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Halliday/Resnick/Walker 7e
Chapter 21
- (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to
m a 2 (^) 2 m a 1 (^) 1 m 2
7 6 3 10 7 0 7 9 0
×
= ×
−
.. − .
kg m s m s
kg.
2 2
c hc h
(b) The magnitude of the (only) force on particle 1 is
F m a k
q q r
q = 1 1 = 1 22 = ×^9
2
c8 99.^10 h0 0032. 2.
Inserting the values for m 1 and a 1 (see part (a)) we obtain | q | = 7.1 × 10 –11^ C.
- Eq. 21-1 gives Coulomb’s Law, F = k q^^1 r^2 q 2 , which we solve for the distance:
| 1 || 2 | (^ 8.99^ 10 N m^9 2 C^2 )^ (^ 26.0^10 6 C^ )^ (^ 47.0^10 6 C) 1.39m.
5.70N
r k q^ q F
× ⋅ × −^ × −
- 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. 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 3 q /4 in the final situation. The repulsive force between spheres 1 and 2 is finally
F = kq^2 / r^2
2 2 2
F k q^ q^ k q^ F F r r F
- The force experienced by q 3 is
3 31 32 34 3 1 3 2 3 4 2 2 2 0
1 | || | (^) ˆj | || | (^) (cos45 iˆ (^) sin 45 j)ˆ | || | (^4) ( 2 )
F F F F q^ q^ q^ q^ q^ q
πε a a a
G G G G ˆ
i
(a) Therefore, the x -component of the resultant force on q 3 is
7 2 3 3 2 4 9 2 2 0
| | | | 2 1.0^101
| | 8.99 10 2 0.17N.
x 4 2 2 (0.050) 2 2 F q^ q q
πε a
⎛ ⎞ × − ⎛ ⎞
= ⎜ + ⎟ = × ⎜ + ⎟=
(b) Similarly, the y -component of the net force on q 3 is
7 2 3 2 9 (^3 2 ) 0
| | | | 2 1.0^101
| | 8.99 10 1 0.046N.
y 4 2 2 (0.050) 2 2 F q^ q q
πε a
× −
= ⎛^ − + ⎞^ = × ⎛^ − + ⎞= −
- (a) The individual force magnitudes (acting on Q ) are, by Eq. 21-1,
k
q Q a
k
q Q a (^) a a
1 2
2
2 2
b^ − −^ g =^ b − g^2
which leads to | q 1 | = 9.0 | q 2 |. Since Q is located between q 1 and q 2 , we conclude q 1 and q 2 are like-sign. Consequently, q 1 / q 2 = 9.0.
(b) Now we have
k
q Q a
k
q Q a (^) a a
1 32 2
2
b − − g =^ b −^32 g^2
which yields | q 1 | = 25 | q 2 |. Now, Q is not located between q 1 and q 2 , one of them must push and the other must pull. Thus, they are unlike-sign, so q 1 / q 2 = –25.
- With rightwards positive, the net force on q 3 is
3 13 23 1 3 2 22 3 12 23 23
F F F k q q^ k q q. L L L
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 q 3 by q 1 ) is negative if they are unlike charges, indicating that q 3 is being pulled toward q 1 , and it is positive if they are like charges (so q 3 would be repelled from q 1 ). Setting the net force equal to zero L 23 = L 12 and canceling k , q 3 and L 12 leads to
1 2 1 2
q (^) q q q
- (a) The magnitude of the force between the (positive) ions is given by
F
q q r
k q r
= b gb g=
2
πε^2
We substitute q 2 = Q – q 1 and solve for q 1 using the quadratic formula. The two roots obtained are the values of q 1 and q 2 , since it does not matter which is which. We get and 3.8 × 10
1.2 × 10 −^5 C
–5 (^) C. Thus, the charge on the sphere with the smaller charge is 1.2 × 10 − (^5) C.
- The unit Ampere is discussed in §21-4. Using i for current, the charge transferred is
q = it = ( 2.5 ×10 A 4 )( 20 × 10 −^6 s) =0.50 C.
- Keeping in mind that an Ampere is a Coulomb per second, and that a minute is 60 seconds, the charge (in absolute value) that passes through the chest is
| q | = ( 0.300 Coulombsecond ) ( 120 seconds ) = 36.0 Coulombs.
This charge consists of a number N of electrons (each of which has an absolute value of charge equal to e ). Thus,
N = |^ q |e = (^) 1.6036.0 C x 10 -19 (^) C = 2.25 × 1020.
- If the relative difference between the proton and electron charges (in absolute value) were
q q e
p −^ e = 0 0000010.
then the actual difference would be q (^) p − qe = 16. × 10 − 25 C. Amplified by a factor of 29 × 3 × 1022 as indicated in the problem, this amounts to a deviation from perfect neutrality of
∆ q = c 29 × 3 × 10 22 hc1 6. × 10 − 25 Ch = 014. C
in a copper penny. Two such pennies, at r = 1.0 m, would therefore experience a very large force. Eq. 21-1 gives
F k q r
= b∆^ g = ×
2 2
17. 108 N.
- (a) A force diagram for one of the balls is shown below. The force of gravity mg^ G^ acts downward, the electrical force of the other ball acts to the left, and the tension in the thread
acts along the thread, at the angle θ to the vertical. The ball is in equilibrium, so its acceleration
is zero. The y component of Newton’s second law yields T cos θ – mg = 0 and the x component
yields T sin θ – F
G
Fe
e = 0. We solve the first equation for^ T^ and obtain^ T^ =^ mg /cos^ θ. We substitute
the result into the second to obtain mg tan θ – Fe = 0.
Examination of the geometry of Figure 21-43 leads to
tan θ =.
x L x
2 b 2 g^2
If L is much larger than x (which is the case if θ is very small), we may neglect x /2 in the
denominator and write tan θ ≈ x /2 L. This is equivalent to approximating tan θ by sin θ. The
magnitude of the electrical force of one ball on the other is
F q e (^) x
2
0
4 πε^2
by Eq. 21-4. When these two expressions are used in the equation mg tan θ = Fe , we obtain
mgx L
q x
x q L 2 mg
2 2
2 0
1 3 ≈ ⇒ ≈
F
HG^
I
πε πε KJ
/ .
(b) We solve x^3 = 2 kq^2 L / mg ) for the charge (using Eq. 21-5):
( )(^ )
3 2 3 8 9 2 2
0.010 kg 9.8 m s 0.050 m 2.4 10 C. (^2) 2 8.99 10 N m C 1.20 m
q mgx kL
= = = ± ×^ −
× ⋅
Thus, the magnitude is | q | = 2.4 × 10 −^8 C.
