ch16 h1 problems

gakhar (eg26949) – Ch16-h1 – chiu – (58655) 1

This print-out should have 13 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 pointsA wire 3 m long lying on the X-axis andcentered at the origin has a nearly uni-form charge. The electric field at loca-tion 〈−0.02, 0.01, 0〉 m is 〈0,−67000, 0〉 N/C.What is the total charge on the wire? (Besure to include the sign of the charge.)Answer in units of C

002 10.0 points

A water molecule is a permanent dipolewith a known dipole moment p = qs. Thereis a water molecule in the air a very short dis-tance x from the midpoint of a long glassrod of length L carrying a uniformly dis-tributed positive charge Q. The axis of thedipole is perpendicular to the rod. Note thats << x << L. You may neglect the smallchange in the dipole moment of the watermolecule induced by the rod.Choose the answer that correctly expresses

the magnitude and direction (along the x-axis) of the electric force on the watermolecule. Your final result must be ex-

pressed only in terms of k, Q, p, L, s andx and any constant numerical factors.

1. −kQp

Lx2

2. k2Qp

xL2

3. k2Qp

Ls2

4. kQp

Ls2

5. −k2Qp

xL2

6. −k2Qp

Ls2

7. kQp

Lx2

8. −kQp

Ls2

9. −k2Qp

Lx2

10. k2Qp

Lx2

003 (part 1 of 6) 1.0 pointsA plastic rod 1.8 m long is rubbed all over withwool, and acquires a charge of −4× 10−8 C(see the figure below). We choose the centerof the rod to be the origin of our coordinatesystem, with the x axis extending to the right,the y axis extending up, and the z axis outof the page. In order to calculate the electricfield at location A, which is at

~rA = 〈0.6, 0, 0〉 m,

we divide the rod into 8 pieces, and approxi-mate each piece as a point charge located atthe center of the piece.


b A

1

2

3

4

5

6

7

8

What is the length of one of these pieces?Use the value of k = 8.98755× 109 N ·m2/C2.Answer in units of m

004 (part 2 of 6) 1.0 pointsWhat is the y coordinate of the center of piecenumber 2?Answer in units of m

005 (part 3 of 6) 2.0 pointsHow much charge is on piece number 2? (Re-member that the charge is negative.)Answer in units of C

006 (part 4 of 6) 2.0 pointsApproximating piece 2 as a point charge, whatis the electric field at location A due onlyto piece 2? This is a vector with x and ycomponents,

~E(A) = 〈E(A)x, E(A)y, 0〉 N/C,

so begin by finding the x component, E(A)x.Answer in units of N/C

007 (part 5 of 6) 2.0 pointsFind E(A)y.Answer in units of N/C

008 (part 6 of 6) 2.0 pointsTo get the net electric field at location A, wewould need to calculate ∆~E due to each ofthe 8 pieces, and add up these contributions.If we did that, which arrow (I-VIII) on thefollowing diagram would best represent the

direction of the net electric field at locationA?

III

III

IV

V

VI

VII

VIII

1. III

2. VI

3. I

4. V

5. II

6. VIII

7. IV

8. VII

009 10.0 pointsConsider a thin glass rod of length L lyingalong the x-axis with its left end at origin.The rod carries a uniformly distributed posi-tive charge Q.

O

L

+Qa

What is the electric field due to the rod atposition a on the x-axis, a > L, as shown inthe figure above?


1. E =1

4 π ǫ0

Q

L

L∫

0

d x

(a− x)2

2. E =1

4 π ǫ0

Q

L

a∫

0

d x

x2

3. E =1

4 π ǫ0Q

L∫

0

d x

x2

4. E =1

4 π ǫ0Q

a∫

0

d x

(a− x)2

5. E =1

4 π ǫ0

Q

L

a∫

0

d x

(a− x)2

6. E =1

4 π ǫ0Q

L∫

0

d x

(a− x)2

7. E =1

4 π ǫ0

Q

L

L∫

0

d x

x2

8. E =1

4 π ǫ0Q

a∫

0

d x

x2

010 (part 1 of 3) 3.0 pointsConsider a uniformly charged non-conductingsemicircular arc with radius r and total nega-tive charge Q. The charge on a small segmentwith angle ∆θ is labeled ∆q.

x

y

−−−−−−−−−−−−−−−−−−

∆θ

θ rx

y

III

III IV

B

A

O

What is ∆q?

1. ∆q = πQ

2. ∆q =Q∆θ

2π

3. ∆q = 2 πQ

4. ∆q =Q

π

5. ∆q =Q

2 π

6. ∆q =2Q

π

7. ∆q =Q∆θ

π

8. ∆q =2Q∆θ

π

9. None of these

10. ∆q = Q

011 (part 2 of 3) 3.0 pointsWhat is the magnitude of the x-component ofthe electric field at the center due to ∆q?

1. ∆Ex = k |∆q| r2

2. ∆Ex = k |∆q| (sin θ) r2

3. ∆Ex =k |∆q| sin θ

r

4. ∆Ex =k |∆q| sin θ

r2

5. ∆Ex = k |∆q| (sin θ) r

6. ∆Ex =k |∆q| cos θ

r

7. ∆Ex = k |∆q| (cos θ) r2

8. ∆Ex = k |∆q| (cos θ) r

9. ∆Ex =k |∆q| cos θ

r2

10. ∆Ex =k |∆q|

r2

012 (part 3 of 3) 4.0 pointsDetermine the magnitude of the electric fieldatO .The total charge is−59.8 µC, the radiusof the semicircle is 121 cm, and the Coulombconstant is 8.98755× 109 N ·m2/C2.Answer in units of N/C


013 10.0 points

Consider a uniformly charged thin rod withtotal charge Q and length L. It is alignedalong the y-axis and centered at the origin(see fig 8-3hw). We wish to determine thefield at P due to the charges on the rod.

Because the rod is centered at the origin,symmetry tells us the electric field at P mustpoint in the r̂ direction. Based on the differ-ential form

dEr = k(Qdy/L)

ρ2sinα ,

determine the integrated expression for Er atP .

{Hint: use the math identity dy/ρ2 = dα/r.This identity can be derived using the geo-metric relation tanα = r/(−y) (1), and thecalculus identity d tanα/dα = sec2 α = y2/ρ2

(2).}

1.kQ

r(cosα2 − cosα1)

2.kQ

Lr(cosα2 − cosα1)

3.kQ

L(cosα2 − cosα1)

4.kQ

Lr(cosα1 − cosα2)

5.kQ

r(cosα1 − cosα2)

6.kQ

L(cosα1 − cosα2)

ch16 h1 problems

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