Example 16.6: Photocopy machine. A photocopy machine works by arranging positive charges (in the pattern to be copied) on the surface of a drum, then gently sprinkling negatively charged dry toner (ink) particles onto the drum. The toner particles temporarily stick to the pattern on the drum (Fig. 16-25) and are later transferred to paper and “melted” to produce the copy. Suppose each toner particle has a mass of 9 .0 ´ 10−16

kg and carries an average of 20 extra electrons to provide an electric charge. Assuming that the electric force on a toner particle must exceed twice its weight in order to ensure sufficient attraction, compute the required electric field strength near the surface of the drum.APPROACH The electric force on a toner particle of charge q = 20e is F = qE, where E is the needed electric field. This force needs to be at least as great as twice the weight (mg) of the particle.SOLUTION The minimum value of electric field satisfies the relation

qE=2 mgwhere q =20e. Hence

E=2 mgq

=2 ( 9. 0 ´ 10−16 kg ) ( 9 .8 m / s2 )20 (1 .6 ´ 10−19C )

=5 .5 ´ 103 N /C .

Example 16.7: Electric field of a single point charge. Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = −3 .0´ 10−6 C . APPROACH The magnitude of the electric field due to a single point charge is given by Eq. 16-4. The direction is found using the sign of the charge Q. SOLUTION The magnitude of the electric field is:

E=kQr2

=(9 .0´ 109 N⋅m2 /C2) (3 . 0 ´ 10−6 C )

(0 .30 m)2=3 . 0´ 105 N /C .

The direction of the electric field is toward the charge Q, to the left as shown in Fig. 16-26a, since we defined the direction as that of the force on a positive test charge which here would be attractive. If Q had been positive, the electric field would have pointed away, as in Fig. 16-26b.NOTE There is no electric charge at point P. But there is an electric field there. The only real charge is Q.

Example 16.8: E at a point between two charges. Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 microCoulombs and the other +50 microCoulombs. (a) Determine the direction and the magnitude of the electric field at a point P between the two charges that

is 2.0 cm from the negative charge (Fig. 16-27a). (b) If an electron (mass = ) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?

APPROACH The electric field at P will be the vector sum of the fields created separately by and . The field due to

the negative charge points toward , and the field due to the positive charge points away from . Thus both fields point to the left as shown in Fig. 16-27b, and we can add the magnitudes of the two fields together algebraically, ignoring the signs of the charges. In (b) we use Newton’s second law (F = ma) to determine the acceleration, where F=qE (Eq. 16-5).

SOLUTION (a) Each field is due to a point charge as given by Eq. 16-4, E = kQ/ . The total field is

E=kQ1

r12 +k

Q 2

r22 =k (Q1

r12 +

Q2

r22 )

¿ (9.0 × 109 N ∙ m2/C2 )( 25× 10−6 C

(2.0 ×10−2m )2+ 50 ×10−6C

( 8.0× 10−2 m )2 )¿6.3 ×108 N /C(b) The electric field points to the left, so the electron will feel a force to the right since it is negatively charged. Therefore the acceleration a = F/m (Newton’s second law) will be to the right. The force on a charge q in an electric field E is F = qE (Eq. 16-5). Hence the magnitude of the electron’s initial acceleration is

a= Fm

=qEm

=(1.60×10−19C ) (6.3 ×108 N /C )

9.11× 10−31kg=1.1× 1020m / s2

NOTE By carefully considering the directionsof each field ¿ and ) before doing any calculations, we made sure our calculation could be done simply and correctly.

Example 16.9(1): E above two point charges. Calculate the total electric field (a) at point A and (b) at point B in Fig. 16-28 due to both charges, Q1 and Q2.APPROACH The calculation is much like that of Example 16-4, except now we are dealing with electric fields instead of force. The electric field at point A is the vector sum of the fields E⃗A 1 due to Q1, and E⃗A 2 due to Q2. We find the magnitude of the field produced by each point charge, then we add their components to find the total field at point A. We do the same for point B.SOLUTION (a) The magnitude of the electric field produced at point A by each of the charges Q1 and Q2 is given by E = kQ/r2, so

EA 1=(9.0 ×109 N ∙m2/C2)(50 × 10−6 C)

(0.60 m)2 =1.25 ×106 N /C,

EA 2=(9.0 ×109 N ∙m2/C2)(50 × 10−6 C)

(0.30 m)2 =5.0 ×106 N /C.

The direction of EA1 points from A towards Q1 (negative charge) whereas EA2 points from A away from Q2, as shown; so the total electric field at A, E⃗A, has components

EAx=EA 1cos 30°=1.1× 106 N /C,

EAy=EA2−EA1 sin 30°=4.4 × 106 N /C.

Thus the magnitude of E⃗A is

EA=√ (1.1 )2+(4.4)2× 106 N /C=4.5 ×106 N /C ,

And its direction is Φ = 76o.(b) Because B is equidistant (40 cm by the Pythagorean theorem) from the two equal charges, the magnitudes of EB1 and EB2 are the same; that is,

EB1−EB2=kQr 2 =

(9.0 ×109 N ∙ m2/C2 ) (50 × 10−6 C )(0.40 m )2

¿2.8 ×106 N /C .Also, because of the symmetry, the y components are equal and opposite, and so cancel out. Hence the total field EB is horizontal and equals EB1 cos θ + EB2 cos θ = 2EB1 cos θ. From the diagram, cos θ = 26 cm / 40 cm = 0.65 Then

EB=2 EB1cosθ=2 (2.8 ×106 N /C ) (0.65 )¿3.6 ×106 N /C ,

And the direction of E⃗B is along the + x direction.NOTE We could have done part (b) the same way we did part (a). But symmetry allowed us to solve the problem with less effort.

Example 16.9(2): Repeated. Calculate the total electric field at point B in Fig. 16-28 due to both charges, Q1 and Q2.APPROACH and SOLUTION1. Draw a careful diagram. The directions of the electric fields E⃗B1 and E⃗B2, as well as the net field E⃗B, are shown in Fig.

16-28. E⃗B2 points away from the positive charge Q2; E⃗B1 points toward the negative charge Q1.2. Apply Coulomb’s law to find the magnitudes of the contributing electric fields. Because B is equidistance (40 cm by the

Pythagorean theorem) from the two equal charges, the magnitudes of EB1 and EB2 are the same; that is

EB1=EB2=kQr 2 =

(9.0 ×109 N ∙ m2/C2 ) (50 × 10−6 C )(0.40 m )2

¿2.8 ×106 N /C .

3. Add vectorially, and use symmetry when possible. The y components of E⃗B1 and E⃗B2 are equal and opposite. Because of this symmetry, the total field EB is horizontal and equals EB1 cos θ + EB2 cos θ = 2 EB1 cos θ. From Fig. 16-28, cos θ = 26 cm / 40 cm = 0.65. Then

EB=2 EB1cosθ=2 (2.8 ×106 N /C ) (0.65 ) ¿3.6 ×106 N /C ,

And the direction of E⃗B is along the + x direction.NOTE Part (a) of Example 16-9 exhibited no useful symmetry.

Conceptual Example 17.1: A Negative ChargeSuppose a negative charge, such as an electron, is placed near the negative plate in Fig. 17-1, at point b, shown here in Fig. 17-2. If the electron is free to move, will its electric potential increase of decrease? How will the electric potential change?RESPONSE: An electron released at point b will move toward the positive plate. As the electron moves toward the positive plate, its potential energy decreases as its kinetic energy gets larger. But note that the electron moves from point b at low

potential to point a at high potential: . (The potentials Va and Vb are due to the charges on the plates, not due to the electron.)NOTE: A positive charge placed near the negative plate at b would not be accelerated. A positive charge tends to move from high potential to low.

Example 17.2: Electron in TV TubeSuppose an electron in the picture of a television set is accelerated from rest through a potential differenceV b−V a=V ba=+5000 V

(Fig. 17-4). (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 X 10-31 kg) as a result of this acceleration?APPROACH: The electron, accelerated toward the positive plate, will decrease in potential energy by an amountΔ PE=qV ba (Eq. 17-3). The loss in potential energy will equal its gain in kinetic energy (energy conservation).SOLUTION: (a) The charge on an electron is q = -e = -1.6 X 10-16 C. Therefore its change in potential energy is

Δ PE=qV ba=(−1.6×10−16C )(+5000V )=−8.0×10−16 J .The minus sign indicates that the potential energy decreases. The potential difference, Vba, has a positive sign since the final potential Vb is higher than the initial potential Va; negative electrons are attracted toward a positive electrode and repelled away from a negative electrode.

(b) The potential energy lost by the electron becomes kinetic energy KE. From conservation of energy, Δ KE+ Δ PE=0 so Δ KE=−Δ PE12

mv2−0=−q (V b−V a )=−qV ba,

where the initial kinetic energy is zero since we are given that the electron started from rest. We solve for v:

v=√−2qV ba

m=√−2(−1. 6×10−19C )(5000 V )

9 .1×10−31 kg=4 . 2×107 m /s

NOTE: The potential energy doesn’t depend on the mass, only on the charge and voltage. The speed does depend on m.

Example 17.4: Potential Energy due to a Positive or a Negative ChargeDetermine the potential at a point 0.50 m (a) from a +20 C point charge, (b) from a -20 C point charge.APPROACH: The potential energy due to a point charge is given by Eq. 17-5, V=kQ/r.SOLUTION: (a) at a distance of 0.50 m from a positive 20C point charge, the potential is

V=kQr=(9 .0×109 N⋅m2/C2)(20×10−6C

0 .50 m )=3.6×105V.

(b) For the negative charge,

V=kQr=(9 .0×109 N⋅m2/C2)(−20×10−6C

0 .50 m )=−3 .6×105V.

NOTE: Potential can be positive or negative. In contrast to calculations of electric field magnitudes, for which we usually ignore the sign of the charges, it is important to include a charge’s sign when we find electric potential.

Example 17.5: Work done to bring two positive charges close together.What minimum work must be done by an external force to bring a charge q=3.00μC from a great distance away (taker=∞) to a point 0.500m from a chargeQ=20.0 μC?

Approach: To find the work we cannot simply multiply the force times distance because the force is not constant. Instead we can set the change in potential energy equal to the (positive of the) work required of an external force (Chapter 6), and

W =∆ PE=q(V b−V a). We get the potentials V b and V a usingV=kQr

.

Solution: The work required is equal to the change in potential energy:

W =q (V b−V a )=q( kQrb

− kQra

),

Where rb=0.500 mand ra=∞. The right-hand term within the parentheses is zero (1/∞=0) so

W =(3.00 ×10−6C )((8.99× 109 N ∙m2

C2 ) (2.00× 10−5C )

0.500 m)=1.08 J

.

Note: we could not use Eqs. 17-4 here because they apply only to uniform fields. But we did use Eq. 17-3(W =∆ PE=q(V b−V a)) because it is always valid.

Conceptual Example 17.9: Inserting a dielectric at constant V.An air-filled capacitor consisting of two parallel plates separated by a distance d is connected to a battery of voltage V and acquires a charge Q. While it is still connected to the battery, a slab of dielectric material with K=3 is inserted between the plates of the capacitor. Will Q increase, decrease, or stay the same?

Response: Since the capacitor remains connected to the battery, the voltage stays constant and equal to the battery voltage V.

The capacitance C increase when the dielectric material is inserted because K in C=K ϵ 0Ad

has increased. From the relation

Q=CV, if V stays constant, but C increases, Q must increase as well. As the dielectric is inserted, more charge will be pulled from the battery and deposited onto the plates of the capacitor as its capacitance increases.

Conceptual Example 17.10: Inserting a dielectric into an isolated capacitor.Suppose the air-filled capacitor of Example 17.9 is charged (to Q) and then disconnected from the battery. Next a dielectric is inserted between the plates. Will Q, C or V change?

Response: The charge Q remains the same—the capacitor is isolated, so there is nowhere for the charge to go. The

capacitance increases as a result of inserting the dielectric (C=K ϵ 0Ad

). The voltage across the capacitor also changes—it

decreases because, by Q=CV, so V=Q/C; if Q stays constant and C increases (it is in the denominator), then V decreases.

Example 18-8: Headlights. Calculate the resistance of a 40-W automobile headlight designed for 12V.APPROACH We are given the power and the potential difference across the headlight, so we solve Eq. 18-6b for R.SOLUTION Given P=40W and V=12V, and solving Eq. 18.6b for R, we obtain

R=V ²P

=(12 V ) ²(40W )

=3.6 Ω

NOTE This is the resistance when the bulb is burning brightly at 40W. when the bulb is cold, the resistance is much lower, as we saw in ρT=ρ0[1+α ( T−T0 )]. Since the current is high when the resistance is low, light bulbs burn out most often when first turned on.

Example 18-9: Electric heater. An electric heater draws a steady 15.9 A on a 120-V line. How much power does it require and how much does it cost per month (30 days) if it operates 3.9 h per day and the electric company charges 9.2 cents per k Wh?APPROACH Given the current and voltage, we use Eq. 18-5 to find the power. We must multiply the power (in k W) by the time (h) used in a month to find the energy transformed in a month, and then multiply by the cost per energy unit. $0.092 per k Wh, to get the cost per month.SOLUTION The power is

P−IV =(15.0 A )(120 V )¿1800 W

Or 1.80 k W. The time (in hours) the heater is used per month is (3.0 h/d)(30d) = 90 h, which at 9.2 ₵/k Wh would cost (1.80 k W)(90 h)($0.092 /k Wh) = $15.NOTE Household current is actually alternating (ac), but our solution is still valid assuming the given values for V and I are the proper averages (rms) as discussed in Section 18-7.

Example 18-10: Lightning bolt. Lightning is spectacular example of electric current in a natural phenomenon (Fig. 18-18). There is much variability to lightning bolts, but a typical event can transfer 109 J of energy across a potential difference of perhaps 5 × 107 V during a time interval of about 0.2 s. Use this information to estimate (a) the total amount of charge transferred between cloud and ground, (b) the current in the lightning bolt, and (c) the average power delivered over the 0.2 s.APPROACH We estimate the charge Q, recalling that potential energy change equals the potential difference Vba times the charge Q, Eq. 17-3. We equate ∆PE ≈ 109 J. Next, the current I is Q/t (Eq. 18-1) and the power P is energy/time.SOLUTION (a) From Eq. 17-3, the energy transformed is ∆PE = QVba. We solve for Q:

Q=∆ PEV ba

≈109 J

5×107 V=20 coulombs.

(b) The current during the 0.2 s is about

I=Qt

≈109 J0.2 s

=100 A.

(c) The average power delivered is

P= energytime

=109 J0.2 s

=5 × 109W =5GW .

We can also use Eq. 18-5:P = IV = (100 A)(5 × 107 V) = 5 GW.

NOTE Since most lightning bolts consist of several stages, it is possible that individual parts could carry currents much higher than the 100 A calculated above.

Example 18-11: Will a fuse blow? Determine the total current drawn by all the devices in the circuit of Fig. 18-20.APPROACH Each device has the same 120-V voltage across it. The current each draws from the source is found from I=P/V , Eq. 18-5.SOLUTION The circuit is Fig. 18-20 draws the following currents: the lightbulb draws I=P/V=100 W/120 V=0.8 A ; the heater draws 1800 W/ 120 V=15. 0 A ; the stereo draws a maximum of 350 W/ 120 V=2. 9 A ; and the hair dryer draws 1200 W/ 120 V=10. 0 A . The total current drawn, if all devices are used at the same time, is

0 . 8 A+15. 0 A+2. 9 A+10 .0 A=28 .7 A .NOTE The heater draws as much current as 18 100-W lightbulbs. For safety, the heater should probably be on a circuit by itself.

Example 18-12: Hair dryer. (a) Calculate the resistance and the peak current in a 1000-W hair dryer connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain?

APPROACH We are given P andV rms, so I rms=P/V rms, and I 0=√2 I rms. Then we find R fromV=IR.

SOLUTION (a) We solveP=I rmsV rmsfor the rms current:

I rms=P

V rms

=1000W120 V

=8.33 A

ThenI 0=√2 I rms=11.8 AThe resistance is

Rrms=V rms

I rms

= 120 V8.33 A

=14.4 Ω

The resistance could equally well be calculated using peak values:

R=V 0

I 0

= 170V11.8 A

=14.4 Ω

(b) When connected to a 240-V line, more current would flow and the resistance would change with the increased temperature. But let us make an estimate of the power transformed based on the same 14.4-Ω resistance. The average power would be

P=V 2

rms

R=

(240 V )2

14.4 Ω=4000 W

This is four times the dryer’s power rating and would undoubtedly melt the heating element or the wire coils of the motor.

Example 18-13: Stereo power. Each channel of a stereo receiver is capable of an average power output of 100 W into an 8−Ω loudspeaker (see Fig. 18-14). What are the rms voltage and the rms current fed to the speaker (a) at the maximum power of 100 W, and (b) at 1.0 W when the volume is turned down?APPROACH We assume that the loudspeaker can be treated as a simple resistance (not quite true – see Chapter 21) with R=8 .0Ω . We are given the power P, so we can determine V rms and I rms using the power equations, Eqs. 18-9.

SOLUTION (a) We solve Eq. 18-9c for V rms and set P=100 W (at the maximum):

V rms=√ P R=√ (100 W ) (8. 0Ω)=28 V .

Next we solve Eq. 18-9b for I rms and obtain

I rms=√ PR

=√100 W8.0Ω

=3. 5 A .

Or we could use Ohm’s law (V=IR ):

I rms=V rms

R=28 V

8 . 0Ω=3 .5 A .

(b) At P=1 . 0 W ,

V rms=√ (1. 0 W ) (8 . 0Ω )=2 .8 V

I rms=2.8 V8.0Ω

=0 .35 A .

Example 18-14: Electron Speeds in a Wire. A copper wire, 3.2 mm in diameter, carries a 5.0-A current. Determine the drift speed of the free electrons. Assume that one electron per Cu atom is free to move (the others remain bound to the atom).APPROACH We can apply Eq. 18-10 to find the drift speed if we can determine the number n of free electrons per unit volume. Since we assume there is one free electron per unit atom, the density of free electrons, n, is the same as the density of Cu atoms. The atomic mass of Cu is 63.5 u (see Periodic Table inside the back cover), so 63.5 g of Cu contains one more or

6.02 X 1023 free electrons. We then use the mass density of copper (Table 10-1), ρD = 8.9 X 103 kg/m3, to find the volume of

this amount of copper, and then n = N/V. (We use ρD to distinguish it here for ρ for resistivity.)

SOLUTION The mass density ρD = m/V is related to the number of free electrons per unit volume, n = N/V, by

n=NV

=Nm /ρD

=N (1mole )n(1 mole )

ρD=((6 .02×1023 electrons

63.5×10−3 kg ) (8 .9×103 kg /m3 )

¿8 .4×1028 m−3

The cross-sectional area of the wire is

A=πr 2=(3. 14 )(1. 6×10−3 m)2=8.0×10−6 m2

Then, by Eq. 18-10, the drift speed is

vd=I

neA= 5 .0 A

(8. 4×1028m−3 )(1. 6×10−19 C )(8 .0×10−6 m2 )=4 .7×10−5 m /s

which is only about 0.05 mm/s.NOTE We can compare this drift speed to the actual speed o free electrons bouncing around inside the metal like molecules

in a gas, calculated to be around 1 .6×10−6 m /s at 20°C.

Example 19-5: Current in one branch What is the current through the 500 ohm resistor in Fig. 19-8a?APPROACH We need to find the voltage across the 500 ohm resistor, which is the voltage between points b and c in Fig. 19

– 8a, and we call it . Once is known, we can apply Ohm’s law, V=IR, to get the current. First we find the voltage

across the 400 ohm resistor , since we know that 17mA passes through it.

SOLUTION can be found using V=IR:

Since the total voltage across the network of resistors is =12.0V, then must be 12.0V-7.0V=5.0V. Then Ohm’s law applied to the 500 ohm resistor tells us that the current I1 through that resistor is

This is the answer we wanted. We can also calculate the current I2 through the 700 ohm resistor since the voltage across it is also 5.0V:

Example 19-6: Bulb brightness in a circuit. The circuit shown in Fig. 19-9 has three identical lightbulbs, each of resistance R. (a) When switch S is closed, how will the brightness of bulbs A and B compare with that of bulb C? (b) What happens when switch S is opened? Use a minimum of mathematics in your answers.RESPONSE (a) With switch S closed, the current that passes through bulb C must split into two equal parts when it reaches the junction leading to bulbs A and B. It splits into equal parts because the resistance of bulb A equals that of B. Thus, bulbs A and B each receive half of C’s current; A and B will be equally bright, but they will be less bright than bulb C.(b) When the switch S is open, no current can flow through bulb A, so it will be dark. We now have a simple one-loop series circuit, and we expect bulbs B and C to be equally bright. However, the equivalent resistance of this circuit (=R+R ) is greater than that of the circuit with the switch closed. When we open the switch, we increase the resistance and reduce the current leaving the battery. Thus, bulb C will dim when we open the switch. Bulb B gets more current when the switch is open (you may have to use some mathematics here), and so it will be brighter than with the switch closed, and B will be as bright as C.

Example 19-7: Analyzing a circuit. A 9.0 V battery whose internal resistance r is 0.50Ω is connected in the circuit. (a) How much current is drawn from the battery? (b) What is the terminal voltage of the battery? (c) hat is the current in the 6.0Ω resistor?APPROACH To find the current out of the battery, we first need to determine the equivalent resistance Req of the entire circuit, including r, which we do by identifying and isolating simple series or parallel combinations of resistors. Once we find

I from Ohm’s law, I=ℇ

Req,we get the terminal voltage using vab=ℇ−Ir .For c we apply Ohm’s law to the 6.0Ω resistor.

SOLUTION a) We want to determine the equivalent resistance of the circuit. But where do we start? We note that the 4.0 Ω and 8.0 Ω ressitors are in parallel, and so have an equivalent resistance Req1 given by

1R eq

= 18.0 Ω

+ 14.0 Ω

= 38.0 Ω

so Req1=2.7Ω. This 2.7 Ω is in series with the 6.0 Ω resistor. The net resistance of the lower arm of the circuit is thenReq 2=6.0 Ω+2.7 Ω=8.7 ΩThe equivalent resistance Req 3 of the 8.7Ω and 10.0Ω resistances in parallel is given by

1Req3

= 110.0Ω

+ 18.7 Ω

=0.21 Ω−1

so Req 3=( 1

0.21Ω−1 )=4.8 Ω. This 4.8Ω is in series with the 5.0Ω resistor and hence the 0.50Ωinternal resistance of the

battery, so the total equivalent resistance Req=4.8 Ω+5.0 Ω+0.50 Ω=10.3Ω. Hence the current drawn is

I= EReq

= 9.0 V10.3 Ω

=0.87 A

b) The terminal voltage of the battery is vab=ℇ−Ir=9.0 V −(0.87 A ) (0.50 Ω )=8.6 V .c) Now we can work back and get the current in the 6.0Ωresistor. It must be the same as the current through the 8.7Ω. The voltage across that 8.7Ω will be the emf of the battery minus the voltage drops across r and the 5.0Ω resistor: v8.7=9.0 V −(0.87 A ) (0.50 Ω+5.0 Ω ) .Applying Ohm’s law, we get the current (I’)

I '=9.0V −(0.87 A ) (0.50 Ω+5.0 Ω )

8.7 Ω=0.48 A .

Example 19-9: Jump starting the car. A good car battery is being used to jump start a car with a weak battery. The good battery has an emf of 12.5V and internal resistance of 0.020Ω. Suppose the weak battery has an emf 10.1 V and internal resistance of 0.10Ω. Each copper jumper cable is 3.0 m long and 0.50 cm in diameter, and can be attached as shown in Fig. 19-15. Assume the starter motor can be represented as a resistor R s=0.15 Ω. Determine the current through the starter motor (a) if only the weak battery is connected to it, and (b) if the good battery is also connected, as shown in Fig. 19-15.APPROACH We apply Kirchhoff’s rules, but in (b) we will fist need to determine the resistance of the jumper cables using their dimensions and the resistivity (ρ=1.68× 10−8 Ω·m for copper) as discussed in Section 18-4.SOLUTION (a) The circuit with only the weak battery and no jumper cables is simple: an emf of 10.1V connected to two resistances in series,0.10 Ω+0.15 Ω=0.25 Ω.Hence the current is I=V / R=(10.1 V )/(0.25Ω)=40 A.(b) We need to find the resistance of the jumper cables that connect the good battery. From Eq. 18-3, each has resistance

RJ=ρLa

=(1.68×10−8 Ω∙m ) (3.0m)

π (0.25×10−2m )2=0.0026 Ω. Kirchhoff’s loop rule for the full outside loop gives

12.5 V−I 1 ( 2 RJ+r1 )−I 3 R s=0

12.5 V−I 1 ( 0.025 Ω )−I 3(0.15 Ω)=0

since (2 RJ+r )=(0.0052 Ω+0.020 Ω )=0.025 Ω.The loop rule for the lower hoop, including the weak battery and the starter, gives10.1 V−I 3 (0.15 Ω )−I 2 (0.10 Ω )=0The junction rule at point B gives

I 1+ I 2=I 3

We have three equations in three unknowns. From Eq. (iii), I 1=I3−I2and we substitute this into Eq. (i): 12.5 V−¿¿12.5 V−I 3 (0.175 Ω )−I 2(0.025 Ω)=0.Combining this last equation with Eq. (ii) gives I 3=71 A ,quite a bit better than in the result of part (a), only 40A. The other currents are I 2=−5 A is in the opposite direction from that assumed in Fig. 19-15. The terminal voltage of the weak 10.1-V

battery is thus V BA=10.1 V−(−5 A )(0.10 Ω)=10.6 V .

NOTE The circuit shown in Fig. 19-15, without the starter motor, is how a battery can be charged. The stronger battery pushes charge back into the weaker one.

Example 19-10: Equivalent Capacitance. Determine the capacitance of a single capacitor that will have the same effect as the combination shown in Fig. 19-19a. Take C1 = C2 = C3

= C.APPROACH First we find the equivalent capacitance of C2

and C3 in parallel, and then consider that capacitance in series with C1.SOLUTION Capacitors C2 and C3 are connected in parallel, so they are equivalent to a single capacitor having capacitance.C23 = C2 + C3 = 2C.This C23 is in series with C1, Fig. 19-19b, so the equivalent capacitance of the entire circuit, Ceq, is given by

1Ceq

= 1C1

+ 1C23

= 1C

+ 12 C

= 32C

.

Hence the equivalent capacitance of the entire combination is Ceq =

23 C, and it is smaller than any of the contributing

capacitors, C1 = C2 = C3 = C.

Example 19-11: Charge and Voltage on Capacitors. Determine the charge on each capacitor in Fig. 19-19a of Example 19-10, and the voltage across each, assuming C = 3.0F and the battery voltage is V = 4.0 V.APPROACH We have to work “backward” through Example 19-10. That is, we find the charge Q that leaves the battery, using the equivalent capacitance. Then we find the charge on each separate capacitor and the voltage across each,. Each step uses Eq. 17-7, Q = CV.

SOLUTION The 4.0-V battery “thinks” it is connected to a capacitance Ceq =

23 C =

23 (3.0F) = 2.0 F. Therefore the

charge Q that leaves the battery, by Eq. 17-7, is Q=CV =(2 .0 μF )( 4 .0 V )=8 . 0 μC .From Fig. 19-19a, this charge arrives at the negative plate of C1, so Q1 = 8.0 C. The charge Q that leaves the positive plate is split evenly between C2 and C3 (symmetry: C2 = C3) and is Q2 = Q3 = ½Q = 4.0 C. Also, the voltages across C2 and C3 have to be the same. The voltage across each capacitor is obtained using V = Q/C. SoV 1=Q1/C1=(8 .0μC )/ (3.0 μF )=2 .7VV 2=Q2 /C2=( 4 .0 μC )/ (3 .0 μF )=1 .3VV 3=Q3 /C3=( 4 .0μC )=(3 .0μF )=1.3V

Example 27-8: X-ray Scattering. X-rays of wavelength 0.140 nm are scattered from a very thin slice of carbon. What will be the wavelengths of X-rays scattered at (a) 0°, (b) 90°, (c) 180°?APPROACH This is an example of the Compton effect, and we use Eq. 27-7 to find the wavelengthsSOLUTION (a) For Φ = 0°, cos Φ = 1 and 1- cos Φ = 0. Then Eq. 27-7 gives λ’ = λ = 0.140 nm. This makes sense since for Φ = 0°, there really isn’t any collision as the photon goes straight through without interacting. (b) For Φ = 90°, cos Φ = 0, and 1- cos Φ = 1. So

λ '=λ+ hm0 c

=0.140 nm+ 6.63 × 10−34 J ∙ s(9.11×10−31 ) ( 3.00× 108 m /s )

¿0.140 nm+2.4 ×10−12m=0.142nm;that is, the wavelength is long by one Compton wavelength (= 0.0024nm for an electron). (c) For Φ = 180°, which means the photon is scattered backward, returning in the direction from which it came (a direct “head-on” collision), cos Φ = -1, and 1 – cos Φ = 2. So

λ '=λ+ hm0 c

=0.140 nm+2 (0.0024 nm )=0.145 nm.

NOTE The maximum shift in wavelength occurs for backward scattering, and it is twice the Compton wavelength.

Example 27-9: Pair production.

(a) What is the minimum energy of a photon that can produce an electron-position pair? (b) What is this photon’s wavelength?

APPROACH: The minimum photon energy E equals the rest energy (m0 c2) of the two particles created, via Einstein’s

famous equation E=m0c2 (Eq. 26-8). There is no energy left over, so the particles produced will have ZERO KE. The

wavelength is λ=c / f where E=hf for the original photon.

SOLUTION: (a) Because E=m0c2, and the mass created is equal to two electron rest masses, the photon must have energy

E=2 ( 9 .11 ´ 10−31 kg ) (3 . 0´ 108 m/s)2=1. 64 ´ 10−13J=1 . 02 MeV(1 MeV=106 eV=1 .60 ´ 10-13 J ) . A photon with less energy cannot undergo pair production.

(b) Since E=hf =hc /λ , the wavelength of a 1.02-MeV photon is

λ=hcE

=(6 . 63 ´ 10−34 J⋅s) (3 .0 ´ 108 m/s)

(1. 64 ´ 10−13 J )=1 .2 ´ 10−12 m,

which is 0.0012 nm. Such photons are in the gamma-ray (or very short X-ray) region of the electromagnetic spectrum (Fig. 22-8).NOTE: Photons of higher energy (shorter wavelength) can also create an electron-position pair, with the excess energy becoming kinetic of the particles.

Example 27-10: Wavelength of a ball. Calculate the de Broglie wavelength of a .20 kg ball moving with a speed of 15 m/s. APPROACH We simply use Eq. 27-8.

SOLUTION λ= hp= h

mv=

( 6.6 X 10−34 J∗s)

(0.20 kg )(10ms )

=2.2 X 10−34m

Example 27-11: Wavelength of an electron. Determine the wavelength of an electron that has been accelerated through a potential difference of 100 V. APPROACH if the kinetic energy is much less than the rest energy, we can use classical KE= 1/2mv2. For an electron, m0c2 = 0.511 MeV. We then apply conservation of energy: the KE required by the electron equals its loss in PE. After solving for v, we use eq 27-8 to find the de Broglie wavelength. SOLUTION change in KE equals loss in PE, so KE = 100 eV. The ratio KE/ m0c2 = 100eV/(0.511x106 eV) = 10-4, so relativity is not needed. Thus12

m v2=eV

and

v=√ 2 eVm

=√ (2 ) ( 1.6× 10−19 C ) (100 V )( 9.1× 10−31 kg )

=5.9 ×106 m

s.

Then

λ= hmv

= 6.63×10−34 Js

(9.1×10−31kg )(5.9×106m

s)=1.2 ×10−10m

or 0.12 nm.

Example 27-12: Wavelength of a Lyman line. Use Fig. 27-27 to determine the wavelength of the first Lyman line, the transition from n= 2 to n=1. In what region of the electromagnetic spectrum does this lie?APPROACH We use Eq. 27-10, hf = Eu – Et , with the energies obtained from the figure to find the nergy and the wavelength of the transition. The region of the electromagnetic spectrum is found using the EM spectrum in Fig, 27-8.SOLUTION In this case, hf = Eu – Et = {-3.4 eV – (-13.6eV)} = 10.2 eV = (10.2eV)(1.60 X 10^-19 J/eV) = 1.63 X 10-18 J. Since λ = c/f, we have

λ= cf= hc

E2−E1

=( 6.63 X 10−34 J∗s)(3.00 X 108 m

s )1.63 X 10−18 J

=.22 X 10−7 m.

Or 122 nm, which is in the UV region of the EM spectrum, Fig. 22-8. See also fig. 27-23.NOTE An alternate approach would be to use Eq. 27-16 to find λ, and it gives the same results.

Example 27-13: Wavelength of a Balmer line. Determine the wavelength of light emitted when a hydrogen atom makes a transition from the n = 6 to the n = 2 energy level according to the Bohr model. APPROACH We can use eq 27-16 or 27-9 with R= 1.097x107m-1. SOLUTION We find1λ=(1.097 × 107 m−1) ( 1

4− 1

36 )=2.44 ×106 m−1

So λ=1/2.44 ×106 m−1= 4.10 x 10-7m or 410 nm. This is the fourth line in the Balmer series, and is violet in color.

Example 27-14: Absorption wavelength. Use Figure 27-27 to determine the maximum wavelength that hydrogen in its ground state can absorb. What would be the next smaller wavelength that would work?APPROACH Maximum wavelength corresponds to minimum energy, and this would be the jump from the ground state up to the first excited state (Fig. 27-27). The next smaller wavelength occurs for the jump from the ground state to the second excited state. In each case, the energy difference can be used to find the wavelength. SOLUTION The energy needs to jump from the ground state to the first excited state is 13.6eV – 3.4eV = 10.2eV; the required wavelength, as we saw in Example 27-12, is 122nm. The energy to jump from the ground state to the second excited state is 13.6 eV – 1.5 eV = 12.1 eV, which corresponds to a wavelength

λ= cf=hc

hf= hc

E3−E1

¿(6.63 x10−34 J ∙ s )(3.00 x108 m /s )

(12.1 eV )(1.60 x10−19 J /eV )

Example 30-6: Uranium decay energy release. Calculate the disintegration energy when U (mass=232.037146 u)92

232 decays to Th (228.028731 u )90228 with the emission of a

αparticle. (As always, masses are for neutral atoms.)APPROACH We use conservation of energy as expressed in Eq. 30-2. U92

232 is the parent, Th90228 is the daughter.

SOLUTION Since the mass of the He24 is 4.002603 u(Appendix B), the total mass in the final state is

228.028731 u+4.002603 u=232.031334 u .The mass lost when the U92

232 decays is

232.037146 u−232.031334 u=0.005812u .Since1 u=931.5 MeV , the energy Q released is

Q= (0.005812u )(931.5MeV

u )≈ 5.4 MeV ,

And this energy appears as kinetic energy of the α particle and the daughter nucleus.NOTE Using conservation of momentum, it can be shown that the α particle emitted by a U92

232 nucleus at rest has a kinetic energy of about 5.3MeV. Thus, the daughter nucleus—which recoils in the opposite direction from the emitted α particle—has about 0.1MeV of kinetic energy. See the next Example and/or Problem 65.

Example 30-7: KE of the α in 23292U decay.

For the 32392U decay of Example 30-6, how much of the 5.4-MeV disintegration energy will be carried off by the α particle?

APPROACH In any reaction, momentum must be conserved as well as energy.

SOLUTION Before disintegration, the nucleus can be assumed to be at rest, so the total momentum was zero. After disintegration, the total vector momentum must still be zero so the magnitude of the α particle’s momentum m equal the magnitude of the daughter’s momentum (Fig.30-6):

mαvα = mDvD.

Thus vα = mDvD/mα and the α’s kinetic energy is

KEα=12

mα vα2=1

2mα (mD vD

mα)

2

=12

mD v D2 (mD

mα)=(mD

mα)KED

¿( 228.028731 u4.002603 u )KED=57 KE D.

The total disintegration energy is Q = KEα + KED = 57KED + KED = 58 KED.Hence

KEα=5758

Q=5.3 MeV .

The lighter α particle carries off (57/58) or 98% of the total KE.

Example 30-8: Energy release in C614 decay.

How much energy is released when C614 decays to N7

14 by β emission?

APPROACH We find the mass difference before and after decay, Δm. The energy released isE=(∆ m ) c2. The masses given in Appendix B are those of the neutral atom, and we have to keep track of the electrons involved. Assume the parent nucleus has six orbiting electrons so it is neutral; its mass is 14.003242 u. The daughter is this decay N6

14 is not neutral since it has the same six orbital electrons circling it but the nucleus has a charge of+7e. However, the mass of this daughter with its six electrons, plus the mass of the emitted electron (which makes a total of seven electrons), is just the mass of a neutral nitrogen atom.SOLUTION The total mass in the final state is

(mass of N714 +6 electrons )+(mass of 1electron)

And this is equal tomass of neutral N (includes7 electrons)7

14

which, from Appendix B is a mass of 14.003074 u. So the mass difference is 14.003242 u−14.003074 u=0.000168 u, which is equivalent to an energy change ∆ mc2=(0.000168u ) (931.5 MeV /u )=0.156 MeV or156keV .NOTE The neutrino doesn’t contribute to either the mass or the charge balance since it has q=0andm ≈ 0.

Example 30-9: Sample Activity. The isotope C6

14 has a half-life of 5730 yr. If at some time a sample contains 1.00 x 10 caron-14 nuclei, what is the activity of the sample?APPRAOCH We first use the half-life to find the decay constant (Eq. 30-6), and use that to find the activity, Eq. 30-3b. the number of seconds in a year is (60)(60)(24)(3654

1) = 3.156 x 107 s.SOLUTION the decay constant λ from Eq. 30-6 is

λ=0.693

T 21

= 0.693

(5730 yr )(3.1566 x107 s

yr )=3.838 x10−12

s−1

From Eq. 30-3b, the magnitude of the activity or rate of decay isΔNΔt

= λN=(3.83 x10−12 s−1)(1.00 x 1022)

= 3.83 x 1010 decays/s.Notice that the graph of Fig. 30-10b stats at this value, corresponding to the original value of N = 1.0 x 1022 nuclei in Fig. 30-10a.NOTE the unit “decays/s” is often written simply as s-1 since “decays” is not a unit but refers only to the number. This simple unit of activity is called the Becquerel: 1 Bq = 1 decays/s, as discussed in chapter 31.

Example 30-10: Safety: activity versus half-life.

One might think that a short half-life material is safer than a long half-life material because it will not last as long. Is this an accurate representation of the situation?RESPONSE No. A shorter half-life means the activity is higher and thus more “radioactive” and dangerous. On the other hand, a shorter half-life means the material will all decay to a low level sooner. For the same sample size N, a shorter half-life material is more radioactive but for a shorter time.

Example 30-11: A sample of radioactive 137

N .

A lab has 1.49 ug of pure 137

N , which has a half-life of 10.0 min (600sec). (a) How many nuclei are present initially? (b)

What is the activity initially? (c) What is the activity after one hour? (d) After approximately how long will the activity drop to less than one per second?APPROACH We use the definition of the mole and Avogadro’s number to find the number of nuclei. For (b) we get λ from the given half-life and use the equation for the activity. For (c) and (d) we use the equation, and/or make a Table of the times.SOLUTION (a) the atomic mass is 13.0, so 13.0g will contain 6.02 X 1023 nuclei. Since we have only 1.49 X 10-6 g, the number of nuclei N that we have initially is given by the ratio

N o

6.02 X 1023=1.49 X 10−6

13.0 gSo N subzero = 6.9 X 1016 nuclei(b) From the equation, λ = (.693)/(600s) = 1.16 X 10-3 per second. Then, at t = 0,∆ N∆ t

=λ N0=(1.16 X 10−3 s−1 ) (6.9 X 1016 )=8.00 X 1013 decayss

.

(c) The half-life is 10.0 min, so the decay rate decreases by half every 10.0 min. We can make the Tale of activity after given periods of time. After 1.0 hr, the activity is 1.25x1012 decay/s.

Easy Alternate Solution (c) 60 minutes is 6 half lives, so the activity will decrease to ( 12 )( 1

2 )( 12 )( 1

2 )( 12 )(1

2 )=( 12 )

6

= 164

of

its original value, or 8.00 ×1013

64=1.25 ×1012 per second.

General Alternate Solution (c) The general way to find the activity, which works even when the time is not a perfect multiple of T1/2, is to use Eq. 30-5. We set t = 60.0 min = 3600 s:∆ N∆ t

=¿

NOTE The slight discrepancy in results arises because we kept only three significant figures.

(d) We want to determine the time t when ∆ N∆ t

=1.00 s−1 . From Eq. 30-5, we have

e λt=¿We take the natural log (ln) of both sides (remember lne− λt=−λt) and divide by λ to find

t=−ln (1.25 ×10−14 )

λ=2.76 ×104 s=7.67 h .

Example 30-12: Decay Chain. The decay chain starting with U92

234 in Fig. 30-11 has four successive nuclides with half-lives of 250,000 yr, 75,000 yr, 1600 yr, and a little under 4 days. Each decay in the chain has an alpha particle of a characteristic energy, and so we can monitor the radioactive decay rate of each nuclide. Given a sample that was pure U92

234 a million years ago, which alpha decay would you expect to have the highest activity rate in the sample?RESPONSE The first instinct to say that the process with the shortest half-life would show the highest activity. Surprisingly, however, the activity rates in this sample are the same! The reason is that in each case the decay of the parent acts as a bottleneck to the decay of the daughter. Compared to the 1600-yr half-life of Ra88

226 , for example, its daughter Rn86222 decays

almost immediately, but it cannot decay until it is made. (this is like an automobile assembly line: if worker A takes 20 minutes to do a task and then worker B takes only 1 minute to do the next task, worker B still does only one car every 20 minutes).

Example 30-13: An ancient animal The mass of carbon in an animal bone fragment found in an archeological site is 200g. If the bone registers an activity of 16 decays/s, what is the age?APPROACH First we determine how many Carbon-14 atoms there were in our 200-g sample when the animal was alive,

given the known fraction of carbon-14, . Then we use Eq. 30 – 3b to find the activity back then, and Eq. 30 – 5 to find out how long ago that ws by solving for the time t.SOLUTION The 200g of carbon is nearly all carbon-12; 12.0g of carbon-12 contains 6.02 x 10^23 atoms, so 200 g contains

.When the animal was alive, the ratio of carbon-14 to carbon-12 in the bone was 1.3 x 10^-12. The number of carbon-14 nuclei at that time was

.From Eq. 30 – 3b the magnitude of the activity when the animal was alive (t=0) was

where as we calculated in Example 30 – 9. So the original activity was

From Eq. 30 – 5

where is given as 16 s^-1. Then

or

.We take natural logs of both sides to obtain

which is the time elapsed since the death of the animal.