21 30 solutions

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21.13. IDENTIFY: Apply Coulomb’s law. The two forces on3

q must have equal magnitudes and opposite directions.

SET UP: Like charges repel and unlike charges attract.

EXECUTE: The force2

F that2

q exerts on 3q has magnitude

2 3

2 2

2

q qF k

r and is in the + x direction.

1 F must be

in the x direction, so 1q must be positive. 1 2

F F gives1 3 2 3

2 2

1 2

q q q qk k

r r .

2 2

11 2

2

2.00 cm(3.00 nC) 0.750 nC4.00 cm

r q qr

.

EVALUATE: The result for the magnitude of 1q doesn’t depend on the magnitude of

2q .

21.24. IDENTIFY: Apply2

k qqF

r to find the force of each charge on q . The net force is the vector sum of the

individual forces.

SET UP: Let 12.50 Cq and 2 3.50 Cq . The charge q must be to the left of 1q or to the right of

2q in

order for the two forces to be in opposite directions. But for the two forces to have equal magnitudes, q must be

closer to the charge 1q , since this charge has the smaller magnitude. Therefore, the two forces can combine to give

zero net force only in the region to the left of 1q . Let q be a distance d to the left of 1

q , so it is a distance

0.600 md from 2q .

EXECUTE: 1 2F F gives

1 2

2 2( 0.600 m)

kq q kq q

d d .

1

2

( 0.600 m) (0.8452)( 0.600 m)q

d d d q

. d must

be positive, so(0.8452)(0.600 m)

3.27 m1 0.8452

d . The net force would be zero when q is at 3.27 m x .

EVALUATE: When q is at 3.27 m x ,1

F is in the x direction and2

F is in the + x direction.

21.96. IDENTIFY: Divide the semicircle into infinitesimal segments. Find the electric field d E due to each segment

and integrate over the semicircle to find the total electric field.

SET UP: The electric fields along the x-direction from the left and right halves of the semicircle cancel. The

remaining y-component points in the negative y-direction. The charge per unit length of the semicircle isQ

aand

2

k dl k d dE

a a.

EXECUTE: sinsin y

k d dE dE

a. Therefore,

22

0 20

2 2 2 2sin [ cos ] y

k k k kQ E d

a a a a,

in the -direction y .

EVALUATE: For a full circle of charge the electric field at the center would be zero. For a quarter-circle of

charge, in the first quadrant, the electric field at the center of curvature would have nonzero x and y components.

The calculation for the semicircle is particularly simple, because all the charge is the same distance from point P.

22.4.IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge.

SET UP: ˆ ˆ( 5.00 N/C m) (3.00 N/C m) x z E = i+ k . The area of each face is 2 L , where 0.300 m L .

EXECUTE:1 11

ˆˆ ˆ 0s S A n = j E n .

2 2

2

2ˆˆ ˆ (3.00 N C m)(0.300 m) (0.27 ( N C) m)S S A z z n = k E n .

2

2(0.27 (N/C)m)(0.300 m) 0.081 (N/C) m .

3 33ˆˆ ˆ 0S S A n = j E n .

4 44ˆˆ ˆ (0.27 (N/C) m) 0 (since 0).S S A z z n = k E n

5 5

2

5ˆ

ˆ ˆ ( 5.00 N/C m)(0.300 m) (0.45 (N/C) m) .S S A x x n = i E n

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2

5 (0.45 (N/C) m)(0.300 m) (0.135 (N/C) m ).

6 66ˆˆ ˆ (0.45 (N/C) m) 0 (since 0).S S A x x n = i E n

(b)Total flux: 2 2

2 5 (0.081 0.135)(N/C) m 0.054 N m /C. Therefore,13

0 4.78 10 C.q P

EVALUATE: Flux is positive when E is directed out of the volume and negative when it is directed into the

volume.

23.48. IDENTIFY: Apply Eq.(21.19).

SET UP: Eq.(21.7) says2

0

1ˆ

4

q

r E r

Pis the electric field due to a point charge q.

EXECUTE: (a) 2 2 2 3 2 32 2 2

.( )

x

V kQ kQx kQx E

x x x y z r x y z

Similarly,3 3

and . y z

kQy kQz E E

r r

(b) From part (a),2 2

ˆ ˆ ˆ

ˆ,kQ x y z kQ

E r r r r r

i j k r which agrees with Equation (21.7).

EVALUATE: V is a scalar. E is a vector and has components.

24.18. IDENTIFY: For capacitors in parallel the voltages are the same and the charges add. For capacitors in series, the

charges are the same and the voltages add. / C Q V .

SET UP: 1

C and 2C are in parallel and 3

C is in series with the parallel combination of 1

C and 2C .

EXECUTE: (a) 1 2andC C are in parallel and so have the same potential across them:

6

2

1 2 6

2

40.0 10 C13.33 V

3.00 10 F

QV V

C . Therefore, 6 6

1 1 1 (13.33 V)(3.00 10 F) 80.0 10 CQ V C . Since 3C

is in series with the parallel combination of 1 2andC C , its charge must be equal to their combined charge:6 6 6

3 40.0 10 C 80.0 10 C 120.0 10 CC .

(b) The total capacitance is found from6 6

tot 12 3

1 1 1 1 1

9.00 10 F 5.00 10 FC C C and tot

3.21 FC .

6

tot

6

tot

120.0 10 C37.4 V

3.21 10 Fab

QV

C .

EVALUATE: 6

3

3 6

3

120.0 10 C

24.0 V5.00 10 F

Q

V C . 1 3abV V V .

24.67. (a) IDENTIFY: The conductor can be at some potential V , where V = 0 far from the conductor. This potential

depends on the charge Q on the conductor so we can define C = Q/V where C will not depend on V or Q.

(b) SET UP: Use the expression for the potential at the surface of the sphere in the analysis in part (a).

EXECUTE: For any point on a solid conducting sphere 0 / 4 if 0 at .V Q R V r P

00

44

Q RC Q R

V Q

PP

(c)12 6 4

04 4 8.854 10 F/m 6.38 10 m 7.10 10 F 710 F.C RP

EVALUATE: The capacitance of the earth is about seven times larger than the largest capacitances in this range.

The capacitance of the earth is quite small, in view of its large size.

25.1.IDENTIFY: / I Q t .

SET UP: 1.0 h 3600 s

EXECUTE: 4(3.6 A)(3.0)(3600s) 3.89 10 C.Q It

EVALUATE: Compared to typical charges of objects in electrostatics, this is a huge amount of charge.

25.7.IDENTIFY and SET UP: Apply Eq. (25.1) to find the charge dQ in time dt . Integrate to find the total charge in the

whole time interval.

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8/4/2019 21 30 Solutions


SET UP: Since the straight sections produce no field at P, the field at P is 0

4

I B

R.

EXECUTE: 0

4

I B

R. The direction of B is given by the right-hand rule: B is directed into the page.

EVALUATE: For a quarter-circle section of wire the magnetic field at its center of curvature is 0

8

I B

R.

28.31. IDENTIFY: Calculate the magnetic field vector produced by each wire and add these fields to get the total field.

SET UP: First consider the field at P produced by the current1

I in the upper semicircle of wire. See Figure

28.31a.

Consider the three parts of this wire

a: long straight section,

b: semicircle

c: long, straight section

Figure 28.31a

Apply the Biot-Savart law 0 0

2 3

ˆ

4 4

Id Id d

r r

l r l r B = = to each piece.

EXECUTE: part a See Figure 28.31b.

0,d l r =

so 0dB

Figure 28.31b

The same is true for all the infinitesimal segments that make up this piece of the wire, so B = 0 for this piece.

part c See Figure 28.31c.

0,d l r =

so 0 and 0dB B for this piece.

Figure 28.31c

part b See Figure 28.31d.

d l r is directed into the paper for all infinitesimal

segments that make up this semicircular piece, so B

is directed into the paper and B dB(the vector sum

of the d B is obtained by adding their magnitudes

since they are in the same direction).

Figure

28.31d

sin .d r dll r The angle between and is 90 and ,d r Rl r the radius of the semicircle. Thus d R dll r

0 0 1 0 1

3 3 24 4 4

I d I R I dB dl dl

r R R

l r

0 1 0 1 0 1

2 2( )

4 4 4

I I I B dB dl R

R R R

(We used that dl is equal to , R the length of wire in the semicircle.) We have shown that the two straight

sections make zero contribution to , B so 1 0 1 / 4 B I R and is directed into the page.

For current in the direction shown in

Figure 28.31e, a similar analysis gives

2 0 2 / 4 , B I R out of the paper

Figure 28.31e

1 2and B B are in opposite directions, so the magnitude of the net field at P is0 1 2

1 2.

4

I I B B B

R

EVALUATE: When 1 2 , 0. I I B

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28.37. IDENTIFY: Apply Ampere’s law.

SET UP: To calculate the magnetic field at a distance r from the center of the cable, apply Ampere’s law to a

circular path of radius r . By symmetry, (2 )d B r B l ú for such a path.

EXECUTE: (a) For 0

encl 0 0, 2 .

2

μ I a r b I I d μ I B πr μ I B

πr B l ú

(b) For ,r c the enclosed current is zero, so the magnetic field is also zero.

EVALUATE: A useful property of coaxial cables for many applications is that the current carried by the cable

doesn’t produce a magnetic field outside the cable. 28.38. IDENTIFY: Apply Ampere's law to calculate . B

(a) SET UP: For a < r < b the end view is shown in Figure 28.38a.

Apply Ampere's law to a circle of radius r ,

where a < r < b. Take currents 1 2and I I to

be directed into the page. Take this direction

to be positive, so go around the integration

path in the clockwise direction.

Figure

28.38a

EXECUTE: 0 encld I B l =ú

encl 1(2 ),d B r I I B l =ú

Thus 0 1

0 1(2 ) and

2

I B r I B

r

(b) SET UP: r > c: See Figure 28.38b.

Apply Ampere's law to a circle of

radius r , where r > c. Both

currents are in the positive

direction.

Figure

28.38b

EXECUTE: 0 encld I B l =ú

encl 1 2(2 ),d B r I I I B l =ú

Thus 0 1 2

0 1 2

( )(2 ) ( ) and

2

I I B r I I B

r

EVALUATE: For a < r < b the field is due only to the current in the central conductor. For r > c both currents

contribute to the total field.

29.67. IDENTIFY: Use Eq.(29.10) to calculate the induced electric field at each point and then use .q F = E

SET UP: Apply Bd

d dt

E l =ú to a concentric circle of

radius r , as shown in Figure 29.67a. Take A to

be into the page, in the direction of . B

Figure

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29.67a

EXECUTE: B increasing then gives 0, so Bd d

dt E l ú is negative. This means that E is tangent to the circle

in the counterclockwise direction, as shown in Figure 29.67b.

(2 )d E r E l =ú

2 Bd dB

r dt dt

Figure

29.67b

2 1

2(2 ) so

dB dB E r r E r

dt dt

point a The induced electric field and the force on q are shown in Figure 29.67c.

1

2

dBF qE qr

dt

F is to the left

( F is in the same direction as E since

q is positive.)Figure 29.67c

point b The induced electric field and the force on q are shown in Figure 29.67d.

1

2

dBF qE qr

dt

F is toward the top of the page.

Figure

29.67d

point c r = 0 here, so E = 0 and F = 0.

EVALUATE: If there were a concentric conducting ring of radius r in the magnetic field region, Lenz’s law tells

us that the increasing magnetic field would induce a counterclockwise current in the ring. This agrees with the

direction of the force we calculated for the individual positive point charges.

30.1. IDENTIFY and SET UP: Apply Eq.(30.4).

EXECUTE: (a) 412

(3.25 10 H)(830 A/s) 0.270 V;di

M dt

E yes, it is constant.

(b) 21

;di

M dt

E M is a property of the pair of coils so is the same as in part (a). Thus1 0.270 V.E

EVALUATE: The induced emf is the same in either case. A constant / di dt produces a constant emf.

21 30 solutions

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