** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #8
(20 Points Total)
Chapter 6
Questions for Thought
1. Explain why a balloon that has been rubbed sticks to a wall for awhile.
I feel that the correct expression of the answer to this type of question is fundamental to your
demonstrating that you understand the concepts of static electricity and charging by induction.
(although this is not a charging by induction problem.) The balloon picks up excess electrons
through the rubbing action. How this occurs or why I am not entirely sure of… However, the
effect is to give the balloon a negative charge. These charges are NOT free to move around on
the balloon because the balloon is made of an insulating material (rubber). The wall is neutrally
charged with equal number of positive and negative charges, as most matter is. However, when
you bring the negatively charged balloon near the wall the charges in the wall slightly rearrange.
Positive charges are pulled closer to the balloon and negative charges are pushed slightly farther
from the balloon. Since the force between the balloon and the wall charges depends on distance
the closer positive charges will have a slightly greater attractive force than the repulsive force
from the slightly more distant negative wall charges. The net force will be attractive and th4e
balloon will be held to the wall. In the parlance of physics we say that the wall becomes
“polarized”. We observed this phenomenon in our study of electroscopes. You might want to
check out this quick summary of charging an electroscope by induction at
http://www.physicsclassroom.com/mmedia/estatics/esn.cfm . The first phase of charging by
induction is to establish a polarized electroscope.
Problems
1. An inflated balloon is rubbed with a wool cloth until an excess of a billion electrons is on
the balloon. What is the magnitude of the charge on the balloon?
This is a charge conversion problem. We need to convert 1 billion electrons into the equivalent
number of Coulombs. The important number here is that 1 e = 1.6x10-19C
1.6 10 19 Coulombs
1109 electrons  1109 electrons  
1 electrons


10
  1.6 10 Coulombs

There is 1.6 x 10-10 Coulombs of charge associated with 10 billion electrons. This is a good
reminder that 1 Coulomb is a HUGE amount of charge. Even 1 billon electrons only amount to
less than a billionth of a Coulomb.
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #8
(20 Points Total)
2. What is the force between 2 balloons with a negative charge of 1.6 x 10-10C if the
balloons are 5 cm apart?
This is a Coulomb’s Law problem where we need to use the electric force equation
qq
FElectric  k 1 2 2 . The only curve ball in this problem is that all length units must be in meters.
r
So we have to convert 5 cm to 0.05 m.
FElectric  k



2
q1q2
1.6  10 10 C 1.6  10 10 C
9 Nm

9

10

 9.22  10 8 N
2
r2
C2
0.05 m
There would be 9.22 x 10-8 Newtons of force between the two balloons.
3. How much energy is available from a 12 volt storage battery that can transfer a total
charge equivalent to 100,000 C?
This is an energy problem where we will use that fact that a Volt is a Joule per coulomb, or
equivalently W  q  V
W  q  V  1105 C 12V  1.2 106 J
The energy available from the 12 Volt battery is 1.2 million Joules.
4. A wire carries a current of 2 A. At what rate is the charge flowing?
This question is a bit of a trick question. It’s very simple answer relies on the definition of the
Ampere. The Ampere is the unit of electric current that is defined as the RATE of flow of
CHARGE (.i.e. coulombs per second). Thus 2 A is the equivalent to 2 Coulombs of charge
flowing per second.
The answer is them just 2 Coulombs per second is the rate of flow of charge.
5. What is the magnitude of the least possible current that could theoretically exist?
I was fooled by this question for about half of the time it took to grade the assignments. I
believed at first that the answer was 1.6 x 10-19 A because that corresponds to the flow of 1
fundamental charge (electron or proton) per second. Since the fundamental charge cannot be
divided into smaller units, it made sort of sense that was the answer the publisher wanted.
However, I began reading answers that stated the theoretical current is 0 A. That got me
thinking… As I explained in class, I now believe that the correct answer is 0 A and that, unlike
charge, current is not quantized.
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #8
(20 Points Total)
6. There is a current of 0.83 A through a light bulb connected to a 120 V circuit. What is
the resistance of the light bulb?
This is an Ohm’s Law problem: V  I  R .
V  I R
V 120 Volts
R 
 144.6 Ohms
I 0.83 Amps
The resistance of the light bulb is 144.6 Ohms.
7. What is the voltage across a 60  resistor with a current of 31/3 amps?
This is another Ohm’s Law problem: V  I  R .
V  I  R  3.33 A  60   199.8 V
The voltage across a 60  resistor with a current of 31/3 amps is 199.8 volts.
8. A 10  light bulb is connected to a 12 V battery.
(a) What current flows through the bulb?
Guess what. This is another Ohm’s Law problem: V  I  R .
V  I R
V 12V
I 
 1.2 A
R 10
The current flowing through the bulb is 1.2 A
(b) What is the power of the bulb?
Now, something different. We use the power equation P  V  I
P  V  I  12V 1.2 A  14.4W
The power of the bulb is 14.4 Watts. (Note, you should be able to prove to yourself (and me)
that the units of volts times amperes results in watts.)
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #8
(20 Points Total)
9. A light bulb designed to operate in a 120 V circuit has a resistance of 192 . At what
rate does the bulb use electric energy?
I noticed while grading this question that many students thought the question asked for a current.
This was a mistake,. The question s asks for a RATE of ENERGY usage. This is a Watt and
represents electric power.
There are two ways to do this problem. Either you do the algebra first and one calculation or you
do two calculations. I prefer the first method (fewer calculations). We need to combine Ohm’s
Law with the equation for electric power.
V  I R
I
P V I
V
R
2
120 V   75 Watts
V  V
P  V  I  V   

192 
R R
2
The bulb uses energy at a rate of 75 Watts. (Notice I did not have to first explicitly calculate the
current. I believe that the fewer calculations you do the less likely you are to make mistakes.)
10. What is the monthly energy cost of leaving a 60 W bulb on continuously if electricity
costs $0.10 per KWh?
Last problem…. We want to convert Kilowatt-hours of energy over a period of time into dollars.
I like to do these problems focusing on the units. They will lead the way.
Total Cost in Dollars = (Amount of energy used) x (Cost per unit energy)
or in unit terminology
$ = (KW-h) x ($/KW-h)
The amount of energy used in kilowatt-hours is 0.060 kilowatts x 24hrs/day x 30 day/month) or
43.2 KW-h per month.
The cost for the Month will then be (43.2 KW-h per month) x ($ 0.10/KW-h) =$4.32 per month.
The light bulb costs $4.32 per month to leave on continuously. Today’s electric rates are about
70% higher so the cost today is more realistically closer to $7.34 for one 60 Watt bulb. If you
leave several on in the house continuously the bill can add up fast.