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This is the chapter where physics starts to get abstract. To the extent you can learn each concept thoroughly, you will make your life easier.
1. Two parallel metal plates are separated by 8.0 cm. A uniform field of 6 * 103 nt/C is
generated between the plates by charging them oppositely.
a) What is the force on a 2 * 10-7 C charge between the plates?
b) How much work is required to carry the charge from one plate to the other?
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a) F = qE F = (2 * 10-7C)(6 * 103 nt/C) F = 12.0 * 10-4 nt |
b) W = F * s W = (12 * 10-4 nt)(.08 m) W 9.60 * 10-5 J |
2. Two metal plates, separated by a distance d, contain a uniform electric field, E.
a) What is the force on a charge q between the plates?
b) How much work is required to carry the charge from one plate to the other?
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a) F = qE |
b) W = F * s W = qEd |
3. An electron gun can be constructed by taking two metal plates and drilling a small hole
through the positive one. Electrons released between the plates are attracted to the positive
plate but some pass through the hole making an electron beam. (Such a device is used in
television tubes.) Such a device has a plate separation of 4 cm, and an electric field
of 28,000 nt/C. Use conservation of energy to calculate the speed of electrons as they exit
the gun.
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WE = KE F * = 1/2mv2 (1.6 * 10-19 C)(28.000 nt/C)(.04 m) = (1/2)(9.1 * 10-31 kg)v2 v = 1.98 * 107 m/sec |
4. A block of mass m, bearing charge q is accelerated in an electric field E over a distance d.
It then slides up an incline and comes to rest on the frictional surface. How far does it slide?
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WE - PEg = WF qEd - mgh = µmgs s = (qEd - mgh)/µmg |
5. A sphere of mass m and charge q is released from a spring which has been compressed a
distance x. How far will the mass penetrate into the electric field before it comes momentarily
to rest?
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PEs = WE (1/2)kx2 = qEs s = (kx2)/(2qE) |
6. We know the force between two charged particles Q and q is k(Qq/r2).
How much work is required to carry one of these charges out to infinity? (Hint: recall
that the force of gravity of G(Mm/r2).)
By analogy:
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7. An electron is released 0.3 cm away from an object with a -4 * 10-7 C charge
on it. How fast will the electron be traveling at infinity?
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WE = KE k(Qq/r) = (1/2)mv2 v = (2kQq/rm)1/2 v = [2(9 * 109)(4 * 10-7)(1.6 * 10-19)]/ [(.003)(9.1 * 10-31)] v = 6.50 * 108 m/sec |
8. A block of mass m and charge q is released at a distance r from a fixed charge Q. After
sliding across the frictional surface it goes over the cliff. How far from the base does it land?
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WE - WF = KE k(Qq/r) - µmgL = (1/2)mv2 v = [2(kQq/rm - µgL)]1/2
The time of fall:
s = vt |
9. A block of mass m and charge q is released from the spring, slides down the incline,
through the wind tunnel and finally approaches charge Q. What is their distance of
closest approach?
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Ws + PEg - Wf - Ww = WE (1/2)kx2 + mgh - µmgL1cosØ - FL2 = k(qQ/r) r = (kqQ)/[(1/2)kx2 + mgh - µmgL1cosØ - FL2] |
10. A charged particle of mass m and charge q is accelerated in an electric field and
then approaches a point charge. If the distance of closest approach is r, what is the charge
on the fixed point?
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WE = WE qEd = k(Qq/r) Q = qEdr/kq Q = Edr/k |
11. Three charges are to be moved in an electric field of 7 * 104 nt/C. If the
charges are of 2, 4, and 12 µC, and each must be pushed 8.0 cm against the field,
how much work is required for each?
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W = F * s W = qE * s W = (2 * 10-6)(7 * 104)(.08) W = 1.12 * 10-2 J
W = F * s
W = F * s |
Look at the results of problem 11. Twice the charge requires twice the work. Six times the charge requires six times the work. By dividing the work by the charge, then, we should get a constant number.
12. Using the values of problem 11, calculate the work per charge done in each case.
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a) W/q = (1.12 * 10-2)/(2 * 10-6) W/q = 5.60 * 103 J/C |
b) W/q = (2.24 * 10-2)/(4 * 10-6) W/q = 5.60 * 103 J/C |
c) W/q = (6.72 * 10-2)/(12 * 10-6) W/q = 5.60 * 103 J/C |
13. Two parallel plates separated by a distance d enclose an electric field E.
a) How much work is required to carry charge q from one plate to the other?
b) How much work per unit charge is required?
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a) W = F * s W = qE * d |
b) W/q = (F * s)/q W/q = qEd/q W/q = Ed |
14. A 6 * 10-7 C charge is brought from infinity to within 17 cm of a point charge of
4 * 10-6 C. What is the work per charge done?
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W/q = [k(qq/r)]/q W/q = k(q/r) W/q = (9 * 109)[(4 * 10-6)/(.17)] W/q = 2.12 * 105 J/C |
So there we have it, the work/charge depends on the conditions of the space, not the charge. We'll define a new quantity, then, the electric potential.
Vab = Wab/q Where Vab is the potential between two points a and b, and Wab is the work required to carry q from a to b.
The unit of potential is the joule/coulomb or volt.
15. It takes 25 joules of energy to carry 4 * 10-2 C from one point to another. What is
the potential between the points?
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V = W/q V = (25 J)/(4 * 10-2C) V = 625 J/C V = 625 volt |
16. If there is a 20,000 volts (V) potential between a and b, how much work will it take to carry
an electron from a to b?
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Vab = Wab/q
W = qV |
17. Two oppositely charged flat plates are separated by 0.01 cm and create an
electric field between them of 7 * 104 nt/coul. What is the potential between
the plates?
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V = W/q V = (F * s)/q V = qEs/q V = (7 * 104 nt/C)(.01 * 10-2 m) V = 7.0 V |
18. An electron is released in an electric field and falls through a potential of 140 V.
How much kinetic energy does it gain?
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Work = KE KE = W KE = qV KE = (1.6 * 10-19)(140) KE = 2.24 * 10-17J |
19. The potential between a and b is 7000 V, while the potential between b and c
is 2000 V. What is the potential between a and c?
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Work to go from a to c is the sum of the work from a to b and from b to c.
Wac = Wab + Wbc |
20. Vab = 50 V. What is Vba?
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Vba = =Vab Vba = 50 V |
Up to now I've ignored the directionality of potential. If it takes work to move in one direction, we will get work when we move oppositely. Although people get a little loose with signs, the correct idea is to think of positive test charges. If it takes work to move a positive charge, then you are moving to higher potential.
21. Vac = 240 v, Vcb = -50 V, Vcd = 270 V. Find
Vab, Vad, and Vdb.
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Think of each point as having a potential level measured from a base value.
Vab = 240 - 50
Vad = 240 - 270
Vdb = -270 - 50 |
22. The circuit diagram has symbols with which you may be unfamiliar. Don't worry about
that, just study the potentials. Vad = -36 V, Vcb = 7 V,
Vbd = -21 V. Find Vab, Vcd, Vac.
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Vab = -36 + 21 Vab = -15 V
Vcd = 7 - 21
Vdb = -36 + 21 - 7 note: d has been drawn below a because vad is less than 0. |
23. The 150 v battery is connected as shown. Vbc = 17 V, Vbd = -71 V.
Find Vad, Vab, Vca.
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A battery establishes a potential between its terminals.
Vad = -150 V
Vab = -150 + 71
Vdb = -17 - 71 + 150 |
24. Find Vab.
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Vab = 6 + 6 + 6 Vab = 18 V |
25. Find Vab.
| Vab = 6 V |
26. Draw a qualitative (that is, without numbers) potential diagram for each circuit.
a)
b)
c)
d)
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a) |
b) |
c) |
d) |
27. Now draw qualitative diagrams for some harder circuits.
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a) |
b) |
c) |
d) |
28. You are given five batteries with potentials of 6.0 V, 4.1 V, 7.7 V, 1.5 V, and 3.6 V.
Show how you could connect some or all of them to get the following potentials: (Remember that
by reversing the direction of the battery, potential is decreased).
a) 7.5 V
b) 2.1 V
c) 8.2 V
d) 5.3 V
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a) |
b) |
c) |
d) |
29. Determine the unknown potentials. Be careful of signs.
a)
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b)
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a) VBC = -20V VDB = 32V VAC = -36V |
b) VBA = 9V VEF = -6V VFG = 0V VCE = -15V |
30. A gun similar to those of previous problems has a potential V
between the two plates. Ions of mass m and charge q are accelerated and pass through the
hole. What is their exit speed in terms of V, q, and m?
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WE = KE qV = (1/2)mv2 v = (2qV/m)1/2 |
31. A particle of mass m and charge q is accelerated through a potential V into a
spring of constant k. How far does it compress the spring?
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WE = PEs qV = (1/2)kx2 x = (2qV/k)1/2 |
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The potential at r is just the work/charge required to move from infinity (zero potential)
to r.
V = W/q |
33. What is the potential 3 * 10-9 cm from an electron?
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Vr = kQ/r Vr = [(9 * 109)(1.6 * 10-19)]/(3 * 10-11) Vr = 48.0 V |
34*. Points A and B are 1.7 m and 3.0 m respectively from a point charge
of 4 * 10-6 C. Find VA, VB, and VAB.
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VA = kq/r VA = (9 * 109)[(4 * 10-6)/(1.7)] VA = 2.12 * 104 V
VB = kq/r
VAB = VB - VA |
35. If an electron is released from infinity, how fast will it be traveling after falling
to within 0.1 nm of a proton?
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PE --> KE k(q1q2)/r = (1/2)mv2
v = [2(k(q1q2)/r)/m]1/2 |
36. An electron gun is made with 15,000 V between the plates. At what velocity will
electrons leave the gun?
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KE = W (1/2)mv2 = qV
v = (2qV/m)1/2 |
37. A beam of electrons, traveling at 6.2 * 106 m/sec passes between two
charged plates as shown. The electric field is E = 4 * 10-3 nt/C and the plates
are 4 cm long. How much will the electron rise due to the plates?
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First determine the time between plates.
H)
V) |
38. In problem 37, what will be the vertical velocity attained by the electron
in passing through the plates?
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t will be the same and so will a.
vy = at |
39. A particle of mass m, charge q, and velocity v passes between plates of length l
and field E. What distance does the particle move vertically?
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As in problem #37:
t = s/v
a = F/m
s = (1/2)at2 |
40. A proton travels at 4 * 105 m/sec between two charged plates 5 cm long
and containing an E-field of 2000 nt/C. At what angle does the electron exit from the plates?
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Time in the field: t = l/v t = (.05)/(4 * 105) sec
Vertical velocity: note: vx is not changed
tanØ = vy/vx |
41. An electron falls toward a proton. It starts at r = 1.9 * 10-10 m away and
is observed at r = 0.7 * 10-10 m. If it started from rest, what is its
observed velocity?
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ÇPE = PEfinal - PEinitial ÇPE = k(qQ/r2) - k(qQ/r1) ÇPE = kqQ(1/r2 - 1/r1) ÇPE = (9 * 109)(1.6 * 10-19)2[(1)/ (.7 * 10-10) - (1)/(1.9 * 10-10)] ÇPE = 2.08 * 10-18 J
But: |
One of the major contributions to 20th century physics (in fact, one of the major contributions to science for all time) was the development of quantum mechanics. It has been found that nature behaves very differently from what Newtonian physics led us to believe. These differences are insignificant for large, macroscopic events, but become startlingly evident on the atomic scale.First you need to know what quantized means. In everyday life the cost of goods is quantized that is, the cost of an item cannot be any arbitrary amount, but must be an integral multiple of one penny the smallest unit of money. For large purchases such as a television, a penny difference in cost is not noticeable, but if you're buying a pencil, a 1¢ difference is noticeable and significant.
In physics it is evident that matter is quantized you cannot gather an arbitrary mass of iron, but can collect only integral multiples of the mass of one iron atom. Charge is quantized all charge is a multiple of the elementary electron charge.
These examples are easier to visualize than the next....that angular momentum is quantized. Like mass and charge, it isn't noticeable until we get to the atomic level, but at this level we find that angular momentum comes only in units of h/2 where h is Planck's constant and has the value h = 6.63 * 10-34 joule-sec.
For an individual hydrogen atom, this has enormous implications. It was found that the angular momentum of a particle of mass m and charge q orbiting a massive particle of charge Q would be L = (kqQmr)1/2. In the case of the hydrogen atom, q = Q = e, the elementary charge, then we get L = (ke2mr)1/2.
42. Given the restriction that angular momentum comes only in units of h/2(3.14), what radii
are permissible in a hydrogen atom?
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From the argument above, Ln = (ke2mrn)1/2 where
Ln and rn are the angular momentum and radius of the nth orbit.
So Bohr's constraint says:
(ke2nrn)1/2 = (nh)/[2(3.14)] |
43. Using the result of problem 42, what is the potential energy of an electron in the
nth orbit of a hydrogen atom?
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PEn = -k(e2/rn) PEn = -k[ (e2)/[(n2h2)/ (4(3.14)2ke2m)] ] PEn = -[(4(3.14)2k2e4m)/(n2h2)] |
44. Again using problem 42, what is the kinetic energy of an electron in the
nth orbit of hydrogen?
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Recall that since: mv2/r = ke2/r2 mv2 = ke2/r
KE = (1/2)mv2 |
45. What is the total energy of an electron in the nth orbit?
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TEn = KEn + PEn TEn = [(2(3.14)2k2e4m)/(n2h2)] - [(4(3.14)2k2e4m)/(n2h2)] TEn = -[(2(3.14)2k2e4m)/(n2h2)] |
46. What is the difference between energies of an electron in the ith orbit
and an electron in the jth orbit? (i and j are integers)
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TEj - TEi = -[(2(3.14)2k2e4m)/
(j2h2)] - [(-2(3.14)2k2e4m)/
(i2h2)] TEj - TEi = [(2(3.14)2k2e4m)/(h2)] * [(1/i2) - (1/j2)] |
47. Given that Planck's constant is 6.63 * 10-34 joule-sec, what is the energy of the jump
from level 3 to level 2 in a hydrogen atom? Find a numerical solution.
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E3,2 = [(2(3.14)2k2e4m)/(h2)] *
[(1/22) - (1/32)] E3,2 = [2(3.14)2(9 * 109)2 (1.6 * 10-19)4(9.1 * 10-31)]/[(6.63 * 10-34)2] E3,2 = 2.17 * 10-18 J |
The model of an atom we are discussing is called the Bohr model after Neils Bohr who proposed it early in this century. It was used by him to describe and predict the spectrum of hydrogen.Bohr's ideas was that when an electron dropped form a high level to a lower one, the lost energy would reappear as a light quantum (a photon, if you prefer that terminology). All that is needed now is to find the relationship between energy of a photon, and frequency. This was provided by Einstein in 1905 with his restatement of Planck's equation: E = hv. That is, the energy of a photon is equal to Planck's constant times the frequency, n, of the light.
48. What is the frequency of a photon whose energy is 5.0 * 10-19 J?
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E = hv v = E/h E = (5 * 10-19)/(6.63 * 10-34) E = 7.54 * 1014 hz |
49. What is the frequency of the light given off when an electron jumps from level 3 to
level 2 in hydrogen? (see problem 41)
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v = E/h v = (2.17 * 10-18)/(6.63 * 10-34) v = 3.27 * 1015 hz |
50. Using Einstein's relation E = hn, derive an equation for the frequency of light emitted
for an electron jump between levels i and j. Use problem 47.
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Ej,i = [(2(3.14)2k2e4m)/
(h2)] * [(1/i2) - (1/j2)] Ej,i = hv v = [(2(3.14)2k2e4m)/(h2)] * [(1/i2) - (1/j2)] |
51. Recall from wave motion that f£ = v, that is, frequency times wavelength equals velocity. In light terms, ln = c. What, then, is the frequency of a 4000 A light wave?
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v = c/£ v = (3 * 108 m/sec)/(4000 * 10-10) v = 7.50 * 1014 hz |
52. What is the expression for the wavelength resulting from a jump between the ith and jth orbits? (Use #50)
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£ = c/v £ = c/[ [(2(3.14)2k2e4m)/(h2)] * [(1/i2) - (1/j2)] ] (ch3) / [ (2(3.14)2k2e4m) * [(1/i2) - (1/j2)] ] |
53**. What is the wavelength of light emitted from a jump between the
4th and 2nd orbits?
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£4,2 = (ch3)/ [ (2(3.14)2k2e4m) *
[(1/22) - (1/42)] ] £4,2 = [(3 * 108)(6.63 * 10 -34)3]/ [2(3.14)2(9 * 104)2 (1.6 * 10-19)4 (9.1 * 10-31)(1/4 - 1/16)] £ = 4.89 * 10-7 m £ = 489 nm |
The predictions of the Bohr model work well for the hydrogen atom. For example, the wavelength derived in problem 53 is the pale blue line emitted by hydrogen. The bright red line comes from a jump between the 3rd and 2nd orbits.
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