MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2013
Problem Set 1 Solutions
Problem 1: Classical Model of the Hydrogen
In a simple model of the hydrogen atom, an electron moves in a circular orbit about a
proton, where the radius of the orbit is 0.529 1010 m .
a) Find the magnitude of the electric force between the two.
Using Coulomb’s law, the magnitude of the force between the proton and the electron
is
F
1 e2
(1.60  1019 C)2
9
2
2

(8.99

10
N

m
C
)
 8.22  108 N
2
10
2
4 0 r
(0.529  10 m)
b) If this force causes the centripetal acceleration of the electron, what is the speed
of the electron?
mv 2
Using F 
, the speed of the electron is
r
v
Fr

m
(0.529  1010 m)(0.529  1010 m)
 2.19  106 m/s .
9.11  1031 kg
c) What is the ratio of the magnitudes of the electrical force and the gravitational
force between the proton and electron?
The ratio of the forces is given by
r
Felec
ke2 / r 2
ke2
(9.0  109 N  m 2  C-2 )(1.6  1019 C)2



r
2
Gme mp (6.67  1011 N  m 2  kg-2 )(1.67  1027 kg)(9.11 1031 kg)
Fgrav Gme mp / r
 2.27  1039
.
This is a very large ratio indicating how much stronger electric forces are than
gravitational forces.
Problem 2 An electron of mass me and charge q  e is injected horizontally midway
between two very large oppositely charged plates. The upper plate has a uniform
positive charge per unit area  and the lower plate has a uniform negative charge per
unit area  . You may ignore all edge effects. There is a uniform electric field
between the plates given by
r

E φ
j.
0
r
The particle has an initial velocity v 0  v0φ
i . What is the y-component of the position of
the particle when the particle reaches the plane defined by x  L .
r
r
Solution: We apply Newton’s Second Law, F  mea , where the force on the electron
r
r
r
r
is given by F  (e)E . Thus Newton’s Second Law becomes (e)E  mea . Using our
r
result for the electric field we have that (e)(( / )φ
j)  m a and so the acceleration of
e
the electron is
e φ
r
a
j.
me0
The initial velocity of the electron is in the positive x -direction. Let t  0 denote the
instant that the electron enters the plates at x  0 and t f denote the instant the electron
leaves the plate at x f  L . Because there is no acceleration in the x -direction, the
electron maintain constant speed in the x-direction and hence L  v0t f . The electron
has constant acceleraton in the positive y -direction. Therefore at t f , the y -component
of the electron when it leaves the plates is given by
1
1 e L2
2
.
y f  ayt f 
2
2 me0 v0 2
Problem 3: Five Charges
Five identical charges Q are equally spaced on a semi-circular circle of radius R, as
shown in the figure. Find the force on a charge q located at the center of the semicircle.
Solution:
By considering the symmetry of the array of charges we can see that


the force exerted on q by the +Q charge at (0, R) cancels the force by the +Q
charge at (0, R), and
the y component of the force on q is zero.
We can apply Coulomb’s law and the principle of superposition of forces to find the net
force acting on q.
The net force is
r
Fq 


1 Qq φ
1 Qq
1 Qq
i  2
cos 45φ
i=
1 2 φ
i
2
2
4 0 R
4 0 R
4 0 R2
PS01-3
Problem 4: Electroscope
An electroscope can be made from two small conducting balls of mass m hanging on
long strings of length L . If a charge 2q is transferred to the system (so each ball
acquires the same charge q ), they will repel.
(a) Neglecting gravitational attraction between the balls, by what distance x do they
move apart when charged? (You must still consider the gravitational interaction
with the Earth.)
(b) Is it reasonable to ignore this gravitational attraction between the two objects?
More precisely, if we were to put a very small charge, say one electron, on each of
the balls, how light would the balls have to be before we could ignore gravitational
attraction between them? Could we make the balls that light?
Solution
(a) To do this analysis we draw a free body diagram for the left mass, where the forces
are gravity (down), electrostatic force (left) and tension in the string (up and to the left):
Since the ball is not accelerating, we know that the forces in both the horizontal and
vertical directions must cancel:
Fy  T cos   Fg  0  T 
Fg
cos 

mg
cos 
PS01-4
1
q2
mg sin 
2 x

T
sin



mg
tan


mg
2
x
cos 
L
Fx  T sin   Fq  0  Fq  k
where we have used the small angle approximation that tan   sin  , which is valid
when the length of the string L is long (relative to the separation x ), which we are told
it is. Therefore
2q2 L
x k
mg
3
13
 2kq 2 L 
x

 mg 

13
 q2 L 


 2 0mg 
(b) Is it reasonable to ignore this gravitational attraction? More precisely, if we
were to put a very small charge, say one electron, on each of the balls, how light
would the balls have to be before we could ignore gravitational attraction
between them? Could we make the balls that light?
This question is a little vague (intentionally). How small does something need to be so
that we can ignore it? A 1% effect? 0.1%? It really depends on what we are trying to
do. Let’s just calculate for a fraction f of the electrostatic force, and we can go from
there.
FG  fFq 
Gm 2
kq 2

f
m
x2
x2
f
kq 2
G
For a single electron charge we find the mass would need to obey:
 9  10 Nm C 1.6 10 C
f
 6.67  10 Nm kg 
9
m
2
19
-2
11
2
-2
2

f  3.5  1018 kg 2  1.9  109 kg  f
So, for a 1% effect (f = 0.01) we’d need m  2  109 kg  10-2  2  1010 kg  200 ng
That is pretty small. We probably aren’t justified in ignoring gravity in this case. But
put on a million electrons so we are up near a gram and this becomes reasonable.
PS01-5
Problem 5: Tetrahedron Model of Methane
r r
Consider a regular tetrahedron whose vertices are given by the position vectors r1  0 ,
r
r
φ , and rr  a(φ
φ , where a is the length of the side of the
r  a(φ
i+φ
j) , r  a(φ
i + k)
j + k)
2
3
4
cube formed by the vertices of the tetrahedron.
(a) Verify that each face of a tetrahedron is an equilateral triangle by showing that the distances
between any pair of vertices are the same.
(b) We define the unit vector pointing from point A to point B by the expression
r
r r
rAB
rB  rA
rφAB  r
 r
.
| rAB | | rAB |
Find the unit vectors r̂21 which points from v2 to v1 , r̂31 which points from v3 to v1 and r̂41
which points from v4 to v1 .
(c) In the methane molecule, CH4 , each hydrogen atom is at the vertex of a tetrahedron with the
carbon atom at the center. In the coordinate system for the four vertices given in part (a), what is
the position vector of the carbon atom?
r
(d) Find the vector rC1 from the carbon atom to the hydrogen atom located at v1 , and the vector
r
rC2 from the carbon atom to the hydrogen atom located at v2 .
(e) The angle  between the bonds between the carbon atom and hydrogen atoms located at
r r
r r
vertices v1 and v2 , is given by the expression   cos1 rC1  rC2 / rC1 rC2 . What is the angle


between these two bonds?
PS01-6
(f) Suppose identical objects each of charge q are placed at each vertex of the tetrahedron shown
in part (a). Find the electric force acting on the charged object located at v1 due to the interaction
with other three charged objects.
Solutions
(a) One can easily verify that the distances between the vertices are
r12  r13  r14  r23  r24  r34  a 1  1  a 2
(b) The unit vectors are
r
r21
φ
r21  r 
| r21 |
r
r31
rφ31  r 
| r31 |
r
r41
rφ41  r 
| r41 |
r r
r1  r2
1

( φ
iφ
j)
r
| r21 |
2
r r
r1  r3
1
φ

( φ
i  k)
r
| r31 |
2
r r
r1  r4
1
φ

( φ
j  k)
r
| r41 |
2
a
r
φ.
(c) The position vector of the carbon atom is rC  ( φ
i+φ
j + k)
2
r
(d) The vector rC1 from the carbon atom to the hydrogen atom located at v1 is given by
a
r
r r
φ . The magnitude is rr  a 3 .
rC1  r1  rC   (φ
i+φ
j + k)
C1
2
2
r
The vector rC2 from the carbon atom to the hydrogen atom located at v2 is given by
a
r
r r
φ  a (φ
φ . The magnitude is rr  a 3 .
rC2  r2  rC  a(φ
i+φ
j)  (φ
i+φ
j + k)
i+φ
j  k)
C2
2
2
2
(e) The angle  between the bonds between the carbon atom and hydrogen atoms located at
vertices v1 and v2 , is given
r
r
r
r
  cos1 rC1  rC2 / rC1 rC2

 a
φ  ( a (φ
φ / rr rr 
 cos 1   ( φ
i+φ
j + k)
i+φ
j  k))
C1 C2 
2
 2





 cos 1 ( a 2 / 4) / (3a 2 / 4)  cos 1 1 / 3  109.5o
PS01-7
(f) The force on the charge at the origin is given by the expression
r i 4 q 2
q2
q2
q2
q2
F1   ke 2 rφi1  ke 2 rφ21  ke 2 rφ31  ke 2 rφ41  ke
rφ + rφ  rφ
2 21 31 41
ri1
r21
r31
r41
i 2


Using our result from part (b) for the unit vectors, we have that the electric force on the object
located at the origin due to the other three is
r
q2
F1  ke 2 rφ21 + rφ31  rφ41
r21


q2  1
1
φ  1 ( φ
φ
 ke 2 
( φ
iφ
j) 
( φ
i  k)
j  k)

2a  2
2
2
 ke



2
q2 1
φ
φ
φ
φ  k q

2
i

2
j

2
k
iφ
j  kφ
e 2
2a 2 2
a 2

PS01-8
Problem 6 Electric field for a Distribution of Point Charges
A right isosceles triangle of side a has objects with charges q , 2q and 3q arranged on its
vertices, as shown in the figure below. (a) What is the magnitude and direction of the electric
field at point P due to the charges in the figure, midway between the line connecting the q and
2q charges? (b) Find a vector expression for the electric dipole moment associated with this
charge distribution.
Answer: Let’s call 1 the object located at (0, a) with q1  q , 2 the object located (0,0) with
q2  3q , and 3 the positively charged object with q3  2q located at (a,0) . We begin by
drawing the three contributions to the electric field.
The electric field at the point P  (a / 2, a / 2) is the superposition of these three fields
r
r
r
r
E(P)  E1 (P)  E2 (P)  E3 (P) .
We start with the field due to object 1 the located at (0, a) with q1  q . The electric field is
given by the expression
r
kq1 r
E1 (P) 
r .
(r1, P )3 1, P
PS01-9
r
Recall that the vector r1, P is the vector that starts at the object 1 and ends at the point P. From the
figure above, we can write this vector as
r
r1, P  (a / 2)φ
i  (a / 2)φ
j.
The magnitude of this vector is
r q , P  r q , P  ((a / 2) 2  (a / 2) 2 )1/ 2  a / 2 .
Thus
r
kq1 r
kq
2kq
E1 (P) 
r 
((a / 2)φ
i  (a / 2)φ
j)  2 (φ
iφ
j) .
3 1, P
3
(r1, P )
a
(a / 2)
The electric field due to the object 2 is:
r
kq2 r
E2 (P) 
r .
(r2, P )3 2, P
r
The vector r2, P is the vector that starts at object 2 located at the origin and ends at the point P.
From the figure above we can write this vector as
r
r2, P  (a / 2)φ
i  (a / 2)φ
j.
The magnitude of this vector is
r
r2, P  r2, P  ((a / 2)2  (a / 2)2 )1/ 2  a / 2 .
Thus
r
E2 (P) 
q2
(r2, P )
3
r
r2, P 
3kq
3 2kq φ φ
((a / 2)φ
i  (a / 2)φ
j) 
( i  j) .
a2
(a / 2)3
The electric field due to object 3 the located at (a,0) with q3  2q is
r
kq3 r
E3 (P) 
r .
(r3, P )3 3, P
r
The vector r3, P is the vector that starts at the object 3 and ends at the point P. From the figure
above, we can write this vector as
r
r3, P  (a / 2)φ
i  (a / 2)φ
j.
The magnitude of this vector is
PS01-10
r
r3, P  r3, P  ((a / 2)2  (a / 2)2 )1/ 2  a / 2 .
Thus
r
r
kq3 r
k2q
2 2kq φ φ
E3 (P)  E3 (P) 
r 
((a / 2)φ
i  (a / 2)φ
j) 
( i  j) .
3 3, P
(r3, P )
a2
(a / 2)3
Thus the vector sum is
r
r
r
r
E( P)  E1 ( P)  E2 ( P)  E3 (P)

2kq φ φ 3 2kq φ φ 2 2kq φ φ
( i  j) 
( i  j) 
( i  j)
a2
a2
a2

2kq
((1  3  2)φ
i  (1  3  2)φ
j)
a2

2kq
(4φ
i  2φ
j)
a2
.
3
r
r
r
(b) The electric dipole moment about a point P is defined as p =  qi rPi where rPi is the vector
i1
form the point P to the ith object. When the total charge of the objects is zero, the dipole
moment is the same about any point (independent of the point P , can you prove this?) so we
will choose the origin O for the point P . Then
r r
r
r
rO1  aφ
j ; rO2  0 ; rO3  a φ
i .
So the dipole moment becomes
r
r
r
r
p  q1rO1  q2rO2  q3rO3  qaφ
j  2qa φ
i  qa(2φ
iφ
j) .
PS01-11
Problem 7: Dipole Moment Water Molecule
The water molecule, (see figure below), consists of two hydrogen atoms and one oxygen
arranged such that the lines joining the center of the O atom with each H atom make an angle of
1040. A distance d  9.6 1011 m separates the centers of each hydrogen and oxygen pair. Each
pair of hydrogen and oxygen atoms have dipole moments p1 and p 2 . The entire molecule has a
dipole moment p , which is the sum of the dipole moments p  p1  p 2 . The magnitude of the
r
total dipole moment is p  6.1  1030 C  m . What is the effective charge on each hydrogen
atom?
Solution
Let q be the effective charge on the H atom. The dipole moment of one H-O pair is then qd ,
where d  9.6 1011 m . The two dipoles are at an angle of 104 degrees or a half-angle of 52
r
degrees. The total dipole moment is thus p  2qd cos(52o)  6.1  1030 C  m . If we use the
value of d given above, we find that q  5.16  1020 C , or about 1/3 of the charge of an
electron.
PS01-12
Problem 8: Five Charges
Four identical particles, each having charge +q, are fixed at the corners of a square of side L. A
fifth point charge –Q lies a distance z along the line perpendicular to the plane of the square and
passing through the center of the square, as shown in the figure.
(a) Find the force exerted on –Q by the other four charges. Be sure to specify the magnitude and
direction.
x
–Q
z
+q
+q
L/2
L/2
+q
+q
The distance from each corner to the center of the square is
r1  (L / 2)2  (L / 2)2 
L
2
Thus, the distance from each positive charge to Q is then r  (z 2  L2 / 2)1/ 2 . Each positive
charge exerts a force directed along the line joining q and Q , of magnitude
F1 
keQq
r2

keQq
z 2  L2 2
where ke  1 / 4 0 . By symmetry argument, one may show that only the z-component of the
force is non-zero. With
PS01-13


k Qqz
z
 2 e 2 3/ 2


2
2
z  L 2  z 2  L2 2  (z  L 2)
keQq
F1z  F1 cos 
The net force exerted on Q by the four charges is
r
φ 
F  4F1z (k)
4keqQz
(z  L / 2)
2
2
3/ 2
kφ = 
qQz
kφ
 0 (z  L2 / 2)3/ 2
2
Note that this force is directed toward the center of the square, whether z is positive (–Q above
the square) or negative (–Q below the square).
(b) If z is small compared with L, show that the motion of –Q is simple harmonic. What is the
period of this motion if the mass of –Q is m?
For z  L , the magnitude of this force may be approximated as
Fz  
4ke qQ
L3
z  maz  m
d 2z
dt 2
Therefore, the object’s vertical acceleration is of the form az   2 z with
2 

4 2
32
keQq
3
mL

keQq 128
mL3
Since the acceleration of the object is always oppositely directed to its excursion from
equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion
with a period given by
T
2


2
128
14
mL3

 14
keQq
8

mL3
.
keQq
PS01-14