Physics II
Problem Set 1
Due: January 30, 2025
This problem set is due on Brightspace by 12:30 pm on January 30, 2025.
You may collaborate with others on this problem set and consult external sources, subject to course policies. However, you must write your own solutions and list your
collaborators and sources for each problem.
Read Chapter 21 in Young & Freedman, which discusses Coulomb’s Law and some of its ramifications. Review the Chapter Summary on page 706. Then answer the following questions
and upload your answers to Brightspace.
1. Young & Freedman, Exercise 21.7 (page 709)
An average human weighs about 650 N. If each of two average humans could carry 1.0 C
of excess charge, one positive and the other negative, how far apart would they have to
be for the electric attraction between them to equal their 650 N weight?
2. Young & Freedman, Exercise 21.21 (page 710)
A proton is placed in a uniform electric field of magnitude 2.75 → 103 N C→1 . Calculate
(a) the magnitude of the electric force felt by the proton.
(b) the proton’s acceleration.
(c) the proton’s speed after 1.00 µs in the field, assuming it starts from rest.
3. Young & Freedman, Exercise 21.33 (page 710)
A very long line of charge, with charge per unit length 8.00 µC m→1 lies along the x-axis,
and its midpoint is at x = 0. A second very long line of charge with charge per unit
length ↑4.00 µC m→1 is parallel to the x-axis at y = 10.0 cm, and its midpoint also is
at x = 0. At what point on the y-axis is the resultant electric field of the two lines of
charge equal to zero?
4. Young & Freedman, Problem 21.57 (page 712)
Four identical charges, Q, are placed at the corners of a square of side L.
(a) In a free-body diagram, show all of the forces that act on one of the charges.
(b) Find the magnitude and direction of the total force exerted on one charge by the
other three charges
5. Young & Freedman, Problem 21.69 (page 713)
A charge +Q is located at the origin. A second charge, +4Q, is located a distance d away
along the x̂ axis. Where should a third charge, q, be placed so that all three charges are
in mechanical equilibrium? A suitable answer includes an expression for q in terms of
Q, d and the coordinates of the charge’s equilibrium position.
Hint: You might want to investigate Earnshaw’s theorem.
6. Young & Freedman, Exercise 21.53 (page 712)
Torque on a dipole: An electric dipole with dipole moment p is in a uniform electric
field, E.
Prof. David Grier
Page 1 of 2
Physics II
Problem Set 1
Due: January 30, 2025
(a) Find the orientations of the dipole for which the torque on the dipole is zero.
(b) Which of the orientations discovered in part (a) is stable, and which is unstable.
(Hint: Consider a small rotation away from the equilibrium position and see what
happens.)
(c) Show that for the stable orientations in part (b), the dipole’s own electric field tends
to oppose the external field.
Prof. David Grier
Page 2 of 2
Given
,
W
650N
=
-
12
=
92
12
=
Coulombs Law
According to
,
here,
KIq1 gal
·
F
=
K
V2
and
K .2
pr
=>
=
8 99 X109
=
.
F
,
Nm252
w
=
F
K
=
8
=
.
.
1
·
82
99x18Nm212
GSON
3718 97m
=
.
know ,
We
magnitude
here ,
F
,
.
g
=
-
E
(1 6x10C)
=>
.
·
(2 75x103N/2)
q
E
.
1
=
=
.
6 x10'C
2
.
75 x103N/C
4 4x106N
.
F
now ,
=
a
=>
here ,
ma
=
=>
-
m
F/m
MXI y
2 63x10" m52
.
Calculating speed when
v
=
=
a
.
1000s
t
(2 63x10" m52) (x10's)
.
-
v
=
,
2
.
63x105m51
where Vo
,
=
o
=
1
.
67x1527kg
Given
,
>
1st line -
-
0
=
0 Im
.
equation for both
this
J -4 Mem
=
,
=*
E
We know ,
J - Brc/m
,
line-by-10cm
and
setting
y
lines
,
2
=
Vi
!
>
By
>
4y
-
-
=>
0 8
-
y
4y
=
.
0
=
=
.
8
0 2
.
electric field resultant
...
when , y
=
0 2 m
.
=
-
o
20cm
or
112
⑧
%3
82x +
E
o
·4
II
My + 92]
V
Figure
free
now ,
:
body diagram
for adjacent changes
for
=
'L
=
,
Fady =F-
[ a b2 ]
+
diagonal changes, ding
changes
,
therefore
the
direction
away
from
square
.
resultant force,
would
the
be
center
facing
of the
F
(Aret
- (H)
=> +Y
=
=
-
-
F,
equilibrium ,
for
=
Fe
=>
I
a
=
Q(d x)2
>
-
-
40X2
=
> -
d
=
-
X
d
=
=
2X
3x
=
x
·:
=
d/z
negative.
be in equilibrium , the change must be
to
therefore
we
=
,
know ,
Horque T pEsig
=
,
for nonzero
...
or
psE
0-8 , stable
,
-
is
O
ifsino-o
aligned with field
0-180 ; unstable and opposite to field
at 0
which
or
stable
and
so
,
dipole is
orientation
orientation
position, the opposite changes
:·
either
parallel or antiparallel.
potential energy
:1
create ther
"
own
counteracts the appliedf. This reduces resultant E
.
is
minimum
"maximum.
electric field