Homework 1 Solutions
Problem 1.1 (practice): Dimensions When physicists calculate the value
of a physical quantity from an equation, they pay particular attention to the
units involved. A force of 2 is ill-defined, a force of 2 Newtons is clear. When
physicists want to check the plausibility of an equation, without worrying exactly
about which set of units will be used (e.g. Newtons vs. pounds vs. dynes), they
often look at the “dimensions” of the physical quantities involved. “Dimension”
refers to the powers of the basic physical quantities: length (L), time (T ), mass
(M ), and charge (C), that make up the physical quantity. For example, since
force is mass times acceleration, the dimensions of force are M L/T 2 . Find
the dimensions of electrostatic potential energy. Also, find the dimensions of
electrostatic potential.
Solution: For electrostatic potential energy (note: the brackets indicate that
we’re talking about the dimensions of the quantity):
[U ] =
M L2
T2
[V ] =
M L2
QT2
For electrostatic potential:
Problem 1.2: Potential vs. Energy In this course, two of the primary
examples we will be using are the force due to gravity and the force due to an
electric charge. Both of these forces vary like 1/r2 , so they will have many,
many similarities. Most of the calculations we do for the one case will be true
for the other. But there are some extremely important differences:
a) Find the value of the electric potential energy of a system consisting of a
hydrogen nucleus and an electron separated by the Bohr radius. Find the
value of the gravitational potential energy of the same two particles at the
same radius. Use the same system of units in both cases. Compare and
the contrast the two answers.
Solution: Bohr radius: a0 ≈ 0.0529 nm
Proton charge: Q ≈ 1.60 × 10−19 C
2
2
1
9
4π0 ≈ 9.0 × 10 Nm /C
Gravitational constant: G ≈ 6.67 × 10−11 Nm2 /kg2
Proton mass: M ≈ 1.67 × 10−27 kg
Electron charge: q ≈ −1.60 × 10−19 C
Electron mass: m ≈ 9.11 × 10−31 kg
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Therefore:
Uelec
=
qVelec
1 qQ
4π0 a0
−4.35 × 10−18 J
(1)
mVgrav
mM
= −G
a0
≈ −1.92 × 10−57 J
(4)
=
≈
Ugrav
=
(2)
(3)
(5)
(6)
The potential energy due to gravity in an atom is 28 orders of magnitude
smaller than the electromagnetic force!
b) Find the value of the electric potential due to the nucleus of a hydrogen
atom at the Bohr radius. Find the gravitational potential due to the
nucleus at the same radius. Use the same system of units in both cases.
Compare and contrast the two answers.
Solution:
See constants above. Therefore:
1 Q
≈ 27.2 J/C
4π0 r0
Velec
=
Vgrav
= −G
M
≈ −2.11 × 10−27 J/kg
r0
(7)
(8)
Even though these are measured in the same system of units, they are not
in the same units and cannot be compared. (Notice that qVelec is not the
binding energy of the electron. The electron in the Bohr model also has
kinetic energy.)
c) Think of and briefly discuss at least one other fundamental difference
between electromagnetic and gravitational systems. Hint: Why are we
bound to the earth gravitationally, but not electromagnetically?
Solution: One difference is that charges of both signs exist, so that
electromagnetic forces can cancel each other. The earth is essentially
neutral, so that we do not feel an electromagnetic attraction to the earth.
Gravitational forces, however, can only add, so that the total force due to
lots of mass can be very large.
Another difference is that masses always attract, so that the gravitational
potential is negative. Whereas, positive charges repel each other so that
the electric potential of a positive charge is positive.
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Problem 1.3: Trig parameters Make sketches of the following functions,
by hand, all on the same axes. Briefly describe, using good scientific writing
that includes both words and equations, the role that the number two plays in
the shape of each graph:
y
=
sin x
y
=
2 + sin x
(10)
(9)
y
=
sin(2 + x)
(11)
y
=
2 sin x
(12)
y
=
sin 2x
(13)
3
2
1
K3
K2
K1
0
1
2
3
x
K1
K2
Solution:
Color
Red
Green
Gold
Blue
Purple
Function
sin x
2 + sin x
sin(2 + x)
2 sin x
sin 2x
Description
the original function
shifts the function 2 units up along the vertical axis.
shifts the function left 2 units along the horizontal axis.
increases the amplitude of the function by two units.
squishes the function to that the frequency of the function
is twice as large as before.
Problem 1.4: The θ function The function θ(x) (the Heaviside or unit step
function) is a defined as:
1
for x > 0
θ(x) =
0
for x < 0
(see p. 352 of Riley, Hobson & Bence). This function is discontinuous at x = 0
and is generally taken to have a value of θ(0) = 1/2.
Make sketches of the following functions, by hand, on axes with the same
scale and domain. Briefly describe, using good scientific writing that includes
both words and equations, the role that the number two plays in the shape of
3
3
3
2
2
y
K3
K2
2
y
1
K4
3
K1
y
1
0
1
2
3
x
4
K4
K3
K2
K1
1
0
1
2
3
4
x
K4
K3
K2
K1
0
1
2
3
4
x
K1
K1
K1
(a)
(b)
(c)
Figure 1: Solution to Problem 4. (a) in red, θ(x), the original function, and
in green, 2 + θ(x), shifts the function 2 units up along the vertical axis. (b) in
gold, θ(2 + x), shifts the function left 2 units along the horizontal axis. (c) in
blue, 2θ(x), increases the amplitude of the function by two units, and in purple,
θ(2x), nothing happens.
each graph:
Solution:
y
y
=
=
θ(x)
2 + θ(x)
(14)
(15)
y
=
θ(2 + x)
(16)
y
=
2θ(x)
(17)
y
= θ(2x)
(18)
See Figure 1.
Problem 1.5: Plotting a vector relation
a) Make a sketch of the graph
|~r − ~a| = 2
(19)
for each of the following values of ~a:
~a = ~0
~a =
~a =
(20)
2ı̂ − 3̂
(21)
points due east and is 2 units long
(22)
Solution: These are the equations for a circle of radius 2 centered at (i)
the origin, (ii) (2, 3) and (iii)(2, 0), respectively.
b) Derive a more familiar equation equivalent to
|~r − ~a| = 2
4
(23)
for arbitrary ~a, by expanding ~r and ~a in rectangular coordinates. Simplify
as much as possible. (Ok, ok, I know this is a terribly worded question.
What do I mean by “more familiar”? What do I mean by “simplify as
much as possible”? Why am I making you read my mind? Try it anyway.
Real life is not full of carefully worded problems. Bonus points to anyone
who can figure out a better way of wording the question that doesn’t give
the point away.)
Solution:
Let
~r
= rx î + ry ĵ
~a = ax î + ay ĵ
then
p
(~r − ~a) · (~r − ~a)
q
=
[(rx î + ry ĵ) − (ax î + ay ĵ)] · [(rx î + ry ĵ) − (ax î + ay ĵ)]
q
(rx − ax )2 + (ry − ay )2
=
|~r − ~a| =
So:
|~r − ~a| = 2
q
→ (rx − ax )2 + (ry − ay )2 = 2
(rx − ax )2 + (ry − ay )2
=
4
which is the equation for a circle written in Cartesian coordinates.
c) Write a brief description of the geometric meaning of the equation
|~r − ~a| = 2
(24)
Solution: This is the equation for a circle centered at the tip of vector
~a with radius 2 units.
Problem 1.6: Dipole symmetries Consider a system consisting of one positive and one negative point charge, with equal magnitude charge. They are not
at the same position in space.
a) Choose a coordinate system for this problem, and sketch the system.
b) Describe all symmetry operations for this system.
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Solution:
a) One reasonable coordinate system choice would be to put the positive
charge on the +z axis, and the negative charge on the −z axis, with
the point halfway between them at the origin. I’ll use rectangular (or
Cartesian) coordinates for this problem.
b) There are a lot of symmetry operations.
1. Vertical mirror plane x → −x, y → −y or any similar mirror plane,
which contains the z axis.
2. Any rotation about the z axis.
3. A horizontal mirror plane, reflecting z → −z, combined with an
inversion of the sign of charge Q → −Q.
4. Inversion through the origin, ~r → −~r, combined with an inversion of
the sign of charge Q → −Q.
5. Inversion through the origin, ~r → −~r, combined with any rotation
around the z axis, and also combined with an inversion of the sign
of charge Q → −Q.
6. A horizontal mirror plane, reflecting z → −z, combined with an
inversion of the sign of charge Q → −Q, as well as any rotation
around the z axis.
7. A horizontal mirror plane, reflecting z → −z, combined with an
inversion of the sign of charge Q → −Q, as well as reflection through
any vertical mirror plane (as in 2a).
8. A π rotation around any horizontal axis, combined with an inversion
of the sign of charge. This is actually identical to 2g, but I list it
separately, since either answer is equally valid.
Problem 1.7 (challenge): Triangle function Consider the function:
f (x) = 3x θ(x) θ(1 − x) + (6 − 3x) θ(x − 1) θ(2 − x)
Make sketches of the following functions, by hand, on the axes with the same
scale and domain. Briefly describe, using good scientific writing that includes
both words and equations, the role that the number two plays in the shape of
each graph:
Solution:
y
=
f (x)
(25)
y
=
2 + f (x)
(26)
y
=
f (2 + x)
(27)
y
=
2f (x)
(28)
y
=
f (2x)
(29)
See Figure 2.
6
6
6
6
5
5
5
4
4
4
3
K4
K3
K2
3
3
2
2
2
1
1
1
K1
y
0
1
2
3
4
K4
K3
K2
K1
0
K1
K1
(a)
(b)
1
2
x
3
4
K4
K3
K2
K1
0
1
2
3
4
K1
(c)
Figure 2: Solution to Problem 7. (a) in red, f (x), the original function, and
in green, 2 + f (x), shifts the function 2 units up along the vertical axis. (b) in
gold, f (2 + x), shifts the function left 2 units along the horizontal axis. (c) in
blue, 2f (x), increases the amplitude of the function by two units, and in purple,
f (2x), nothing happens.
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