# data[].attributes.body
1. The magnitude of the wool's charge will be the same as the rod, because of the law of conservation of charge.
2. It revealed that electron have a negative charge.
3. 10.0c/(1.603x10^-1.9)= 6.24x10^19 in
4. When the tape is pulled off the desk it is negatively charged and it is attracted to other parts of the desk.
5. This is due to plastic being electrically insulating

1. 8.98x10^9(-8.0x10^6)^2/(.05^2)=230 N
2. a) 8.98x10^9(12x10^-9)-18x10^-9/.3^2 = 2.158 x10^-5 N
b) 8.98x10^9(-3x10^9)^2/.3^2= 8.98x10^-7 N
3. 175=8.98x10^9(60x10^-6)(50x10^-6)/r^2= 39 cm

# data[].attributes.body
1.) The charge of the rubber will become positive because the charge of the wool is transferred.
2.) The Oil experiment allowed Millikan to discover the fact that protons and electrons have the same magnitude.
3.) 10/ 1.6021 * 10 ^ (-19))
6.24 * 10 ^19
4.) Once the tape is removed from an area that area will have a negative charge. This area will not attract positive charges.
5.)Metals transfer electrons easily because they are conductors. Plastics cannot because they are poor conductors.
1. F = (9.0*10^9)*(8.0*10^-6)
(8.0*10^-6)/(.05^2)
230 N
2.) K(12*10^-9)
(18*10^9)/(0.30^2)
F= 9.0*10^-7
3) 175 = (k(60.0*10^-6)
(50.0*10^-6)/(D^2)
d = 0.393m

# data[].attributes.body
1. The rubber rods charge becomes positive because the of the wool's magnitude.
2. Milikan's Oil Experiment discover that the magnitude of  a proton and an electron are the same.
3. 10/(1.602x10^-19) = 6.24x10^19
4. The spot where the tape was removed has a negative charge. Attracts objects with positive charge, opposites attract.
5. Metals are good conductors which allow them to flow and transfer electrons unlike plastics which are not conductors.
1. F=((9.0x10^9)(8.0x10^-6)(8.0x10^-6))/(.05^2)
F=230 N
2a F=(k(12x10^-9)(18x10^/-9))/(0.30^2)
F= 2.2x10^-5 N
2b F=(k(3x10^-9)(3x10^-9))/(0.30^-2)
F=9.0x10^-7 N
3 175 N=(k(60.0x10^-6)(50.0x10^-6))/(d^2)
d= 0.393m

# data[].attributes.body
1. F=kq1q2/r^2
=8.99*10^9(8*10^6)^2/0.0025
=230 N
2a. F=kq1q2/r^2
=8.99*10^9(12*10^-9)(-18*10^-9)/(0.3^2)
=2.2*10^-5 N
2b. F=kq1q2/r^2
k=8.99*10^9
r^2= .09 m
q1=3*10^-9 C
q2=3*10^-9 C
F=9*10^-7 N
3. F=kq1q2/r^2
F=175 N
q1=6*10^-5 C
q2=5*10^-5 C
k=8.99*10^9
175r^2=kq1q2
r=0.393 m

# data[].attributes.body
1. The magnitude of the wool's charge will be positive and greater after the rubbing process.
2. Millikan's oil process showed that the charge of protons and electrons would be the same.
3. 10/(1.602x10^-19) = 6.24x10^19
4. When the tape is removed, it attracts objects from the opposite charge because it has a negative charge.
5. Metals are able to conduct electrons, which allow them to flow. Unlike metals, plastics are not conductive.
1) F=((9.0x10^9)(8.0x10^-6)(8.0x10^-6))/(.05^2)
F=230 N
2a) F=(k(12x10^-9)(18x10^/-9))/(0.30^2)
F= 2.2x10^-5 N
b) F=(k(3x10^-9)(3x10^-9))/(0.30^-2)
F=9.0x10^-7 N
3) 175=(k(60.0x10^-6)(50.0x10^-6))/(d^2)
d= 0.393m

# data[].attributes.body
1. The charge is transferred from the wool to the rubber, when the rubber makes contact with the wool.
2. The Oil experiment quantified the charge.
3.10c/(-1.602*10^19 )= 6.25*10^19 electrons
4. The taped removed spot has a negative charge. The tape repels, and goes towards other parts.
5. Plastic  do not have free-flowing electrons,  where as metals do.
Columb's Prlmbs
1) F=((9.0x10^9)(8.0x10^-6)(8.0x10^-6))/(.05^2)
F=230 N
2a) F=(k(12x10^-9)(18x10^/-9))/(0.30^2)
F= 2.2x10^-5 N
2b) F=(k(3x10^-9)(3x10^-9))/(0.30^-2)
F=9.0x10^-7 N
3) 175=(k(60.0x10^-6)(50.0x10^-6))/(d^2)
d= 0.393m

# data[].attributes.body
b) (8.99*10^9)((30*10^-9)^2/0.3^2)(0.3^2)= 9.0 x 1^-7N
3)  (8.99*10^9)(((50*10^-6)(60*10^-6))/(D^2))=175
D=0.393

# data[].attributes.body
1) The magnitude of the wool's charge after the rubbing process is greater than the rod's charge and is positive
2) Millikan's oil experiment showed that the magnitude of the electric charge is the same for protons and electrons
3) 10 C/-1.602*10^19= 6.25*10^19 electrons.
4) When we pull the tape off quickly it emits the same charge as the sticky tape. The tape repels from the table and is attracted to other parts of the table.
5) Plastics is not conductive like metals.

1) a. F= (9.0x10^9)(-8

# data[].attributes.body
1.  When the rubber is rubbed with wool the charge transfers from the wool to the rubber.
2.   Millikan's oil drop experiment found the magnitude of the charge.
3. 1C = 1.602*10^19
10* 1.602*10^19=6.25*10^19
4.  When the tape is torn away from the table it has a negative charge  making it attract objects with an opposite charge.
5. Metals are conductors making them better at transferring electrons unlike plastic.
1. k e = 8.99×10⁹ N m² C−2
(8.99×10⁹ )(-8.0*10^-6)(-8.0*10^6)/0.05^2 = 230N
2.  a) (8.99×10⁹ )(12*10^-9)(-18*10^-9)/0.3^2 = 2.2*10^-5
b) 3*10*-8

# data[].attributes.body
1. It equalizes to the same magnitude of the rod's charge.
2. Milikan's Oil drop experiment reveals that charge is quantified.
3. 10 divided by the charge on 1 electron
(1.6021 × 10^(–19) )
=6.241 x 10^19 electrons
4. The spot where the tape is pulled off has a negative charge. It attracts other things on the desk, because it has an opposite charge of the desk surface.
5. Plastics are electrical insulators meaning the charge cannot move through the material. While copper and silver are conductors so the charge moves through the material.
6. (8.99x10^9)(-8.0x10^-6)(8.0x10^-6)/(1/2)^2
=-230.14
the absolute magnitude is 230N that is attractive
7. (12.0x10^9

# data[].attributes.body
1.  The wool will have a positive charge.
2. Millikan's oil experiment showed that the magnitude electric charge and mass of the electrion are the same.
3.

# data[].attributes.body
1. If the rubber rod becomes negative the wool's charge would become positive even out the field.
2. Millikan's oil drop experiment revealed that the charge of a proton and electron would both be 1.6x10^-19 coulombs.
3. There are about 10/(1.602x10^-19)=6.24x10^19
4. The tape would be attracted to the desk areas that are not charged since the sticky parts of the tape are adhesive which has the same charge as certain areas of the desk.
5. Metal materials are known to be conductive while non metals are not.
1.-8.0c
electric force
f=k1*q1*q2/d2
(9.0*10^9))(-8*10^-6(8*10^-6)/.5^2
=230N

2.a (9*10^9)(12*10^-9)/(.3^2)
-2.2*10^-5
b. k(12*10^-9)(18*10^-9/.3^2)
9.0*10^-7
3. 175N=k(60.0*10^-6*50*10^-6/d^2-=.393m

# data[].attributes.body
5) Metal conducts electricity, so they can be charged by induction while nonmetals cannot.

Calculations
1) F = 9.0 x 10^9*(-8 x 10^-6)(8 x 10^-6)/(0.05^2) = -230 N (attractive)

2)a. F = (9.0 x 10^9)(12 x 10^-9)(-18 x 10^9)/(0.3^2) = -2.2 x 10^-5 N (attractive)
b. F = (9.0 x 10^9)(15 x 10^-9)(-15 x 10^-19) / (0.3^2) = ?

3)175 N = (9.0 x 10^9)(60 x 10^-6)(50 x 10^-6)/ (r^2)
r = 0.39 m

# data[].attributes.body
1) The magnitude of the wool's charge, if it begins neutral, will increase, since the electrons have moved from the wool onto the rubber, thus making the wool more negative.
2) Millikan's Oil Experiment revealed that electric field has an effect similar to the gravitational field (that it is a field force).
3) 6.24*10^19 electrons
4)  This happens since the tape is adhesive, and is sticky, which combats any electric charge it has.
5)  Metals have free-flowing electrons, while plastics do not. Since metals have these free electrons, they can move more freely.
Coulomb's Law Problems
1) F=((9.0x10^9)(8.0x10^-6)(8.0x10^-6))/(.05^2)
F=230 N
2a) F=(k(12x10^-9)(18x10^/-9))/(0.30^2)
F= 2.2x10^-5 N
2b) F=(k(3x10^-9)(3x10^-9))/(0.30^-2)
F=9.0x10^-7 N
3) 175=(k(60.0x10^-6)(50.0x10^-6))/(d^2)
d= 0.393m

# data[].attributes.body
1. The magnitude of the wool would have a positive charge if it was rubbed against a rubber rod.
2. Millikan's oil drop experiment showed that the proton and electron have the same magnitude.
3.  10/(-1.602x10^-19)= 6.25x10^19
4. If you were to pull transparent tape off of a table quickly, it would be attracted to other areas due to the charge of the adhesive being the same as the charge of the place the tape was.
5. Metals are conductors and therefore electricity can flow through them, whereas plastics are insulators thus blocking the flow.

1. F=kqakqb/r^2
(8.99x10^9)((8x10^-6)^2/0.05^2) = 230 N
2. a. (8.99x10^-9)(((12x10^-9)(18x10^-9))/0.3^2)= 2.2x10^-5
b. (8.99x10^9)((30x10^-9)^2/0.3^2)(0.3^2)= 9.0 x 1^-7N
3. (8.99x10^9)(((50x10^-6)(60x10^-6))/(x^2))=175
x=0.393

# data[].attributes.body
1. The wool must be absolutely positively charged.

2. Millikan's experiment showed the magnitude of the  electrical charge and mass of an electron.

3. 10C / 1.602 X10^-19 C/e. = 6.24*10^19

4. The tape adhesive has no charge and so it is attracted to things with no charge. Like attracts like.

5. Metals can conduct electrons and allow them to flow. Plastic cannot. Metalloids have free electrons which allow movement, plastics do not have free electrons.

6.  1. F=kq1q2/r^2
=8.99*10^9(8*10^6)^2/0.0025
=230 N

7. a.)F=kq1q2/r^2
=8.99*10^9(12*10^-9)(-18*10^-9)/(0.3^2)
=2.2*10^-5 N
b.) deltaC = 3X10^-9 C
F=8.99X10^9(3X10^-18)/.3^2 = 9 X10^-7 N.

8. 175=8.99*10^9((60*50)*10^-6)/r^2
r=.393 m

# data[].attributes.body
1) The wool would have a greater charge than the rubber rod.
2) Millikan's Oil drop experiment determined the size of the charge on an electron.
3)  10/-1.602*10^-19
=6.25*10^19
4) Because the desk and the tape have opposite charged that attract each other.
5) Plastic can not conduct electricity. To charge a material by induction the charge must be able to flow freely throughout the material.

1)F=9.0*10^9 -8.0q2/25
=230N

# data[].attributes.body
1)The magnitude of the wool is greater than the rubber positive charge.
2)The experiment shows that the charge of a proton is equal to the charge of a electron.
3)10/-1.602*10^-19
=6.25*10^19
4)This happens because the tape and the spot on the desk have the same charge meaning that they will repel each other.
5)Metals can conduct electricity easily while plastic does not conduct.
Coulomb's law
1)F=9.0*10^9-4*2/25
=230N
2)8.99*10^9*12*10^9*18*10^9/.3^2= 2.2*10^5
2b)
3)175=9.0*10^9*60*50/d^2
distance=.393m

# data[].attributes.body
Review Questions
1. The wool would have a positive charge because that's how insulators become charged.
2. Millikan's oil experiment revealed that the magnitude of the electric charge is the same for a proton and an electron/
3. Electrons=Charge/(Charge/Electron)
=1*10^1 C/(1.602*10^-19 C)= 6.24*10^19 electrons
4. The tape has adhesive, which is not charged, causing attraction.
5. Plastic is an insulator and is not conductive.

Coulomb's Law Questions
1. F=kq1q2/r^2
=8.99*10^9(8*10^6)^2/0.0025
=230 N
2a. F=kq1q2/r^2
=8.99*10^9(12*10^-9)(-18*10^-9)/(0.3^2)
=2.2*10^-5 N
2b. F=kq1q2/r^2
k=8.99*10^9
r^2= .09 m
q1=3*10^-9 C
q2=3*10^-9 C
F=9*10^-7 N
3. F=kq1q2/r^2
F=175 N
q1=6*10^-5 C
q2=5*10^-5 C
k=8.99*10^9
175r^2=kq1q2
r=0.393 m

# data[].attributes.body
1) If a rubber rod is rubbed with wool, the magnitude of the wool's charge would have a positive charge.
2) Millikan's oil drop experiment showed that the magnitude of the electric charge for a proton and electron would be the same.
3) 10/ (-1.602x10^-19)=
6.25x10^19
There are 6.25x10^19 amount of excess electrons in a lightning bolt.
4) If I quickly pull off transparent tape, the tape is attracted to other areas of the desk that is not charged because tape has adhesive which is sticky. The adhesive also has no charge.
5)Plastic materials are not conductive.
1. F=kq1q2/r^2
((8.99x10^9)(8x10^6))^2)/(.05^2) =
230 N
2.a. (8.99x10^9)(12x10^-9)(-18x10^-9)/(.09)=
2.15x10^-5
3. (8.99x10^9)(50x10^6)(60x10^-6)/ (x^2)=175
x= .393m

# data[].attributes.body
1) Since positively charged rubber becomes negatively charged after rubbed with wool, the magnitude of the wool's charge is greater than the rubber rod's charge.

2) Millikan's oil experiment revealed that the proton and electron have equal magnitudes.

3) Excess electrons of a lightning bolt = 10 C/ (-1.602*10^19) = 6.25 *10^19 electrons.

4) When we quickly pull off transparent tape from the table, every other spot attracts the tape because the spot where the tape pulled off emits the same charge as the adhesive of the tape. They repel each other. The tape now is attracted to other parts of the table.

5)