PHY–309 L.
Solutions for Problem set # 1.
Question Q.4 at the end of chapter 12:
(a) When a piece of glass is rubbed with a nylon cloth, some electrons move from the glass
to the nylon. Since the electrons carry negative charges, this means that the nylon becomes
negatively charged while the glass becomes positively charged.
Subsequently, when the glass rod touches the first pith ball, the rod recovers some of its
missing electrons by ‘stealing’ them from the ball. In this way, some of the rod’s positive
electric charge gets transferred to the ball, so the first pith ball becomes positively charged.
Likewise, when the nylon cloth touches the second pith ball, some of the nylon’s extra
electrons move to the ball. Thus, some of the nylon’s negative charge is transferred to the
ball, so the second ball becomes negatively charged.
(b) The two pith balls have opposite electric charges — positive for the first ball and negative
for the second — so they attract each other.
Question Q.14 at the end of chapter 12:
Initially, the pith ball has no net electric charge. Although the molecules of cellulose, lignin,
etc., in the pith naturally contain both positively and negatively charged atoms, their charges
cancel each other and the net charge is zero. Moreover, both types of charge are distributed
uniformly throughout the pith, so every little piece of the ball is neutral and not just the
whole ball.
But when a charged rod is brought near the ball, the pith becomes polarized. Suppose
the rod is positively charged as in figure 12.9 on page 242 of the textbook. This charge
attract the negatively charged atoms in the pith so they move a tiny distance closer to the
rod. On the other hand, the positively charged atoms in the pith are repelled by the rod’s
charge so they move a tiny distance away from the rod. Consequently, the two types of
charges cancel each other only in the middle of the pith ball but not on its surface! On the
side of the ball facing the rod the surface becomes negatively charged because the negatively
charged atoms move in while the positively charged atoms move out. Similarly, on the other
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side of the ball the surface becomes positively charged because the positive atoms move in
while the negative atoms move out.
Altogether, the net charge of the pith ball remains zero, but the side of the ball facing
the rod acquires a negative charge while the opposite side gets an opposite positive charge.
Consequently, the side of the ball facing the positively charged rod is attracted to it while
the other side of the ball is repelled by the rod. But since the ball’s side attracted to the
rod is closer to it than the repelled side, the attractive force is stronger than the repulsive!
Thus, the net force on the neutral pith ball is towards the charged rod.
Now suppose the rod is negatively charged. In this case, the polarization of the pith
works in the opposite direction: the positively charged atoms move a tiny distance towards
the rod while the negatively charged atoms move a tiny distance away from it. Consequently,
the surface of the pith ball facing the rod becomes positively charged — which is attracted
to the negatively charged rod — while the surface on the other side of the ball becomes
negatively charged and is repelled by the rod. The two charges are equal and opposite so the
net charge of the ball is zero, but the forces between the two sides of the ball and the rod
do not cancel. Instead, because of un-equal distances from the rod, the attractive force on
the near side of the ball is stronger than the repulsive force on the far side, so the net force
of the neutral pith ball is towards the charged rod, regardless of rod’s charge being positive
or negative!
Thus, until the rod touches the pith ball, they attract each other because of the ball’s
polarization. But once they touch, some of the rod’s charge is transferred to the ball. After
that, the ball and the rod have like-sign charges — both positive or both negative — and
they repel each other. Once the ball picks up enough net charge, this repulsive force becomes
stronger than the attraction due to polarization, so the ball suddenly flies away from the
rod.
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Question Q.23 at the end of chapter 12:
C
A
B
D
On this diagram, the red arrows are the electric fields of the positive charge +q only, the
blue arrows are the electric fields of the negative charge −q only, and the purple arrows are
the net electric fields due to both charges.
Problem E.8 at the end of chapter 12:
The proton has positive charge +e = +1.602176 · 10−19 C while the electron has exactly
opposite charge −e. Like any other pair of opposite-sign electric charges, the electron and
the proton attract each other with a Coulomb force
F =
k(−e)(+e)
ke2
=
−
r2
r2
(1)
where the ‘−’ indicates attractive rather than repulsive force, r is the distance between the
two particles, and
k =
1
= 8.98755 · 109 N · m2 /C2 .
4πε0
(2)
In a hydrogen atom, the average distance between the electron and the proton is about
r = 0.5 Å = 5 · 10−11 m. At this distance, the numerical value of the attractive force is
|F | =
(9 · 109 N m2 /C2 )(1.6 · 10−19 C)2
ke2
=
≈ 9 · 10−8 N.
r2
(5 · 10−11 m)2
(3)
This may not look like a big force, but please remember that the electron has has a very
tiny mass me ≈ 9.1 · 10−31 kg. So the Coulomb force (3) gives it a tremendous acceleration
ae = F/me ≈ 1 · 1023 m/s2 , or 10,000 billion billion gees!
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Problem E.10 at the end of chapter 12:
A charge q in electric field E experiences an electrostatic force
F = qE.
(4)
This is a vector equation, so for a positive charge q the force F has the same direction as the
electric field E, while for a negative charge q the direction of the force F is opposite from
the direction of the field E.
The test charge in question q = +4µC is positive, so the electrostatic force acting on
it has the same direction as the electric field. Since we know the direction of the force is
downward, it follows that the direction of the electric field E is also downward.
As to the magnitude E of the electric field, eq. (4) implies
F = |q| × E
=⇒ E =
F
,
|q|
(5)
hence
E =
12 N
= 3 · 106 N/C ≡ 3, 000, 000 V/m.
4 · 10−6 C
(6)
Problem E.12 at the end of chapter 12:
The force on an electric charge q due to electric field E is
F = qE.
(4)
This is a vector equation, so for a positive charge q the force F has the same direction as the
electric field E, while for a negative charge q the direction of the force F is opposite from
the direction of the field E.
In this problem, the electric field E points to the right while the charge q is negative.
Thus, the direction of the force F is opposite from the direction of E; in other words, the
force on the charge is directed to the left.
As to the magnitude of this force,
F = |q| × E = (3.0 · 10−6 C) × (8.5 · 10+4 N/C) = 25.5 · 10−2 N ≈ 0.25 N.
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(7)
Problem SP.1 at the end of chapter 12:
Note: I am not sure the authors of this problem paid attention to significant figures, so I
am calculating all the answers to 2 significant figures and then rounding them off to just 1
significant figure warranted by the problem data.
(a)
FAB =
kqA qB
(9.0 · 109 N m2 /C2 )(+0.10 C)(+0.02 C)
=
= 4.5·106 N ≈ 5·106 N. (8)
2
2
(2 m)
rAB
(b)
FCB =
kqC qB
(9.0 · 109 N m2 /C2 )(+0.04 C)(+0.02 C)
=
= 7.2·106 N ≈ 7·106 N. (9)
2
(1 m)2
rCB
(c) Since all 3 charges are positive, all forces between them are repulsive. Charge A sits left
from charge B, so its force FAB pushes charge B to the right. On the other hand, charge
C sits right from charge B, so its force FCB pushes charge B to the left. The net force on
charge B is therefore
net
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6
6
6
Fon
B = FAB − FCB = 4.5 · 10 N − 7.2 · 10 N = −2.7 · 10 N ≈ −3 · 10 N. (10)
The magnitude of this force is 2.7 ≈ 3 mega–Newtons, and its direction is to the left.
(d) The net force on the test charge B is
net
Fon
B = qB × E(B)
(11)
where E(B) is the electric field at point B created by all the charges except B itself. Given
the net force obtained in part (c), this electric field is
E(B) =
net
Fon
−2.7 · 106 N
B
= −1.35 · 108 V/m ≈ 1 · 108 V/m.
=
qB
+0.02 C
(12)
In other words, the electric field created by charges A and C at point B has magnitude
135 ≈ 100 mega-Volt per meter and direction to the left.
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Note units: 1 V/m (Volt per meter) is the same as 1 N/C (Newton of force per Coulomb
of charge).
′ = −0.06 C
(e) In this part we change the value of the test charge B from qB = +0.02 C to qB
but we don’t change the charges A or C or the locations of any charges. Consequently, the
electric field E(B) on the test charge B from the other charges A and C remains exactly as
in part (d). Therefore, the force on the new test charge is simply
net
′
Fon
new B = (new qB ) × (old E(B))
= (−0.06 C) × (−1.35 · 108 V/m) = +8.1 · 106 N ≈ +8 · 106 N.
(13)
Note that the direction of this 8 mega–Newton force is to the right rather than to the left.
The non-textbook problem I is postponed to the next homework.
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