Phy213: General Physics III
Chapter 21 Worksheet
2/6/2016
1
Coulomb’s Law & Electric Fields:
1. Consider a fixed point charge of +1.5 C (q1). A 2nd charge of -0.5 C (q2) is placed at a
distance of 0.5 m from q1.
q1 = +1.5 C
q2 = -0.5 C
d = 0.5 m
a. What is the magnitude and direction of the electric force exerted on q2 due to q1?
Ans.
qq
FE= -k 1 2 2 ˆi=- 8.99×109
d

N×m2
C2

1.5×10 C 0.5×10 C  ˆi=-0.0270N ˆi
-6
-6
0.5m
2
b. A 3rd charge of -1.0 C (q3) is placed 0.25 m directly to the right of q2. What is the magnitude
and direction of the electric force exerted on q3?
Ans.
q1 = +1.5 C
 qq
qq 
FE=  -k 21 3 + k 22 3  ˆi =
d2 3 
 d13

9 N×m2
C2
FE= 8.99×10
q2 = -0.5 C
d = .25 m
d = 0.5 m

 
 1.5×10-6C
0.5×10-6C
1.0×10 C  +
2
  0.75m2
0.25m


-6

q3 = -1.0 C
  ˆi =0.0479N ˆi


c. If q3 is exactly halfway between q1 and q2, what is the magnitude and direction of the electric
force exerted on q3?
q1 = +1.5 C
Ans.
d = .25 m
 qq
qq 
FE=  -k 21 3 - k 22 3  ˆi =
d2 3 
 d13

9 N×m2
C2
FE= 8.99×10

 1.5×10-6C
1.0×10 C  2

0.25m


q2 = -0.5 C
q3
-6


d = .25 m
0.5×10 C   ˆi = -0.288N ˆi
-6
-
0.25m
2


d. If q3 is moved 0.3 m downward but still equidistant to q1 and q2, what is the magnitude and
direction of the electric force exerted on q3?
Fq1
Ans.
1
q
 
3
2
q3 
ˆ
ˆ

F
q2
q
cos

+q
cos

i
+
q
cos

+q
cos

j=
 1
 1
1
2
2
1
2
2 
d2 
-6
-6
o
-1.0×10-6C  1.5×10 C+0.5×10 C cos 50.2 ˆi 
9 N×m2


FE= 8.99×10 C2
2
-6
-6
o ˆ

0.391m + 1.5×10 C-0.5×10 C sin 50.2 j
FE= k



FE= -0.075N ˆi + 0.045N ˆj
 







Phy213: General Physics III
Chapter 21 Worksheet
2/6/2016
2
e. Where would q3 need to be placed such that there is no electrical force acting on the charge?
Ans. {Assume all charge positions are referenced to the origin at q3}

qq
qq 
FE=  -k 21 3 - k 22 3  ˆi +
dx 2 3 
 dx 13
in x: k
q1q3
qq
= k 22 3
2
dx 13
dx 2 3

qq
qq 
 -k 21 3 - k 22 3  ˆj= 0 N ˆi + 0.0 N ˆj
dy 2 3 
 dy 13

dx 2 3
d2
= 2x 2 3 =
dx 13
dx 13
q2
= 0.58
q1
when xq1 = -0.75 mˆi then xq2 = -0.43 mˆi
qq
qq
in y: k 21 3 = k 22 3
dy 13
dy 2 3

dy 2 3
dy 13
=
d2y 2 3
2
y 1 3
d
=
q2
= 0.58
q1
when yq1 = +0.75 m ˆj then y q2 = +0.43 m ˆj
2. Consider a simple model of an atomic nucleus. In this case, three protons are present in the
nucleus of a lithium atom, forming an equilateral triangle. The distance, r, between each pair of
protons is 1.5 x 10-15 m.
Lithium Nucleus
+
a. What is the magnitude of the electric force exerted between
any two of the protons?
Ans.
e2
FE = k 2 = 8.99×109
d

1.602×10 C

1.5×10 m
2
-19
2
N×m
C2
-15
2
= 103N
b. What is the magnitude and direction of the total electric force
exerted on a proton (due to its neighboring protons). You will
need to assign an appropriate coordinate system.
r
+
60
ĵ
o
+
î
F1
F2
30o
FE
Ans. Using the bottom charge as a reference and applying the Law of Cosines to the forces:
2
 e2 
k 2 
 d 
FE =
2
 e2 
+ k 2 
 d 
2
 e2 
- 2  k 2  cos150o = 178 N
 d 
Applying the Law of Sines to obtain the direction of the resultant force:
sinF
 sin 120o 
 sinF =103N 
 = 0.501
E
103N
 178 N 
F =sin-1 0.501 =30o south of -ˆi or 210o w/r to ˆi
E
=
sin 120o
178 N

E

c. Express the total electric force vector, in component form.
Ans. Using the bottom charge as a reference:
FE= -k
e2
d2
1 + cos60 ˆi +
o

sin60o ˆj= -155 N ˆi - 89.2 N ˆj
Phy213: General Physics III
Chapter 21 Worksheet
2/6/2016
3
d. What is the net force acting on the subject proton in (b) & (c)? Explain.
Ans. Since the protons are fixed (static equilibrium), there must be at least 1 additional force
acting on each nucleon to oppose the net electric (net) force:
FNet= FE+F??= 0 N ˆi + 0 N ˆj
This is was the justification for proposing a “strong nuclear force” within the nucleus!
3. Consider the “Bohr Model” of the atom, where an electron moves around a proton in a circular
orbit. Assume that the proton and electron are separated by a distance, r, of 5.29x10-11 m.
proton
electron
r = 5.29x10
-11
m
a. What is the magnitude and direction of the electric
force exerted on electron?
Ans.
e2
FE = k 2 = 8.99×109
d

1.602×10 C

5.29×10 m
-19
2
N×m
C2
-11
2
2
= 8.24×10-8N
b. What is the centripetal force exerted on the electron such that it maintains this orbit?
Ans.
Fnet = Fc = FE = 8.24×10-8N
c. What is the speed of the electron?
Ans.
mv2
Fc =
= 8.24×10-8N  v=
r
Fc r
m
8.24×10 N5.29×10
-8
=
-11
-31
9.11×10 kg

m
2
= 2.19×106
d. What is the kinetic and potential energy, respectively, for the electron in J and eV?
Ans.
K=
mv2
= 2.18×10-18 J = 13.6 eV
2



U= FE  r  cos 180o = - 8.24×10-8N 5.29×10-11m = -4.36×10-18 J = -27.2 eV
e. What is the total energy of the electron in J and eV?
Ans.


Etot = K + U = 2.18×10-18J + -4.36×10-18J = -2.18×10-18 J
or
Etot = K + U = 13.6 eV - 27.2 eV = -13.6 eV
f. Calculate the orbital radius for an electron with a total energy (K+U) of -3.40 eV.
m
s
Phy213: General Physics III
Chapter 21 Worksheet
2/6/2016
4
Ans.
Etot = K + U = -3.40 eV = -5.45×10-19 J
 mv2 
mv2
mv2
o
+ FE  r  cos 180o =
+ 
  r  cos 180 =
2
2
r


2
2
mv
mv
Etot =
- mv2 = = -5.45×10-19 J
2
2
2
-18
or mv = 1.09×10 J
Etot =
since
Fc = FE you can solve for r:
mv2
e2
e2
=k 2 r=k
=
r
r
mv2
8.99×10  1.602×10
1.09×10 J
9 Jm2
C2
-19
-18
C

2
= 2.12×10-10m
4. What is the total electric charge of 1.0 moles of electrons?
Ans.

Q = Ne = 6.022×1023
electrons
mol
1.602×10
-19
C
electron

= 96400
5. How many electrons are present in -1.5 C?
Ans.


1.5×10-6C
Q
N=
=
C
e
1.602×10-19 electron

or
1.5×10 C 

= 9.36  1012 electrons
-6
n=
96400
C
mol
= 1.56×10-11 mol
C
mol