q

a2

8. EO =2k q

a2

9. EO = 22k q

a2

10. EO =1

22

k q

a2


010 10.0 pointsA line of charge starts at x = x0, where x0is positive, and extends along the x-axis topositive infinity.If the linear charge density is given by

= 0 x0/x, where 0 is a constant, findthe electric field at the origin. (Here denotesthe unit vector in the positive x direction.)

1.k 02 x0

()

2.k 0

2 x20

()

3.k 0x0

()

4.k 0x0

()

5.k 2

0

2 x0()

6.k 02 x0

()

011 10.0 points1) Two uncharged metal balls, X and Y,stand on glass rods and are touching.

X Y

2) A third ball, carrying a negative charge, isbrought near the first two.

X Y

3) Then the first two balls are separated fromeach other,

X Y

4) and the third ball is finally removed.

X Y

When this is all four steps are done, it isfound out that

1. BallY is negative and ballX is positive.

2. BallY is positive and ballX is negative.

3. Balls Y and X are both negative.

4. Balls Y and X are both positive.

5. Balls Y and X are still uncharged.

012 10.0 pointsAn electron moves at 1.5 106 m/s into auniform electric field of magnitude 990 N/C.The charge on an

electron is 1.60218 1019 C and the massof an electron is 9.10939 1031 kg .The field is parallel to the electrons velocity

and acts to decelerate the electron.How far does the electron travel before it is

brought to rest?Answer in units of cm

013 10.0 pointsConsider the setup shown in the figure.

Determine the force exerted by -Q on thedipole. Choose the correct expression for theforce. + sign is along the positive x-direction.

1. k q sQz3


2. + kq sQ

z2

3. + kq sQ

z3

4. k q sQz2

5. + k2 q sQ

z3

6. + k2 q sQ

z2

7. k 2 q sQz2

8. k 2 q sQz3

014 10.0 pointsThe electric field at a location C points northand its magnitude is 1.2456 106 N/C. Toproduce this field, a proton and an electronare placed at equal distances from C and op-posite each other with respect to C. Find thedistance between each charge and C in unitsof m.Answer in units of m

015 (part 1 of 2) 10.0 pointsSuppose you want to create an electric field= 2,2, 0 N/C at the origin located at0, 0, 0. Where would you place an electronto create this field? Denote the coordinatevector of the electron by ~r = r rDetermine r

1.1, 1, 0

2

2. 2 1, 1, 0

3.1,1, 0

2

4. 1, 1, 0

5.1,1, 0

2

6. 2 1,1, 0

7.1, 1, 0

2

8. 1,1, 0

016 (part 2 of 2) 10.0 pointsFind the magnitude r (in units of m).

1. 2.26 105 m

2. 1.13 106 m

3. 0.00226 m

4. 2.26 107 m

5. 0.000452 m

6. 0.000226 m

7. 2.26 106 m

8. 0.000678 m

017 10.0 pointsConsider the experiment using two tapes Thesetup is shown in figure 1. The Tapes A andB are oppositely charged. The electric attrac-tive force lifts it up to the location shown witha separation d. Following the approximationof the text, the attractive force is estimated asthe attraction between two point charges QAand QB separated by distance d. The upwardpull on the lower tape due to the electric at-traction is balanced by the weight of the lowertape.


Now consider a new case where the lowertape is replaced by a new tape that weighs 2times as much as the original tape, and thecharges are the same as before. Assumingthe same approximation is still valid , whichexpression gives the new separation distanced so that the lower tape is again lifted?

1. d =d3

2. d =d

3.

3. d =2 d

4. d = 2 d

5. d =d

2

6. d = 3 d

7. d =3 d

8. d =d2

018 10.0 pointsA lightweight metal ball hangs from a thread,to the right of an plastic rod. (See Figure 1.)Both are initially uncharged.

The plastic (insulator) case: You rub theleft end of the plastic rod with wool, deposit-ing charged molecular fragments whose totalnegative charge is that of 1 109 electrons.The metal (conductor) case: You perform

a similar experiment with a conducting metalrod. You touch the left end of the rod witha charged metal object, depositing the sameexcess electrons to the left end. You thenremove the object.Statement I : Consider two cases below:(i). There are polarized dipoles inside of

the rod.(ii). There are electrons at the surface.

Ia. (i) is applicable for the insulator. (ii) isapplicable for the conductor.Ib. (ii) is applicable for the insulator. (i) isapplicable for the conductor.

Statement II : Compare the attraction be-tween the rod and the ballIIa. The insulator rod has more attraction.IIb. The conductor rod has more attraction.

Statement III :IIIa. The induced dipole moment for a con-ductor occurs over a macroscopic length scale.IIIb. The induced dipole moment for a con-ductor occurs over an atomic length scale.

1. Ib, IIa, IIIa

2. Ib, IIb, IIIb

3. Ib, IIb, IIIa

4. Ib, IIa, IIIb

5. Ia, IIb, IIIb

6. Ia, IIa, IIIa

7. Ia, IIa, IIIb

8. Ia, IIb, IIIa

019 10.0 pointsTwo large, thin, charged plastic circularplates each of radius R are placed a shortdistance s apart; s is much smaller than thedimensions of a plate. The right-hand platehas a positive charge of +Q evenly distributedover its inner surface (Q is a positive number).


The left-hand plate has a negative charge of2Q evenly distributed over its inner surface.A very thin plastic spherical shell of radiusr is placed midway between the plates (andshown in cross section). It has a uniformlydistributed positive charge of +q. You can ig-nore the contributions to the electric field dueto the polarization of the thin plastic shelland the thin plastic plates.

Without the plastic sphere, the electric fieldat A would be

Ia) the sameIb) smallerIc) larger

Rank the electric field magnitudes at A,Band C.

IIa) EA > EB > ECIIb) EC > EB > EAIIc) EB > EC > EA

1. Ib, IIb

2. Ia, IIb

3. Ia, IIc

4. Ic, IIb

5. Ib, IIc

6. Ic, IIc

7. Ia, IIa

8. Ic, IIa

9. Ib, IIa

020 (part 1 of 2) 10.0 points

A plastic hollow sphere, which is uniformlycharged with negative charge on its surface, isplaced near the center of a horizontal plasticrod, which is uniformly charged with negativecharge. What is the direction of the electricfield vector at P due to the charges on theplastic sphere alone?

1. direction 7

2. Zero magnitude

3. direction 2

4. direction 3


5. direction 5

6. direction 8

7. direction 4

8. direction 1

9. direction 6

021 (part 2 of 2) 10.0 pointsWhat is direction of the resultant field vectorcontributed by the rod plus the sphere?

1. Zero magnitude

2. direction 5

3. direction 7

4. direction 2

5. direction 4

6. direction 6

7. direction 1

8. direction 3

9. direction 8

022 10.0 pointsWhich is the correct integral formulation for

the electric field on the axis of a uniformlycharged disk of total charge Q and radius R?

1. Ez =

pi0

1

4pi0

(Q

piR2

)z

(R2 + z2)3/2d

2. Ez =

2pi0

1

4pi0

(Q

piR2

)z

(R2 + z2)3/2d

3. Ez =

R0

1

20

(Q

2pi

)z

(R2 + z2)3/2r dr

4. Ez =

pi0

1

4pi0

(Q

2pi

)z

(R2 + z2)3/2d

5. Ez =

R0

1

20

(Q

piR2

)z

(r2 + z2)3/2r dr

6. Ez =

R0

1

20

(Q

2pi

)z

(r2 + z2)3/2r dr

7. Ez =

2pi0

1

4pi0

(Q

2pi

)z

(R2 + z2)3/2d

8. Ez =

R0

1

20

(Q

piR2

)z

(R2 + z2)3/2r dr

midterm 1 - turner

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