Physics 213 — Problem Set 1 — Solutions
Spring 1998
1. Reading Assignment
Serway 23.1–23.7
2. Conceptual Questions
These short-answer questions are conceptual rather than computational.
a) A balloon is negatively charged by rubbing and then clings to the wall. Does this mean the wall was
initially positively charged? Why does the balloon eventually fall?
b) Why is it nonsense to think that two electric field lines could cross?
c) When a metal object receives a positive charge, does it mass increase or decrease?
SOLUTION:
a)No, it doesn’t necessarily mean that the wall was initially positively charged. The molecules in the wall
can be polarized due to the negative charges on the ballon and can attract the negatively charged balloon.
The excess elcetrons (negative charges) in the balloon will be slowly discharged through the wall (since it is
not a perfect insulator) and as the balloon becomes neutral, it will fall.
b)The tangent to the electric field line at any point gives the direction of the net force acting on a test
charge at thath point. Corssing a field line would imply different net forces at the same point.
c)Since only the negative charges (electrons) are free to move in a metal, electrons must leave the metal in
order for it to be positively charged. Thus the mass of the metal will decrease (very slightly) if it becomes
positively charged.
3. The nucleus of a helium atom contains two protons and two neutrons. The protons are spaced roughly
1.6 × 10−15 m apart. How big is the repulsive Coulomb force between these protons? Compare this
with the attractive gravitational force between the protons? Have you any idea why the nucleus holds
together?
SOLUTION: The charge of a proton is e = 1.6 × 10−19 C. With d = 1.6 × 10−15 m and k = 9 × 109 Nm2 /C2
the Coulomb force is
F = ke2 /d2 = 90N ' 20 lb
This is huge compared to the attractive gravitational force between the protons, Gm2p /d2 = 7.6 × 10−35
N. What keeps protons and neutrons together in the nucleus is the attraction due to nuclear forces. These
forces are very strong, but only at a short range of roughly the radius of a nucleon, about 10−15 m. Since
nuclear forces and electrostatic forces depend on distance in such different ways, the balance between them
depends on the size and structure of the nucleus. While the electrostatic repulsion between the protons
affects all protons, the nuclear forces can only strongly hold together nucleons that are close to each other.
Thus large nuclei can break up and fly apart due to Coulomb repulsion. This process is called nuclear fission.
4. See Serway 23.8
Two identical point charges −q are fixed in space and separated by a distance d. A third point charge
Q is free to move and lies on a perpendicular bisector of the line connecting the two fixed charges. The
third charge is initially at rest and a distance x0 from the line connecting the fixed charges. Assume that
x0 is small compared with d. (See Fig. P23.8 in Serway).
a) Calculate the force on charge Q due to the fixed charges when it is distance x away from the line
joining the fixed charges. Show that Q’s motion is approximately simple harmonic when x is small
compared to d. What then is the period of the harmonic motion?
b) How fast is Q moving when it is at the midpoint between the two charges?
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SOLUTION:
a)At a position x on the bisector, the magnitude of the force exerted by each charge −q on Q is F =
k (d/2)qQ
2 +x2 . The y-components of these forces cancel. The net force is in the negative x-direction, Fnet =
−2Fx x̂. In terms of x and d
x
qQ
p
Fnet = −2k
(d/2)2 + x2
(d/2)2 + x2
For small displacements, x << d/2, we can neglect the x2 in the denominator. This leaves
Fnet = −2kqQ
x
x
= −16kqQ 3
(d/2)3
d
2
The minus-sign is consistent with the force pointing to the left in the figure. Since Fnet = M ddt2x , where M
is the mass of the point charge Q we have as equation of motion for a small displacement x
d2 x
16kqQ
=−
x
dt2
Md3
(4.1)
which describes a simple harmonic motion of the charge Q. Initially, at t = 0, the charge Q is at x0 . A
solution of (6.1) that satisfies this boundary condition is
x(t) = x0 cos ωt
where
ω2 =
16kqQ
Md3
The period of this motion is
2π
π
T =
=
ω
2
r
Md3
kqQ
b)The velocity of charge Q is, as a function of time
v(t) =
dx
= −x0 ω sin ωt
dt
This has a maximum value of x0 ω which it reaches twice during each period, when it passes the midpoint
between the two charges. In terms of the given quantities
r
vmax = x0 ω = 4x0
kqQ
Md3
Rx
Alternately, one can calculate the work needed to move charge Q from x = 0 to x = x0 : W = 0 0 F (x)dx
with F (x) = 16kqQ dx3 . This is the potential energy stored in the oscillator when Q is at x = ±x0 . We can
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equate it to the kinetic energy when Q is at the origin, 1/2 Mvmax
. Solving for vmax gives the same solution.
5. Serway 23.16A
An object having a net charge Q is placed in a uniform electric field of magnitude E directed vertically.
What is the mass of this object if it “floats” in the field?
SOLUTION: The sign of the charge and the direction of the electric field must be such that the electric
force, QE, is pointing up, i.e. (E up, Q > 0) or (E down, Q < 0). If the electric force is equal in magnitude
to the weight, Mg, of the object, it “floats”. Mg = QE, and
M=
QE
g
6. A point charge q1 is located at (x1 , y1) in the xy plane. A second charge, q2 , is at (x2 , y2 ). Find
expressions for the x and y components of the total electric field at point (x, y) due to the two charges.
(Hint: See Serway 23.18.)
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SOLUTION: We first need to find the electric fields created by each of the charges q1 and q2 ; then add the
field vectors component by component (Principle of Superposition).
E1 =
ke q1
ke q1
=
r12
(x − x1 )2 + (y − y1 )2
is the magnitude of the E-field at the point (x,y) due to the charge q1 , and in terms of its components :
E~1
=
E1 cos θ î + E1 sin θ ĵ
=
ke q1
(x − x1 )2 + (y − y1 )2
"
x − x1
p
#
y − y1
î + p
ĵ
(x − x1 )2 + (y − y1 )2
(x − x1 )2 + (y − y1 )2
Similarly,
"
ke q2
E~1 =
(x − x2 )2 + (y − y2 )2
x − x2
p
y − y2
(x − x2 )2 + (y − y2 )2
#
î + p
ĵ
(x − x2 )2 + (y − y2 )2
Hence the x and y components of the total electric field at the point (x,y) are :
Etotal x
=
Etotal y
=
ke q1 (x − x1 )
[(x − x1 )2 + (y − y1 )2 ]3/2
ke q1 (y − y1 )
[(x − x1 )2 + (y − y1 )2 ]3/2
+
+
ke q2 (x − x2 )
[(x − x2 )2 + (y − y2 )2 ]3/2
ke q2 (y − y2 )
[(x − x2 )2 + (y − y2 )2 ]3/2
7. Serway 23.22
Three point charges, q, 2q, and 3q, are arranged on the vertices of an equilateral triangle. Determine the
magnitude of the electric field at the geometric center of the triangle.
SOLUTION: Let the distance from each charge to the geometric center of the triangle be d. Then, the
electric field due to each charge is :
q
:
2q
:
3q
:
ke q
ke q
E~1 = 2 cos 30o î + sin30o ĵ = 2
d
d
√
3
1
î + ĵ
2
2
√
2ke q
2ke q
3
1
E~2 = 2 − cos 30o î + sin30o ĵ = 2
−
î + ĵ
d
d
2
2
3ke q
E~3 = − 2 ĵ
d
The total electric field found by adding those three vectors is then,
~ total = − ke q
E
d2
√ 3
2
î −
3ke q
ĵ
2d2
with magnitude
s
√
Etotal =
3ke q
2d2
2
+
−3k q 2
e
2d2
√
3ke q
=
d2
8. Serway 23.26A
A charge −q1 is located at the origin, and a charge −q2 is located along the y axis at y = d. At what
position along the y axis is the electric field zero?
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SOLUTION:
We need to find the position y from the origin where the electric fields due to each charge exactly cancel.
The electric field due to charges q1 and q2 are :
E~1
=
E~2
=
~ total
E
=
ke q1
ĵ
y2
ke q2
ĵ
(d − y)2
−
E~1 + E~2 = ke
q1
q2
− 2 +
y
(d − y)2
ĵ = 0
therefore
q1
= q2 (d − y)2
y2
So, we end up with the following quadratic equation :
(q1 − q2 )y 2 − 2q1 dy + q1 d2 = 0
Solving this equation, we find two solutions :
y=
√
√
q1 d ± q1 q2 d
q1
= d√
√
q1 − q2
q1 ∓ q2
Only the solution using the lower sign is physically sensible, (the upper sign would predict, e.g. that if
q1 = q2 then y = ∞) so
√
q1
y = d√
√
q1 + q2
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