1. Do Now: Write down what you would like to review.
Aim: Station Review
Do Now: Write down what you would like to review.
HW: Study for the Test

2. Charge Conservation
Metal sphere A has a charge of -8C and an identical sphere, B, has a charge of -4C. If the spheres are brought together and then separated what is the charge on sphere A and B?
Which Sphere gained electrons?
A Pinball has a charge of +8x10-8C. What is the pinball’s deficiency of electrons?

3. Solutions
-8C + (-4C) = -12C total NET charge C/2 = -6C on each sphere once separated
Sphere B gained electrons because it has a greater negative charge now.
Deficiency of electrons:

4. Station Activity
Work with your group to answer the following questions.

5. Conduction/Induction
An electroscope is charged when in contact with a negative rod. Describe the flow of electrons.
What are all the possible charges on an object attracted to a negatively charged balloon?
A negative rod is brought close (not touching) an electroscope. Electrons flow away from the top of the scope to the leaves, which then diverge. What happens if I ground the electroscope with the rod still close and then take away the ground, followed by the negative rod?

6. Conduction/Induction
Touching the electroscope allows electrons to flow into it by means of conduction. This leaves a negative charge.
If an object is attracted to a negatively charged balloon it may be positive OR neutral.
If the electroscope is grounded it will provide an exit for the electrons to travel through, which will leave the scope with a NET positive charge once the ground followed by rod is taken away.

7. Electric Field & Coulomb’s Law
If I double the charge on two objects and double the distance between them, what is the resulting force? 1. F F F F
Draw the electric field between two unlike charges.
Draw the electric field between two positive charges.
What is the force exerted on a 2x10-6C charge in an electric field of magnitude 2x104N/C?

8. Electric Field and Coulomb’s Law
(2q)(2q)/(2r)2 = F All the numbers cancel!
Draw on the Board!
Fe = qE (2x10-6C)(2x104N/C) = 0.04N

9. Potential Diff. & Parallel Plates
How much energy in Joules is required to move one electron through a potential difference of 1x103V?
Calculate the potential difference across a 6-ohm resistor if 4 x 10-6J of work are required to move 2 μC of charge through the resistor?
How does the force when the electron first enters
compare to when it hits the
plate? e-

10. Potential Diff. & Parallel Plates
PEE = qV PEE= (1.6x10-19C)(1x103V)
PEE = 1.6x10-16J
2. PEE = qV 2μC = 2x10-6C
(2x10-6C)ΔV = 4x10-6J
ΔV = 2V
3. The force of the electron as it eneters is the same for when it hits the plate because between parallel plates we have a uniform electric field. As long as the charge is the same, the force is the same everywhere.