1. (a) With a understood to mean the magnitude of acceleration, Newton’s second and

third laws lead to

c6.3 × 10 kghc7.0 m s h = 4.9 × 10

=ma m =

−7

m2 a2

1 1

2

2

2

9.0 m s

−7

kg.

(b) The magnitude of the (only) force on particle 1 is

2

q q

q

F = m1a1 = k 1 2 2 = 8.99 × 109

.

r

0.0032 2

c

h

Inserting the values for m1 and a1 (see part (a)) we obtain |q| = 7.1 × 10–11 C.

2. The magnitude of the mutual force of attraction at r = 0.120 m is

hc

hc

h

3.00 × 10−6 150

. × 10−6

q1 q2

9

F=k

= 8.99 × 10

= 2.81 N .

r2

0120

. 2

c

3. Eq. 21-1 gives Coulomb’s Law, F = k

k | q1 || q2 |

r=

=

F

(8.99 ×10 N ⋅ m

9

2

q1 q2

r2

, which we solve for the distance:

C2 ) ( 26.0 ×10−6 C ) ( 47.0 ×10−6 C )

5.70N

= 1.39m.

4. The fact that the spheres are identical allows us to conclude that when two spheres are

in contact, they share equal charge. Therefore, when a charged sphere (q) touches an

uncharged one, they will (fairly quickly) each attain half that charge (q/2). We start with

spheres 1 and 2 each having charge q and experiencing a mutual repulsive force

F = kq 2 / r 2 . When the neutral sphere 3 touches sphere 1, sphere 1’s charge decreases to

q/2. Then sphere 3 (now carrying charge q/2) is brought into contact with sphere 2, a total

amount of q/2 + q becomes shared equally between them. Therefore, the charge of sphere

3 is 3q/4 in the final situation. The repulsive force between spheres 1 and 2 is finally

(q / 2)(3q / 4) 3 q 2 3

F' 3

F′ = k

= k 2= F

= = 0.375.

2

r

F 8

8 r

8

5. The magnitude of the force of either of the charges on the other is given by

F=

b

1 q Q−q

r2

4 πε 0

g

where r is the distance between the charges. We want the value of q that maximizes the

function f(q) = q(Q – q). Setting the derivative df/dq equal to zero leads to Q – 2q = 0, or

q = Q/2. Thus, q/Q = 0.500.

6. For ease of presentation (of the computations below) we assume Q > 0 and q < 0

(although the final result does not depend on this particular choice).

(a) The x-component of the force experienced by q1 = Q is

§

·

Q )( Q )

| q |) ( Q ) ¸ Q | q | § Q / | q | ·

(

(

¨

F1x =

−

cos 45° +

2

2

¸ = 4πε a 2 ¨© − 2 2 + 1¸¹

a

4πε 0 ¨¨

0

¸

2a

©

¹

1

(

)

which (upon requiring F1x = 0) leads to Q / | q |= 2 2 , or Q / q = −2 2 = −2.83.

(b) The y-component of the net force on q2 = q is

§

·

| q |) ( Q ) ¸

(

1 ¨ | q |2

| q |2 § 1

Q·

F2 y =

sin

45

°

−

=

−

¨

2

2

2

¸ 4πε a © 2 2 | q | ¸¹

4πε 0 ¨¨ 2a

a

0

¸

©

¹

(

)

which (if we demand F2y = 0) leads to Q / q = −1/ 2 2 . The result is inconsistent with

that obtained in part (a). Thus, we are unable to construct an equilibrium configuration

with this geometry, where the only forces present are given by Eq. 21-1.

7. The force experienced by q3 is

G G

G

G

| q || q | ·

1 § | q3 || q1 | ˆ | q3 || q2 |

F3 = F31 + F32 + F34 =

j+

(cos45°ˆi + sin 45°ˆj) + 3 2 4 ˆi ¸

¨−

2

2

a

a

4πε 0 ©

( 2a )

¹

(a) Therefore, the x-component of the resultant force on q3 is

2 (1.0 ×10−7 ) § 1

| q3 | § | q2 |

·

·

9

F3 x =

+ | q4 | ¸ = ( 8.99 ×10 )

+ 2 ¸ = 0.17N.

¨

2 ¨

2

4πε 0 a © 2 2

(0.050)

¹

©2 2

¹

2

(b) Similarly, the y-component of the net force on q3 is

2 (1.0 ×10−7 ) §

| q3 | §

| q2 | ·

1 ·

9

F3 y =

− | q1 | +

= ( 8.99 ×10 )

−1 +

¸

¨

¸ = −0.046N.

2 ¨

2

4πε 0 a ©

(0.050)

2 2¹

2 2¹

©

2

8. (a) The individual force magnitudes (acting on Q) are, by Eq. 21-1,

k

q1 Q

b−a − g

a 2

2

=k

q2 Q

ba − g

a 2

2

which leads to |q1| = 9.0 |q2|. Since Q is located between q1 and q2, we conclude q1 and q2

are like-sign. Consequently, q1/q2 = 9.0.

(b) Now we have

k

q1 Q

b−a − g

3a 2

2

=k

q2 Q

ba − g

3a 2

2

which yields |q1| = 25 |q2|. Now, Q is not located between q1 and q2, one of them must

push and the other must pull. Thus, they are unlike-sign, so q1/q2 = –25.

9. We assume the spheres are far apart. Then the charge distribution on each of them is

spherically symmetric and Coulomb’s law can be used. Let q1 and q2 be the original

charges. We choose the coordinate system so the force on q2 is positive if it is repelled by

q1. Then, the force on q2 is

Fa = −

1 q1q2

qq

= −k 1 2 2

2

4 πε 0 r

r

where r = 0.500 m. The negative sign indicates that the spheres attract each other. After

the wire is connected, the spheres, being identical, acquire the same charge. Since charge

is conserved, the total charge is the same as it was originally. This means the charge on

each sphere is (q1 + q2)/2. The force is now one of repulsion and is given by

1

Fb =

4 πε 0

q1 + q2

2

q1 + q2

2

d id i = k bq + q g .

2

1

r2

2

4r 2

We solve the two force equations simultaneously for q1 and q2. The first gives the product

b

gb

g

2

0.500 m 0108

.

N

r 2 Fa

q1q2 = −

=−

= −3.00 × 10−12 C 2 ,

9

2

k

8.99 × 10 N ⋅ m C 2

and the second gives the sum

q1 + q2 = 2r

Fb

= 2 0.500 m

k

b

g

0.0360 N

= 2.00 × 10−6 C

9

2

2

8.99 × 10 N ⋅ m C

where we have taken the positive root (which amounts to assuming q1 + q2 ≥ 0). Thus, the

product result provides the relation

q2 =

− ( 3.00 ×10−12 C2 )

q1

which we substitute into the sum result, producing

q1 −

3.00 × 10−12 C 2

= 2.00 × 10−6 C.

q1

Multiplying by q1 and rearranging, we obtain a quadratic equation

c

h

q12 − 2.00 × 10−6 C q1 − 3.00 × 10−12 C 2 = 0 .

The solutions are

q1 =

2.00 × 10−6 C ±

2

c−2.00 × 10 Ch − 4c−3.00 × 10

−6

2

−12

C2

h.

If the positive sign is used, q1 = 3.00 × 10–6 C, and if the negative sign is used,

q1 = −1.00 ×10−6 C .

(a) Using q2 = (–3.00 × 10–12)/q1 with q1 = 3.00 × 10–6 C, we get q2 = −1.00 ×10−6 C .

(b) If we instead work with the q1 = –1.00 × 10–6 C root, then we find q2 = 3.00 ×10−6 C .

Note that since the spheres are identical, the solutions are essentially the same: one sphere

originally had charge –1.00 × 10–6 C and the other had charge +3.00 × 10–6 C.

What if we had not made the assumption, above, that q1 + q2 ≥ 0? If the signs of the

charges were reversed (so q1 + q2 < 0), then the forces remain the same, so a charge of

+1.00 × 10–6 C on one sphere and a charge of –3.00 × 10–6 C on the other also satisfies

the conditions of the problem.

10. With rightwards positive, the net force on q3 is

F3 = F13 + F23 = k

q1q3

( L12 + L23 )

2

+k

q2 q3

.

L223

We note that each term exhibits the proper sign (positive for rightward, negative for

leftward) for all possible signs of the charges. For example, the first term (the force

exerted on q3 by q1) is negative if they are unlike charges, indicating that q3 is being

pulled toward q1, and it is positive if they are like charges (so q3 would be repelled from

q1). Setting the net force equal to zero L23= L12 and canceling k, q3 and L12 leads to

q1

q

+ q2 = 0 1 = −4.00.

4.00

q2

11. (a) Eq. 21-1 gives

( 20.0 ×10 C ) = 1.60N.

qq

F12 = k 1 22 = ( 8.99 ×109 N ⋅ m 2 C2 )

2

d

(1.50m )

−6

2

(b) A force diagram is shown as well as our choice of y axis (the dashed line).

The y axis is meant to bisect the line between q2 and q3 in order to make use of the

symmetry in the problem (equilateral triangle of side length d, equal-magnitude charges

q1 = q2 = q3 = q). We see that the resultant force is along this symmetry axis, and we

obtain

F qI

= 2 G k J cos 30° = 2.77 N .

H dK

2

Fy

2

12. (a) According to the graph, when q3 is very close to q1 (at which point we can

consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the

positive x direction. This is a repulsive force, then, so we conclude q1 has the same sign

as q3. Thus, q3 is a positive-valued charge.

(b) Since the graph crosses zero and particle 3 is between the others, q1 must have the

same sign as q2, which means it is also positive-valued. We note that it crosses zero at r

= 0.020 m (which is a distance d = 0.060 m from q2). Using Coulomb’s law at that point,

we have

q3 q2

q1 q3

=

4πεo r2 4πεo d2

or q2/q1 = 9.0.

2

d

q2 = §¨ 2 ·¸ q1 = 9.0 q1 ,

©r ¹

13. (a) There is no equilibrium position for q3 between the two fixed charges, because it is

being pulled by one and pushed by the other (since q1 and q2 have different signs); in this

region this means the two force arrows on q3 are in the same direction and cannot cancel.

It should also be clear that off-axis (with the axis defined as that which passes through the

two fixed charges) there are no equilibrium positions. On the semi-infinite region of the

axis which is nearest q2 and furthest from q1 an equilibrium position for q3 cannot be

found because |q1| < |q2| and the magnitude of force exerted by q2 is everywhere (in that

region) stronger than that exerted by q1 on q3. Thus, we must look in the semi-infinite

region of the axis which is nearest q1 and furthest from q2, where the net force on q3 has

magnitude

k

q1q3

q2 q3

−k

2

x

d+x

b

g

2

with d = 10 cm and x assumed positive. We set this equal to zero, as required by the

problem, and cancel k and q3. Thus, we obtain

q1

q2

−

2

x

d+x

b

F d + x IJ

= 0G

H x K

g

2

2

=

q2

=3

q1

which yields (after taking the square root)

d+x

d

= 3x=

≈ 14 cm

x

3 −1

for the distance between q3 and q1.

(b) As stated above, y = 0.

14. Since the forces involved are proportional to q, we see that the essential difference

between the two situations is Fa ∝ qB + qC (when those two charges are on the same side)

versus Fb ∝ −qB + qC (when they are on opposite sides). Setting up ratios, we have

qB + qC

Fa

=

Fb

- qB + qC

20.14

1+r

=

-2.877

-1 + r

where in the last step we have canceled (on the left hand side) 10−24 N from the numerator

and the denominator, and (on the right hand side) introduced the symbol r = qC /qB .

After noting that the ratio on the left hand side is very close to – 7, then, after a couple of

algebra steps, we are led to

r=

7 +1 8

= = 1.333.

7 −1 6

15. (a) The distance between q1 and q2 is

r12 =

( x2 − x1 ) + ( y2 − y1 )

2

2

=

( −0.020 − 0.035) + ( 0.015 − 0.005)

2

2

= 0.056 m.

The magnitude of the force exerted by q1 on q2 is

| q q | ( 8.99 ×10

F21 = k 1 2 2 =

r12

9

) ( 3.0 ×10 ) ( 4.0 ×10 ) = 35 N.

−6

−6

(0.056) 2

G

(b) The vector F21 is directed towards q1 and makes an angle θ with the +x axis, where

§ y2 − y1 ·

−1 § 1.5 − 0.5 ·

¸ = tan ¨

¸ = −10.3° ≈ −10°.

© −2.0 − 3.5 ¹

© x2 − x1 ¹

θ = tan −1 ¨

(c) Let the third charge be located at (x3, y3), a distance r from q2. We note that q1, q2 and

q3 must be collinear; otherwise, an equilibrium position for any one of them would be

impossible to find. Furthermore, we cannot place q3 on the same side of q2 where we also

find q1, since in that region both forces (exerted on q2 by q3 and q1) would be in the same

direction (since q2 is attracted to both of them). Thus, in terms of the angle found in part

(a), we have x3 = x2 – r cosθ and y3 = y2 – r sinθ (which means y3 > y2 since θ is negative).

The magnitude of force exerted on q2 by q3 is F23 = k | q2 q3 | r 2 , which must equal that of

the force exerted on it by q1 (found in part (a)). Therefore,

k

q2 q3

qq

q

= k 1 2 2 r = r12 3 = 0.0645 cm .

2

r

r12

q1

Consequently, x3 = x2 – r cosθ = –2.0 cm – (6.45 cm) cos(–10°) = –8.4 cm,

(d) and y3 = y2 – r sinθ = 1.5 cm – (6.45 cm) sin(–10°) = 2.7 cm.

16. (a) For the net force to be in the +x direction, the y components of the individual

forces must cancel. The angle of the force exerted by the q1 = 40 µC charge on

q3 = 20 µ C is 45°, and the angle of force exerted on q3 by Q is at –θ where

θ = tan −1

FG 2.0IJ = 33.7° .

H 3.0 K

Therefore, cancellation of y components requires

k

q1 q3

( 0.02 2 )

2

sin 45° = k

(

| Q | q3

(0.030) + (0.020)

2

2

)

2

sin θ

from which we obtain |Q| = 83 µC. Charge Q is “pulling” on q3, so (since q3 > 0) we

conclude Q = –83 µC.

(b) Now, we require that the x components cancel, and we note that in this case, the angle

of force on q3 exerted by Q is +θ (it is repulsive, and Q is positive-valued). Therefore,

k

q1 q3

( 0.02 2 )

2

cos 45° = k

(

Qq3

(0.030) + (0.020)

from which we obtain Q = 55.2 µC ≈ 55 µ C .

2

2

)

2

cos θ

17. (a) If the system of three charges is to be in equilibrium, the force on each charge

must be zero. The third charge q3 must lie between the other two or else the forces acting

on it due to the other charges would be in the same direction and q3 could not be in

equilibrium. Suppose q3 is at a distance x from q, and L – x from 4.00q. The force acting

on it is then given by

F3 =

4qq3 ·

1 § qq3

¨ 2 −

¸

2

4πε 0 ¨© x

( L − x ) ¸¹

where the positive direction is rightward. We require F3 = 0 and solve for x. Canceling

common factors yields 1/x2 = 4/(L – x)2 and taking the square root yields 1/x = 2/(L – x).

The solution is x = L/3.

The force on q is

Fq =

−1 § qq3 4.00q 2 ·

+

¨

¸.

L2 ¹

4 πε 0 © x 2

The signs are chosen so that a negative force value would cause q to move leftward. We

require Fq = 0 and solve for q3:

q3 = −

q

4qx 2

4

4

= − q 3 = − = −0.444

2

L

9

q

9

where x = L/3 is used. We may easily verify that the force on 4.00q also vanishes:

F4 q =

2

4qq0 ·

1 § 4q 2

1 § 4q 2 4 ( − 4 9 ) q ·

1 § 4q 2 4q 2 ·

=

+

=

−

=0.

¨ 2 +

¸

¨

¸

2

4πε 0 ¨© L

( 4 9 ) L2 ¸¹ 4πε 0 ¨© L2 L2 ¸¹

( L − x ) ¸¹ 4πε 0 ¨© L2

(b) As seen above, q3 is located at x = L/3. With L = 9.00 cm, we have x = 3.00 cm.

(c) Similarly, the y coordinate of q3 is y = 0.

1

18. (a) We note that cos(30º) = 2 3 , so that the dashed line distance in the figure is

r = 2d / 3 . We net force on q1 due to the two charges q3 and q4 (with |q3| = |q4| = 1.60 ×

10−19 C) on the y axis has magnitude

2

| q1q3 |

3 3 | q1q3 |

cos(30°) =

.

2

4πε 0 r

16πε 0 d 2

This must be set equal to the magnitude of the force exerted on q1 by q2 = 8.00 × 10−19 C

= 5.00 |q3| in order that its net force be zero:

3 3 | q1q3 |

| q1q2 |

=

2

16πε 0 d

4πε 0 ( D + d ) 2

§

D = d ¨2

©

·

− 1¸ = 0.9245 d .

3 3

¹

5

Given d = 2.00 cm, then this leads to D = 1.92 cm.

(b) As the angle decreases, its cosine increases, resulting in a larger contribution from the

charges on the y axis. To offset this, the force exerted by q2 must be made stronger, so

that it must be brought closer to q1 (keep in mind that Coulomb’s law is inversely

proportional to distance-squared). Thus, D must be decreased.

19. The charge dq within a thin shell of thickness dr is ρ A dr where A = 4πr2. Thus, with

ρ = b/r, we have

z

q = dq = 4 πb

z

r2

r1

c

h

r dr = 2πb r22 − r12 .

With b = 3.0 µC/m2, r2 = 0.06 m and r1 = 0.04 m, we obtain q = 0.038 µC = 3.8 × 10−8 C.

20. If θ is the angle between the force and the x-axis, then

cosθ =

x

.

x + d2

2

We note that, due to the symmetry in the problem, there is no y component to the net

force on the third particle. Thus, F represents the magnitude of force exerted by q1 or q2

on q3. Let e = +1.60 × 10−19 C, then q1 = q2 = +2e and q3 = 4.0e and we have

Fnet = 2F cosθ =

2(2e)(4e)

4πεo (x2 + d2)

4e2 x

x

=

.

πεo (x2 + d2 )3/2

x2 + d2

(a) To find where the force is at an extremum, we can set the derivative of this expression

equal to zero and solve for x, but it is good in any case to graph the function for a fuller

understanding of its behavior – and as a quick way to see whether an extremum point is a

maximum or a miminum. In this way, we find that the value coming from the derivative

procedure is a maximum (and will be presented in part (b)) and that the minimum is

found at the lower limit of the interval. Thus, the net force is found to be zero at x = 0,

which is the smallest value of the net force in the interval 5.0 m ≥ x ≥ 0.

(b) The maximum is found to be at x = d/ 2 or roughly 12 cm.

(c) The value of the net force at x = 0 is Fnet = 0.

(d) The value of the net force at x = d/ 2 is Fnet = 4.9 × 10−26 N.

21. (a) The magnitude of the force between the (positive) ions is given by

bqgbqg = k q

F=

4 πε 0r 2

2

r2

where q is the charge on either of them and r is the distance between them. We solve for

the charge:

q=r

F

= 5.0 × 10−10 m

k

c

h

3.7 × 10−9 N

= 3.2 × 10−19 C.

9

2

2

8.99 × 10 N ⋅ m C

(b) Let N be the number of electrons missing from each ion. Then, Ne = q, or

N=

q 3.2 × 10−9 C

=

= 2.

. × 10−19 C

e 16

22. The magnitude of the force is

FG

H

e2

N ⋅ m2

F = k 2 = 8.99 × 109

r

C2

−19

2

−10

2

. × 10 Ch

IJ c160

K c2.82 × 10 mh

= 2.89 × 10−9 N .

23. Eq. 21-11 (in absolute value) gives

q 10

. × 10−7 C

n= =

= 6.3 × 1011 .

−19

. × 10 C

e 16

24. (a) Eq. 21-1 gives

. × 10 Ch

c8.99 × 10 N ⋅ m C hc100

F=

. × 10 mh

c100

9

2

−16

2

−2

2

2

= 8.99 × 10−19 N .

(b) If n is the number of excess electrons (of charge –e each) on each drop then

n=−

. × 10−16 C

q

−100

=−

= 625.

160

. × 10−19 C

e

third laws lead to

c6.3 × 10 kghc7.0 m s h = 4.9 × 10

=ma m =

−7

m2 a2

1 1

2

2

2

9.0 m s

−7

kg.

(b) The magnitude of the (only) force on particle 1 is

2

q q

q

F = m1a1 = k 1 2 2 = 8.99 × 109

.

r

0.0032 2

c

h

Inserting the values for m1 and a1 (see part (a)) we obtain |q| = 7.1 × 10–11 C.

2. The magnitude of the mutual force of attraction at r = 0.120 m is

hc

hc

h

3.00 × 10−6 150

. × 10−6

q1 q2

9

F=k

= 8.99 × 10

= 2.81 N .

r2

0120

. 2

c

3. Eq. 21-1 gives Coulomb’s Law, F = k

k | q1 || q2 |

r=

=

F

(8.99 ×10 N ⋅ m

9

2

q1 q2

r2

, which we solve for the distance:

C2 ) ( 26.0 ×10−6 C ) ( 47.0 ×10−6 C )

5.70N

= 1.39m.

4. The fact that the spheres are identical allows us to conclude that when two spheres are

in contact, they share equal charge. Therefore, when a charged sphere (q) touches an

uncharged one, they will (fairly quickly) each attain half that charge (q/2). We start with

spheres 1 and 2 each having charge q and experiencing a mutual repulsive force

F = kq 2 / r 2 . When the neutral sphere 3 touches sphere 1, sphere 1’s charge decreases to

q/2. Then sphere 3 (now carrying charge q/2) is brought into contact with sphere 2, a total

amount of q/2 + q becomes shared equally between them. Therefore, the charge of sphere

3 is 3q/4 in the final situation. The repulsive force between spheres 1 and 2 is finally

(q / 2)(3q / 4) 3 q 2 3

F' 3

F′ = k

= k 2= F

= = 0.375.

2

r

F 8

8 r

8

5. The magnitude of the force of either of the charges on the other is given by

F=

b

1 q Q−q

r2

4 πε 0

g

where r is the distance between the charges. We want the value of q that maximizes the

function f(q) = q(Q – q). Setting the derivative df/dq equal to zero leads to Q – 2q = 0, or

q = Q/2. Thus, q/Q = 0.500.

6. For ease of presentation (of the computations below) we assume Q > 0 and q < 0

(although the final result does not depend on this particular choice).

(a) The x-component of the force experienced by q1 = Q is

§

·

Q )( Q )

| q |) ( Q ) ¸ Q | q | § Q / | q | ·

(

(

¨

F1x =

−

cos 45° +

2

2

¸ = 4πε a 2 ¨© − 2 2 + 1¸¹

a

4πε 0 ¨¨

0

¸

2a

©

¹

1

(

)

which (upon requiring F1x = 0) leads to Q / | q |= 2 2 , or Q / q = −2 2 = −2.83.

(b) The y-component of the net force on q2 = q is

§

·

| q |) ( Q ) ¸

(

1 ¨ | q |2

| q |2 § 1

Q·

F2 y =

sin

45

°

−

=

−

¨

2

2

2

¸ 4πε a © 2 2 | q | ¸¹

4πε 0 ¨¨ 2a

a

0

¸

©

¹

(

)

which (if we demand F2y = 0) leads to Q / q = −1/ 2 2 . The result is inconsistent with

that obtained in part (a). Thus, we are unable to construct an equilibrium configuration

with this geometry, where the only forces present are given by Eq. 21-1.

7. The force experienced by q3 is

G G

G

G

| q || q | ·

1 § | q3 || q1 | ˆ | q3 || q2 |

F3 = F31 + F32 + F34 =

j+

(cos45°ˆi + sin 45°ˆj) + 3 2 4 ˆi ¸

¨−

2

2

a

a

4πε 0 ©

( 2a )

¹

(a) Therefore, the x-component of the resultant force on q3 is

2 (1.0 ×10−7 ) § 1

| q3 | § | q2 |

·

·

9

F3 x =

+ | q4 | ¸ = ( 8.99 ×10 )

+ 2 ¸ = 0.17N.

¨

2 ¨

2

4πε 0 a © 2 2

(0.050)

¹

©2 2

¹

2

(b) Similarly, the y-component of the net force on q3 is

2 (1.0 ×10−7 ) §

| q3 | §

| q2 | ·

1 ·

9

F3 y =

− | q1 | +

= ( 8.99 ×10 )

−1 +

¸

¨

¸ = −0.046N.

2 ¨

2

4πε 0 a ©

(0.050)

2 2¹

2 2¹

©

2

8. (a) The individual force magnitudes (acting on Q) are, by Eq. 21-1,

k

q1 Q

b−a − g

a 2

2

=k

q2 Q

ba − g

a 2

2

which leads to |q1| = 9.0 |q2|. Since Q is located between q1 and q2, we conclude q1 and q2

are like-sign. Consequently, q1/q2 = 9.0.

(b) Now we have

k

q1 Q

b−a − g

3a 2

2

=k

q2 Q

ba − g

3a 2

2

which yields |q1| = 25 |q2|. Now, Q is not located between q1 and q2, one of them must

push and the other must pull. Thus, they are unlike-sign, so q1/q2 = –25.

9. We assume the spheres are far apart. Then the charge distribution on each of them is

spherically symmetric and Coulomb’s law can be used. Let q1 and q2 be the original

charges. We choose the coordinate system so the force on q2 is positive if it is repelled by

q1. Then, the force on q2 is

Fa = −

1 q1q2

= −k 1 2 2

2

4 πε 0 r

r

where r = 0.500 m. The negative sign indicates that the spheres attract each other. After

the wire is connected, the spheres, being identical, acquire the same charge. Since charge

is conserved, the total charge is the same as it was originally. This means the charge on

each sphere is (q1 + q2)/2. The force is now one of repulsion and is given by

1

Fb =

4 πε 0

q1 + q2

2

q1 + q2

2

d id i = k bq + q g .

2

1

r2

2

4r 2

We solve the two force equations simultaneously for q1 and q2. The first gives the product

b

gb

g

2

0.500 m 0108

.

N

r 2 Fa

q1q2 = −

=−

= −3.00 × 10−12 C 2 ,

9

2

k

8.99 × 10 N ⋅ m C 2

and the second gives the sum

q1 + q2 = 2r

Fb

= 2 0.500 m

k

b

g

0.0360 N

= 2.00 × 10−6 C

9

2

2

8.99 × 10 N ⋅ m C

where we have taken the positive root (which amounts to assuming q1 + q2 ≥ 0). Thus, the

product result provides the relation

q2 =

− ( 3.00 ×10−12 C2 )

q1

which we substitute into the sum result, producing

q1 −

3.00 × 10−12 C 2

= 2.00 × 10−6 C.

q1

Multiplying by q1 and rearranging, we obtain a quadratic equation

c

h

q12 − 2.00 × 10−6 C q1 − 3.00 × 10−12 C 2 = 0 .

The solutions are

q1 =

2.00 × 10−6 C ±

2

c−2.00 × 10 Ch − 4c−3.00 × 10

−6

2

−12

C2

h.

If the positive sign is used, q1 = 3.00 × 10–6 C, and if the negative sign is used,

q1 = −1.00 ×10−6 C .

(a) Using q2 = (–3.00 × 10–12)/q1 with q1 = 3.00 × 10–6 C, we get q2 = −1.00 ×10−6 C .

(b) If we instead work with the q1 = –1.00 × 10–6 C root, then we find q2 = 3.00 ×10−6 C .

Note that since the spheres are identical, the solutions are essentially the same: one sphere

originally had charge –1.00 × 10–6 C and the other had charge +3.00 × 10–6 C.

What if we had not made the assumption, above, that q1 + q2 ≥ 0? If the signs of the

charges were reversed (so q1 + q2 < 0), then the forces remain the same, so a charge of

+1.00 × 10–6 C on one sphere and a charge of –3.00 × 10–6 C on the other also satisfies

the conditions of the problem.

10. With rightwards positive, the net force on q3 is

F3 = F13 + F23 = k

q1q3

( L12 + L23 )

2

+k

q2 q3

.

L223

We note that each term exhibits the proper sign (positive for rightward, negative for

leftward) for all possible signs of the charges. For example, the first term (the force

exerted on q3 by q1) is negative if they are unlike charges, indicating that q3 is being

pulled toward q1, and it is positive if they are like charges (so q3 would be repelled from

q1). Setting the net force equal to zero L23= L12 and canceling k, q3 and L12 leads to

q1

q

+ q2 = 0 1 = −4.00.

4.00

q2

11. (a) Eq. 21-1 gives

( 20.0 ×10 C ) = 1.60N.

F12 = k 1 22 = ( 8.99 ×109 N ⋅ m 2 C2 )

2

d

(1.50m )

−6

2

(b) A force diagram is shown as well as our choice of y axis (the dashed line).

The y axis is meant to bisect the line between q2 and q3 in order to make use of the

symmetry in the problem (equilateral triangle of side length d, equal-magnitude charges

q1 = q2 = q3 = q). We see that the resultant force is along this symmetry axis, and we

obtain

F qI

= 2 G k J cos 30° = 2.77 N .

H dK

2

Fy

2

12. (a) According to the graph, when q3 is very close to q1 (at which point we can

consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the

positive x direction. This is a repulsive force, then, so we conclude q1 has the same sign

as q3. Thus, q3 is a positive-valued charge.

(b) Since the graph crosses zero and particle 3 is between the others, q1 must have the

same sign as q2, which means it is also positive-valued. We note that it crosses zero at r

= 0.020 m (which is a distance d = 0.060 m from q2). Using Coulomb’s law at that point,

we have

q3 q2

q1 q3

=

4πεo r2 4πεo d2

or q2/q1 = 9.0.

2

d

q2 = §¨ 2 ·¸ q1 = 9.0 q1 ,

©r ¹

13. (a) There is no equilibrium position for q3 between the two fixed charges, because it is

being pulled by one and pushed by the other (since q1 and q2 have different signs); in this

region this means the two force arrows on q3 are in the same direction and cannot cancel.

It should also be clear that off-axis (with the axis defined as that which passes through the

two fixed charges) there are no equilibrium positions. On the semi-infinite region of the

axis which is nearest q2 and furthest from q1 an equilibrium position for q3 cannot be

found because |q1| < |q2| and the magnitude of force exerted by q2 is everywhere (in that

region) stronger than that exerted by q1 on q3. Thus, we must look in the semi-infinite

region of the axis which is nearest q1 and furthest from q2, where the net force on q3 has

magnitude

k

q1q3

q2 q3

−k

2

x

d+x

b

g

2

with d = 10 cm and x assumed positive. We set this equal to zero, as required by the

problem, and cancel k and q3. Thus, we obtain

q1

q2

−

2

x

d+x

b

F d + x IJ

= 0G

H x K

g

2

2

=

q2

=3

q1

which yields (after taking the square root)

d+x

d

= 3x=

≈ 14 cm

x

3 −1

for the distance between q3 and q1.

(b) As stated above, y = 0.

14. Since the forces involved are proportional to q, we see that the essential difference

between the two situations is Fa ∝ qB + qC (when those two charges are on the same side)

versus Fb ∝ −qB + qC (when they are on opposite sides). Setting up ratios, we have

qB + qC

Fa

=

Fb

- qB + qC

20.14

1+r

=

-2.877

-1 + r

where in the last step we have canceled (on the left hand side) 10−24 N from the numerator

and the denominator, and (on the right hand side) introduced the symbol r = qC /qB .

After noting that the ratio on the left hand side is very close to – 7, then, after a couple of

algebra steps, we are led to

r=

7 +1 8

= = 1.333.

7 −1 6

15. (a) The distance between q1 and q2 is

r12 =

( x2 − x1 ) + ( y2 − y1 )

2

2

=

( −0.020 − 0.035) + ( 0.015 − 0.005)

2

2

= 0.056 m.

The magnitude of the force exerted by q1 on q2 is

| q q | ( 8.99 ×10

F21 = k 1 2 2 =

r12

9

) ( 3.0 ×10 ) ( 4.0 ×10 ) = 35 N.

−6

−6

(0.056) 2

G

(b) The vector F21 is directed towards q1 and makes an angle θ with the +x axis, where

§ y2 − y1 ·

−1 § 1.5 − 0.5 ·

¸ = tan ¨

¸ = −10.3° ≈ −10°.

© −2.0 − 3.5 ¹

© x2 − x1 ¹

θ = tan −1 ¨

(c) Let the third charge be located at (x3, y3), a distance r from q2. We note that q1, q2 and

q3 must be collinear; otherwise, an equilibrium position for any one of them would be

impossible to find. Furthermore, we cannot place q3 on the same side of q2 where we also

find q1, since in that region both forces (exerted on q2 by q3 and q1) would be in the same

direction (since q2 is attracted to both of them). Thus, in terms of the angle found in part

(a), we have x3 = x2 – r cosθ and y3 = y2 – r sinθ (which means y3 > y2 since θ is negative).

The magnitude of force exerted on q2 by q3 is F23 = k | q2 q3 | r 2 , which must equal that of

the force exerted on it by q1 (found in part (a)). Therefore,

k

q2 q3

q

= k 1 2 2 r = r12 3 = 0.0645 cm .

2

r

r12

q1

Consequently, x3 = x2 – r cosθ = –2.0 cm – (6.45 cm) cos(–10°) = –8.4 cm,

(d) and y3 = y2 – r sinθ = 1.5 cm – (6.45 cm) sin(–10°) = 2.7 cm.

16. (a) For the net force to be in the +x direction, the y components of the individual

forces must cancel. The angle of the force exerted by the q1 = 40 µC charge on

q3 = 20 µ C is 45°, and the angle of force exerted on q3 by Q is at –θ where

θ = tan −1

FG 2.0IJ = 33.7° .

H 3.0 K

Therefore, cancellation of y components requires

k

q1 q3

( 0.02 2 )

2

sin 45° = k

(

| Q | q3

(0.030) + (0.020)

2

2

)

2

sin θ

from which we obtain |Q| = 83 µC. Charge Q is “pulling” on q3, so (since q3 > 0) we

conclude Q = –83 µC.

(b) Now, we require that the x components cancel, and we note that in this case, the angle

of force on q3 exerted by Q is +θ (it is repulsive, and Q is positive-valued). Therefore,

k

q1 q3

( 0.02 2 )

2

cos 45° = k

(

Qq3

(0.030) + (0.020)

from which we obtain Q = 55.2 µC ≈ 55 µ C .

2

2

)

2

cos θ

17. (a) If the system of three charges is to be in equilibrium, the force on each charge

must be zero. The third charge q3 must lie between the other two or else the forces acting

on it due to the other charges would be in the same direction and q3 could not be in

equilibrium. Suppose q3 is at a distance x from q, and L – x from 4.00q. The force acting

on it is then given by

F3 =

4qq3 ·

1 § qq3

¨ 2 −

¸

2

4πε 0 ¨© x

( L − x ) ¸¹

where the positive direction is rightward. We require F3 = 0 and solve for x. Canceling

common factors yields 1/x2 = 4/(L – x)2 and taking the square root yields 1/x = 2/(L – x).

The solution is x = L/3.

The force on q is

Fq =

−1 § qq3 4.00q 2 ·

+

¨

¸.

L2 ¹

4 πε 0 © x 2

The signs are chosen so that a negative force value would cause q to move leftward. We

require Fq = 0 and solve for q3:

q3 = −

q

4qx 2

4

4

= − q 3 = − = −0.444

2

L

9

q

9

where x = L/3 is used. We may easily verify that the force on 4.00q also vanishes:

F4 q =

2

4qq0 ·

1 § 4q 2

1 § 4q 2 4 ( − 4 9 ) q ·

1 § 4q 2 4q 2 ·

=

+

=

−

=0.

¨ 2 +

¸

¨

¸

2

4πε 0 ¨© L

( 4 9 ) L2 ¸¹ 4πε 0 ¨© L2 L2 ¸¹

( L − x ) ¸¹ 4πε 0 ¨© L2

(b) As seen above, q3 is located at x = L/3. With L = 9.00 cm, we have x = 3.00 cm.

(c) Similarly, the y coordinate of q3 is y = 0.

1

18. (a) We note that cos(30º) = 2 3 , so that the dashed line distance in the figure is

r = 2d / 3 . We net force on q1 due to the two charges q3 and q4 (with |q3| = |q4| = 1.60 ×

10−19 C) on the y axis has magnitude

2

| q1q3 |

3 3 | q1q3 |

cos(30°) =

.

2

4πε 0 r

16πε 0 d 2

This must be set equal to the magnitude of the force exerted on q1 by q2 = 8.00 × 10−19 C

= 5.00 |q3| in order that its net force be zero:

3 3 | q1q3 |

| q1q2 |

=

2

16πε 0 d

4πε 0 ( D + d ) 2

§

D = d ¨2

©

·

− 1¸ = 0.9245 d .

3 3

¹

5

Given d = 2.00 cm, then this leads to D = 1.92 cm.

(b) As the angle decreases, its cosine increases, resulting in a larger contribution from the

charges on the y axis. To offset this, the force exerted by q2 must be made stronger, so

that it must be brought closer to q1 (keep in mind that Coulomb’s law is inversely

proportional to distance-squared). Thus, D must be decreased.

19. The charge dq within a thin shell of thickness dr is ρ A dr where A = 4πr2. Thus, with

ρ = b/r, we have

z

q = dq = 4 πb

z

r2

r1

c

h

r dr = 2πb r22 − r12 .

With b = 3.0 µC/m2, r2 = 0.06 m and r1 = 0.04 m, we obtain q = 0.038 µC = 3.8 × 10−8 C.

20. If θ is the angle between the force and the x-axis, then

cosθ =

x

.

x + d2

2

We note that, due to the symmetry in the problem, there is no y component to the net

force on the third particle. Thus, F represents the magnitude of force exerted by q1 or q2

on q3. Let e = +1.60 × 10−19 C, then q1 = q2 = +2e and q3 = 4.0e and we have

Fnet = 2F cosθ =

2(2e)(4e)

4πεo (x2 + d2)

4e2 x

x

=

.

πεo (x2 + d2 )3/2

x2 + d2

(a) To find where the force is at an extremum, we can set the derivative of this expression

equal to zero and solve for x, but it is good in any case to graph the function for a fuller

understanding of its behavior – and as a quick way to see whether an extremum point is a

maximum or a miminum. In this way, we find that the value coming from the derivative

procedure is a maximum (and will be presented in part (b)) and that the minimum is

found at the lower limit of the interval. Thus, the net force is found to be zero at x = 0,

which is the smallest value of the net force in the interval 5.0 m ≥ x ≥ 0.

(b) The maximum is found to be at x = d/ 2 or roughly 12 cm.

(c) The value of the net force at x = 0 is Fnet = 0.

(d) The value of the net force at x = d/ 2 is Fnet = 4.9 × 10−26 N.

21. (a) The magnitude of the force between the (positive) ions is given by

bqgbqg = k q

F=

4 πε 0r 2

2

r2

where q is the charge on either of them and r is the distance between them. We solve for

the charge:

q=r

F

= 5.0 × 10−10 m

k

c

h

3.7 × 10−9 N

= 3.2 × 10−19 C.

9

2

2

8.99 × 10 N ⋅ m C

(b) Let N be the number of electrons missing from each ion. Then, Ne = q, or

N=

q 3.2 × 10−9 C

=

= 2.

. × 10−19 C

e 16

22. The magnitude of the force is

FG

H

e2

N ⋅ m2

F = k 2 = 8.99 × 109

r

C2

−19

2

−10

2

. × 10 Ch

IJ c160

K c2.82 × 10 mh

= 2.89 × 10−9 N .

23. Eq. 21-11 (in absolute value) gives

q 10

. × 10−7 C

n= =

= 6.3 × 1011 .

−19

. × 10 C

e 16

24. (a) Eq. 21-1 gives

. × 10 Ch

c8.99 × 10 N ⋅ m C hc100

F=

. × 10 mh

c100

9

2

−16

2

−2

2

2

= 8.99 × 10−19 N .

(b) If n is the number of excess electrons (of charge –e each) on each drop then

n=−

. × 10−16 C

q

−100

=−

= 625.

160

. × 10−19 C

e

## Solution manual fundamentals of physics extended, 8th edition ch01

## Solution manual fundamentals of physics extended, 8th editionch02

## Solution manual fundamentals of physics extended, 8th editionch03

## Solution manual fundamentals of physics extended, 8th editionch04

## Solution manual fundamentals of physics extended, 8th editionch05

## Solution manual fundamentals of physics extended, 8th editionch06

## Solution manual fundamentals of physics extended, 8th editionch07

## Solution manual fundamentals of physics extended, 8th editionch08

## Solution manual fundamentals of physics extended, 8th editionch09

## Solution manual fundamentals of physics extended, 8th editionch10

Tài liệu liên quan