Download Homework #1 Solutions

21.1.
Use the charge of one electron (−1.602 ×10−19 C) to find the number
of electrons required to produce the net charge.
E XECUTE : The number of excess electrons needed to produce net charge q is
(a) I DENTIFY and S ET U P :
q
−3.20 × 10−9 C
=
= 2.00 × 1010 electrons.
−e −1.602 × 10−19 C/electron
(b) I DENTIFY and S ET U P : Use the atomic mass of lead to find the number of lead atoms in
8.00 ×10−3 kg of lead. From this and the total number of excess electrons, find the number of excess
electrons per lead atom.
E XECUTE : The atomic mass of lead is 207 ×10−3 kg/mol, so the number of moles in
mtot
8.00 × 10−3 kg
=
= 0.03865 mol. N A (Avogadro’s number) is the
M
207 × 10−3 kg/mol
number of atoms in 1 mole, so the number of lead atoms is
N = nNA = (0.03865 mol)(6.022 ×1023 atoms/mol) = 2.328 ×1022 atoms. The number of excess
8.00 ×10−3 kg is n =
2.00 × 1010 electrons
= 8.59 × 10−13.
2.328 × 1022 atoms
E VALUATE : Even this small net charge corresponds to a large number of excess electrons. But the
number of atoms in the sphere is much larger still, so the number of excess electrons per lead atom is
very small.
electrons per lead atom is
21.3.
I DENTIFY and S ET U P :
A proton has charge +e and an electron has charge − e, with
e = 1.60 ×10−19 C. The force between them has magnitude F = k
| q1 q2 |
r2
=k
e2
r2
and is attractive
since the charges have opposite sign. A proton has mass mp = 1.67 × 10−27 kg and an electron has
!
!
!
mass 9.11×10−31 kg. The acceleration is related to the net force F by F = ma .
EXECUTE: F = k
proton: ap =
e2
r2
= (8.99 × 109 N ⋅ m2 /C2 )
(1.60 × 10−19 C)2
(2.0 × 10−10 m) 2
= 5.75 × 10−9 N.
F
5.75 × 10−9 N
=
= 3.4 × 1018 m/s 2 .
mp 1.67 × 10−27 kg
electron: ae =
F
5.75 × 10−9 N
=
= 6.3 × 1021 m/s 2
me 9.11× 10−31 kg
The proton has an initial acceleration of 3.4 ×1018 m/s2 toward the electron and the electron has an
initial acceleration of 6.3 ×1021 m/s2 toward the proton.
E VALUATE : The force the electron exerts on the proton is equal in magnitude to the force the
proton exerts on the electron, but the accelerations of the two particles are very different because their
masses are very different.
21.10. IDENTIFY: We need to determine the number of protons in each box and then use Coulomb’s law to
calculate the force each box would exert on the other.
SET UP: The mass of a proton is 1.67 ×10−27 kg and the charge of a proton is 1.60 ×10−19 C. The
distance from the earth to the moon is 3.84 ×108 m. The electrical force has magnitude Fe = k
where k = 8.99 ×109 N ⋅ m2/C2 . The gravitational force has magnitude Fgrav = G
G = 6.67 ×10−11 N ⋅ m2/kg2 .
m1 m2
r2
q1 q2
, where
r2
,
EXECUTE: (a) The number of protons in each box is N =
1.0 × 10−3 kg
1.67 × 10−27 kg
= 5.99 × 1023. The total
charge of each box is q = Ne = (5.99 ×1023 )(1.60 ×10−19 C) = 9.58 ×104 C. The electrical force on
each box is Fe = k
q2
= (8.99 × 109 N ⋅ m2 /C2 )
(9.58 × 104 C)2
= 560 N = 130 lb. The tension in the
r2
(3.84 × 108 m)2
string must equal this repulsive electrical force. The weight of the box on earth is
w = mg = 9.8 ×10−3 N and the weight of the box on the moon is even less, since g is less on the moon.
The gravitational forces exerted on the boxes by the earth and by the moon are much less than the
electrical force and can be neglected.
(b) Fgrav = G
m1 m2
r2
= ( 6.67 ×10−11 N ⋅ m 2/kg 2 )
(1.0 × 10−3 kg) 2
(3.84 ×108 m)2
= 4.5 ×10−34 N.
EVALUATE: Both the electrical force and the gravitational force are proportional to 1/r 2 . But in SI
units the coefficient k in the electrical force is much greater than the coefficient G in the gravitational
force. And a small mass of protons contains a large amount of charge. It would be impossible to put
1.0 g of protons into a small box, because of the very large repulsive electrical forces the protons
would exert on each other.
21.20. I DENTIFY :
Apply F = k
qq ʹ
r2
to each pair of charges. The net force is the vector sum of the forces
due to q1 and q2 .
S ET U P : Like charges repel and unlike charges attract. The charges and their forces on q3 are shown
in Figure 21.20.
E XECUTE :
F2 = k
q2q3
r22
F1 = k
q1q3
r12
= (8.99 × 109 N ⋅ m 2 /C2 )
= (8.99 × 109 N ⋅ m 2 /C2 )
(4.00 × 10−9 C)(6.00 × 10−9 C)
(0.200 m) 2
(5.00 × 10−9 C)(6.00 × 10−9 C)
(0.300 m) 2
= 5.394 × 10−6 N.
= 2.997 × 10−6 N.
Fx = F1x + F2 x = + F1 − F2 = 2.40 × 10−6 N. The net force has magnitude 2.40 × 10−6 N and is in the
+x-direction.
E VALUATE : Each force is attractive, but the forces are in opposite directions because of the placement
of the charges. Since the forces are in opposite directions, the net force is obtained by subtracting their
magnitudes.
Figure 21.20
!
!
21.28. I DENTIFY : The electric force is F = qE .
S ET U P : The gravity force (weight) has magnitude w = mg and is downward.
E XECUTE : (a) To balance the weight the electric force must be upward. The electric field is
downward,
so for an upward force the charge q of the person must be negative. w = F gives mg = q E and
q =
mg (60 kg)(9.80 m/s 2 )
=
= 3.9 C.
E
150 N/C
qq ʹ
= (8.99 × 109 N ⋅ m 2 /C2 )
(3.9 C) 2
= 1.4 × 107 N. The repulsive force is immense and
r2
(100 m) 2
this is not a feasible means of flight.
E VALUATE : The net charge of charged objects is typically much less than 1 C.
(b) F = k
1 q
to calculate the electric field due to each charge and add the two
4πε 0 r 2
field vectors to find the resultant field.
!
S ET U P : For q1, rˆ = ˆj. For q2 , rˆ = cos θ iˆ + sin θ ˆj , where θ is the angle between E2 and the +xaxis.
r
q1 ˆ (9.0 × 109 N ⋅ m2 /C2 )(−5.00 × 10−9 C) ˆ
E XECUTE : (a) E1 =
j=
j = (−2.813 × 104 N/C) ˆj.
4πε 0r12
(0.0400 m)2
!
r
q2
(9.0 × 109 N ⋅ m 2 /C2 )(3.00 ×10−9 C)
4
E
The
angle
of
E2 =
=
=
1
.
080
×
10
N/C.
2 , measured
4πε 0r22
(0.0300 m) 2 + (0.0400 m )2
21.44. I DENTIFY :
Apply E =
⎛ 4.00 cm ⎞
from the x -axis, is 180° − tan −1 ⎜
⎟ = 126.9° Thus
⎝ 3.00 cm ⎠
r
E2 = (1.080 ×104 N/C)( iˆ cos126.9° + ˆj sin126.9°) = (−6.485 ×103 N/C) iˆ + (8.64 ×103 N/C) ˆj.
!
!
(b) The resultant field is E1 + E2 = (−6.485 × 103 N/C) iˆ + (−2.813 × 104 N/C + 8.64 × 103 N/C) ˆj.
!
!
E1 + E2 = (−6.485 × 103 N/C) iˆ − (1.95 × 104 N/C ) ˆj.
!
!
E VALUATE : E1 is toward q1 since q1 is negative. E2 is directed away from q2 , since q2 is
positive.
21.52. (a) I DENTIFY : The field is caused by a finite uniformly charged wire.
S ET U P : The field for such a wire a distance x from its midpoint is
E=
E XECUTE :
E=
(b) I DENTIFY :
1
⎛ 1 ⎞
λ
= 2⎜
.
⎟
2πε 0 x ( x/a) + 1
⎝ 4πε 0 ⎠ x ( x/a) 2 + 1
λ
2
(18.0 × 109 N ⋅ m 2 /C2 )(175 × 10−9 C/m)
2
= 3.03 × 104 N/C, directed upward.
⎛ 6.00 cm ⎞
(0.0600 m) ⎜
⎟ +1
⎝ 4.25 cm ⎠
The field is caused by a uniformly charged circular wire.
S ET U P : The field for such a wire a distance x from its midpoint is E x =
1
Qx
. We first
4πε 0 ( x 2 + a 2 )3/2
find the radius a of the circle using 2πa = l.
E XECUTE : Solving for a gives a = l/2π = (8.50 cm)/2π = 1.353 cm.
The charge on this circle is Q = λl = (175 nC/m)(0.0850 m) = 14.88 nC.
The electric field is
E=
1
Qx
(9.00 × 109 N ⋅ m 2 /C2 )(14.88 × 10−9 C/m)(0.0600 m)
=
3/2
4πε 0 (x 2 + a 2 )3/2
⎡(0.0600 m)2 + (0.01353 m)2 ⎤
⎣
⎦
E = 3.45 × 104 N/C, upward.
E VALUATE : In both cases, the fields are of the same order of magnitude, but the values are different
because the charge has been bent into different shapes.
21.58. I DENTIFY :
Find the vector sum of the fields due to each charge in the dipole.
S ET U P : A point on the x-axis with coordinate x is a distance r = (d/2) 2 + x 2 from each charge.
⎞
q ⎛
1
⎜⎜
⎟,
2
2
4πε 0 r
4πε 0 ⎝ ( d/2) + x ⎟⎠
where d is the distance between the two charges. The x-components of the forces due to the two
charges
are equal and oppositely directed and so cancel each other. The two fields have equal y-components, so
⎞
2q ⎛
1
E = 2E y =
⎜
⎟ sin θ , where θ is the angle below the x-axis for both fields.
4πε 0 ⎜⎝ (d/2) 2 + x 2 ⎟⎠
⎞
⎞⎛
⎛ 2q ⎞ ⎛
d/2
1
d/2
qd
⎟=
⎜
⎟⎜
and Edipole = ⎜
sin θ =
.
⎟
2
3/2
2
⎜
⎟
2
2
2
(d/2) + x
⎝ 4πε 0 ⎠ ⎜⎝ ( d/2 ) + x ⎟⎠ ⎜ ( d/2 ) + x 2 ⎟ 4πε 0 ⎡(d/2)2 + x 2 ⎤
⎝
⎠
⎣
⎦
The field is the −y-direction.
(b) At large x, x 2 ≫ (d/2) 2 , so the expression in part (a) reduces to the approximation
E XECUTE :
Edipole ≈
(a) The magnitude of the field due to each charge is E =
qd
4πε 0 x 3
1
q
2
=
.
E VALUATE : Example 21.14 shows that at points on the +y-axis far from the dipole,
qd
Edipole ≈
.
2πε 0 y 3
The expression in part (b) for points on the x-axis has a similar form.