Ch14 h2 Extra Solutions

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001 (part 1 of 2) 10.0 points

What is the direction, if any, of the electricfield due to this configuration at a pointX >>l >> s?

1. +y correct

2. −y

3. −x

4. +x

5. ~E(X) = 0

Explanation:Since the dipole to the right of the origin is

closer to point X , its electric field will dom-inate; since the moment of the dipole pointsin the − − y direction, the electric field at Xpoints in the +y direction.

002 (part 2 of 2) 10.0 pointsUsing a procedure similar to that used tocalculate the electric field of a dipole, find anapproximate algebraic expression for |~E(X)|,the magnitude of the electric field due to theconfiguration at this point X , X >> l >> s.

1. k|~p|sX4

2. 3k|~p|sX3

3. 6k|~p|lX4

4. 0

5. 3k|~p|lsX5

6. 3k|~p|X4

7. 6k|~p|sX4

8. 3k|~p|lX3

9. 3k|~p|lX4

correct

10. 6k|~p|X3

Explanation:Using the formula for the electric field of a

dipole along its perpendicular axis,

~Edip⊥ = −k~p

r3

and expressing the distance to the left andright dipoles as X + l/2 and X − l/2 respec-tively, we apply the superposition principle atX:

~Enet(X) = k|~p|

(X + l/2)3(−y)+ k

|~p|(X − l/2)3

y

~Enet(X) = k|~p|(

1

(X − l/2)3− 1

(X + l/2)3

)

y

~Enet(X) = k|~p|X3

(

(

1− l

2X

)−3

−(

1 +l

2X

)−3)

y

Making the small argument approximation(since l << X) and taking the magnitude we

obtain an approximate value for |~Enet(X)|:

k|~p|X3

((

1− 3

(

− l

2X

))

−(

1− 3

(

l

2X

)))

therefore,

~Enet(X)≈ k|~p|X3

6l

2X

~Enet(X)≈3k|~p|lX4

003 10.0 points


Consider the setup shown in the figure,where charges Q1, Q2 and Q3 and thepoint A occupy four corners of a squarewith the length a at each side. Given thatQ1 = Q3 = q > 0 and Q2 = −2q. Theresultant electric field at A contributed bycharges Q1 and Q3 is labeled as ~E13. Theelectric field at A contributed by Q2 is la-beled as ~E2. Verify that the vectors ~E13 and~E2 are aligned along a same line. Such an

alignment implies that the ratio~E13

~E2

may be

represented by a number. Here, a positive ra-tio implies that two vectors are pointing inthe same direction and a negative ratio im-plies that they are pointing in the oppositedirection. Determine this ratio.

1. 2

2. -2

3.√8

4. 1

5. -√3

6. -√2 correct

7.√3

8. -1

9. -√8

10.√2

Explanation:Notice that at A, ~E1 is perpendicular to ~E3.

The magnitude of ~E13 is given by

E13 =√

E21 + E2

3 =√2×E1

where E1 =kq

a2, and by symmetry E3 = E1.

By inspection the magnitude of ~E2 is

E2 =k(2q)

(√2a)2

=kq

a2= E1

~E13 and ~E2 are aligned along the same lineand are pointing in the opposite direction.

The ratio~E13

~E2

is a negative number. So, we

have~E13

~E2

= −√2E1

E1

~E13

~E2

= −√2

004 (part 1 of 6) 10.0 pointsA conceptual model of aluminum triflouride(AlF3) is approximately a square with chargesat the corners.

QD= −q

QA= 3 q

QC= −q

QB= −q

Oa

What is the magnitude of the electric fieldE

Oat the center O?

1. EO=

k q

a2

2. EO=

1

3√2

k q

a2

3. EO=

1√2

k q

a2

4. EO=

4 k q

a2

5. EO= 2

√2k q

a2


6. EO= 3

k q

a2

7. EO=

√2k q

a2

8. EO=

8 k q

a2correct

9. EO= 4

√2k q

a2.

10. EO=

1

4√2

k q

a2

Explanation:The distance between each corner and the

center isa√2, so the magnitude of each field is

E = kQ

(

a√2

)2= 2

k Q

a2

The direction of the electric field at the fieldpoint O is the direction exerted on a positive

charge, so negative charges yield a field point-ing toward them from O and positive chargesa field pointing away from them from O:

EB

ED

EC

EA

EB

and ED

cancel; they have the samemagnitude but act in opposite directions.Since ~E

Aand ~E

Care collinear, they add

algebraically:

‖~E‖ = ‖~EA+ ~E

C‖ = E

A+ E

C

=6 k q

a2+

2 k q

a2=

8 k q

a2

directed from O through C.

005 (part 2 of 6) 10.0 pointsWhat is the magnitude of the electric field E

C

at C due to the charges at A, B, and D ?

1. EC= 4

√2k q

a2

2. EC= 3

k q

a2

3. EC=

1

3√2

k q

a2

4. EC=

3

2

k q

a2

5. EC=(

3−√2) k q

a2

6. EC=

9

4

k q

a2

7. EC=

√2k q

a2

8. EC=

(

3

2−

√2

)

k q

a2correct

9. EC= 2

√2k q

a2.

10. EC= 3

√2k q

a2

Explanation:The charge at A is at a distance

√2a, with

a field of magnitude

EA=

3 k q

(√2 a)2

=3 k q

2 a2,

whereas the charges at B and D are at adistance a, with fields of magnitude

EB= E

D=

k q

a2.

EB + EDEB

ED

EA

C

~EBand ~E

Dform a square; the vector sum

is the diagonal:

|~EB+ ~E

D| =

√2E

B=

√2 k q

a2.

The vectors ~EB+ ~E

Dand ~E

Ain the figure

act in opposite directions, and ~EAhas a larger


magnitude, so the magnitude of ~ECis

EC =3 k q

2 a2−

√2 k q

a2=

(

3

2−

√2

)

k q

a2.

006 (part 3 of 6) 10.0 pointsFind the absolute value of tanα, where αis the angle between the horizontal and theelectric field at C due to the three charges atA, B, and D.

1. | tanα| = 1

2√2− 1

2. | tanα| = 2√2− 1

3. | tanα| =√2

4. | tanα| = 2√2 + 1

2√2− 1

5. | tanα| = 1

2√2 + 1

6. | tanα| = 1√2

7. | tanα| = 1 correct

8. | tanα| = 2√2− 1

2√2 + 1

9. | tanα| =√3

10. | tanα| = 2√2 + 1

Explanation:The resultant electric field is in the direc-

tion ~AC (45◦ below the horizontal), so

| tanα| = tan 45◦ = 1 .

007 (part 4 of 6) 10.0 pointsConsider charges in a square again, but thistime with a different assignment of charges:

QD= q

QA= q

QC= −q

QB= q

O

a

Find the magnitude of the electric field atO .

1. EO=

1

3√2

k q

a2

2. EO= 2

√2k q

a2

3. EO= 3

k q

a2

4. EO=

1

5√2

k q

a2

5. EO=

1

4√2

k q

a2

6. EO= 4

k q

a2correct

7. EO=

k q

a2

8. EO=

√2k q

a2

9. EO=

1√2

k q

a2

10. EO= 3

√2k q

a2

Explanation:

EA=

2 k q

a2is the magnitude of each field at

O.

EA

EB EC

ED

The contributions from B and D cancel,whereas the contributions from A and C add:

E = 2 · 2 k qa2

=4 k q

a2.

008 (part 5 of 6) 10.0 pointsFind the electric field E

Cat C due to the

3 charges at A, B, and D for new chargeorientation.

1. EC=

5

2

k q

a2


2. EC=

(√2 +

1

2

)

k q

a2correct

3. EC= 3

k q

a2

4. EC= 4

k q

a2

5. EC= 2

k q

a2

6. EC=

1

3√2

k q

a2

7. EC= 3

√2k q

a2

8. EC=

k q

a2

9. EC=

√2k q

a2

10. EC=

7

4√2

k q

a2

Explanation:The magnitudes are the same as in Part 2,

but the directions are again different.

QAQB

QDC

~EBand ~E

Dact at right angles, so

|~EB+ ~E

D| =

√2 k q

a2

in the same direction as ~EA. The magnitudes

of collinear vectors add, so

E =

√2 k q

a2+

k q

2 a2=

(√2 +

1

2

)

k q

a2.

009 (part 6 of 6) 10.0 pointsFind the absolute value of tanα, where αis the angle between the horizontal and theelectric field at C due to the three charges atA, B, and D.

1. | tanα| = 2√2 + 1

2√2− 1

2. | tanα| = 1

2√2− 1

3. | tanα| = 2√2 + 1

4. | tanα| = 2√2− 1

5. | tanα| = 2√2− 1

2√2 + 1

6. | tanα| = 1 correct

7. | tanα| = 1√2

8. | tanα| =√2

9. | tanα| =√3

10. | tanα| = 1

2√2 + 1

Explanation:The resultant vector is directed 45◦ below

the horizontal, so

tan 45◦ = 1 .

010 10.0 points

Two identical permanent dipoles, each con-sisting of charges +q and −q separated by adistance s are aligned along the x axis andplaced on the x-axis with their centres sepa-rated by a distance of r (r >> s) as shownin the figure. Let F be the magnitude ofthe force exerted by one of the dipoles onthe other. If we change the distance betweenthe dipoles to r′ = 2r, what will be the newmagnitude of force F ′?

1.F

4

2. 2F

3. 4F


4.F

8

5. 8F

6. F

7. 16F

8.F

2

9.F

16correct

Explanation:Let qs = p be the magnitude of dipole mo-

ment of each of the dipoles. We can considerthe field due to one of the dipoles as actingon each of the charges constituting the seconddipole. Magnitude of electric field due to thefirst dipole at a distance of x from it along itsaxis(for x >> s) is given by the formula

E =1

4πǫ0

2p

x3

For the two charges constituting the seconddipole, the corresponding values of x will be

r− s

2and r+

s

2. Since the two charges are op-

posite, the magnitude of net force will be ob-tained by subtracting the force acting on thefarther charge (farther away from the dipole)from the force acting on the closer charge.Thus, we can write

F =2p

4πǫ0

[

q

(r − s/2)3− q

(r + s/2)3

]

This can be simplified to get

F =2pq

4πǫ0

3r2s+1

4s3

(r − s/2)3(r + s/2)3

Now we can use the approximation r >> sto neglect terms of order 2 or more in s innumerator and denominator to get the result

F =2pq

4πǫ0

3r2s

r6=

6p2

4πǫ0r4

where we have made use of the fact that p =qs.Clearly, this force is inversely proportional

to the fourth power of r. Hence, if we increaser by a factor of 2, then we reduce F by afactor of 16.

011 (part 1 of 3) 10.0 pointsConsider the figure below:

+q −q

s

+Q

d ≫ s

~Edipole

~F

If the charge of the point charge in the figurewere −6Q (instead of +Q, by what factorwould there be a change in the magnitudeof the force on the point charge due to thedipole?

Correct answer: 6.

Explanation:This problem is best analyzed from the field

viewpoint–the dipole sets up an electric field~E in the region of the charge Q. This fieldthen produces a force on the charge, accordingto ~F = Q~ETherefore, if we increase the magnitude of

the charge by a factor of 6, we also increasethe magnitude of the force by a factor of 6.

012 (part 2 of 3) 10.0 pointsWould the direction of the force change?

1. The force would stay in the same direc-tion.

2. The new force would be in an unrelateddirection.

3. The force would be in the opposite direc-tion. correct

4. The new force would be zero.

Explanation:


Since the force is related to the field by theformula ~F = Q~E, if we place two oppositelycharged point particles in the same field, theywill have force vectors that point in oppositedirections.

013 (part 3 of 3) 10.0 pointsThe distance between the dipole and the pointcharge in the diagram in the figure is d.If the distance between them were changed

to1

5d, by what factor would the force on the

point charge due to the dipole change?Your answer should be a decimal within five

percent of the correct answer.

Correct answer: 125.

Explanation:

Dipole fields fall off as1

r3. So, if the dis-

tance is reduced by a factor of1

5, the strength

of the field, and thus the force on the chargeincreases by a factor of 53 = 125.

014 (part 1 of 2) 10.0 pointsA charge of 1 nC (1 nC = 1× 10−9 C) and adipole with charges +q and −q separated by0.3 mm contribute to a net field at location Athat is zero, as shown in the following figure.

bA

8 cm

18 cm+

+1 nC

Which end of the dipole is positivelycharged?

1. The left end correct

2. The right end

Explanation:The left end of the dipole must be posi-

tively charged. In order for there to be zeronet electric field at point A, the dipole mustcontribute an overall rightward pointing field

at A to oppose the leftward pointing field dueto the 1 nC charge. The only way that thedipole’s field at A can point toward the rightis if the positive charge is on the left, andthe negative charge is on the right. In thatcase, each of the dipole’s two charges wouldcontribute a rightward-pointing field at pointA.

015 (part 2 of 2) 10.0 pointsHow large is the charge q?

Correct answer: 5.26749× 10−8 C.

Explanation:We begin by writing down expressions for

the electric fields of the charge and the dipoleat point A. For the charge,

∣

∣

∣

~Echarge

∣

∣

∣=

1

4πǫ0

q

(rcharge)2

=1

4πǫ0

1 nC

(18 cm)2

= 274.691 N/C .

(Make sure to convert to correct units.) Forthe dipole, we use the expression given in yourMatter and Interactions textbook:

∣

∣

∣

~Edipole

∣

∣

∣=

1

4πǫ0

qs

(rdipole)3

=1

4πǫ0

q(0.3 mm)

(8 cm)2

= q × 5.21484× 109 N/C2 .

We know that these two fields have to can-cel each other, so let’s set them equal to eachother:

Echarge = Edipole

274.691 N/C = q × 5.21484× 109 N/C2

⇒ q =274.691 N/C

5.21484× 109 N/C2

q = 5.26749× 10−8 C .

016 10.0 points


Given: G(x) =N

D, where

N =√1− x− (1− x)3/2

D =

√

1− (1− x2)3/2 ,

successively apply the small argument expan-sion in order to determine the correct answer.

1.

√

5

3

2.

√

4

3

3.

√

2

3correct

4.

√

1

3

5. 1

Explanation:Applying the small argument approxima-

tion successively,

N =√1− x− (1− x)3/2

= (1− x/2)− (1− 3x/2) = x ,

D =

√

1− (1− x2)3/2

=√

1− (1− (3/2)x2)

=

√

3

2x2 =

√

3

2x .

G =N

D=

x√

3

2x

=

√

2

3

Ch14 h2 Extra Solutions

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