1. Electrostatics

2. Learning Objectives
The electrostatic force (Coulomb’s Law) can be either repulsive or attractive (SOL 12.a)
The interaction of two particles can be described as: the creation of a field by one of the particles and the interaction of the field with the second particle (SOL 12.b).

3. Magnitude of charge on protons and electrons are exactly the same
Protons have a positive charge
Electrons have a negative charge
Neutral atoms contain equal numbers of protons and electrons

4. Insulators and Conductors Need to know
Insulator: electrons are bound very tightly to the nuclei. Wood and rubber are good insulators.
Conductor: electrons are bound very loosely and can move about freely. They are often referred to free electrons. Metals are good conductors.
Semiconductor: very few free electrons (silicon, germanium and carbon)

5. Static Electricity
You have probably experienced a charge lately (comb, dryer, carpet, car seat, …)
An object becomes charged due to a rubbing process and is said to possess a net electric charge
An item containing a net positive charge has lost electrons
An item containing a net negative charge has gained electrons

6. Law of Conservation of Electric Charge Need to know
The net amount of electric charge produced in any process is zero
If one object or one region of space acquires a positive charge, then an equal amount of negative charge will be found in neighboring areas or objects

7. Unlike Charges Attract; Like Charges Repel Need to know

8. 3 Ways to Charge an Object Need to know
Friction: Rubbing two objects together with different electron attachment. Heat generated frees electrons to join object with stronger attachment.

9. 3 Ways to Charge an Object Need to know
2. Conduction: Electrons are transferred from one object to another by touching. Usually it involves moving from one electric potential to another.
John TraVOLTa Demo

10. 3 Ways to Charge an Object Need to know
3. Induction: Rod does not touch sphere. It pushes electrons out of the back side of the sphere and down the wire to ground. The ground wire is disconnected to prevent the return of the electrons from ground, then the rod is removed.

11. Electromagnetism
One of the four fundamental forces of the universe (electromagnetism, gravity, weak nuclear and strong nuclear forces)
The forces that act between atoms and molecules to hold them together are electrical forces
Elastic, normal and contact forces (pushes and pulls) result from electric forces acting at the atomic level

12. Forces resulting from charges
Charges push and pull on one another
Closer the charge the higher the force
The stronger the charge the higher the force

13. Coulomb’s Law Need to know
The magnitude of the force between charge qA and charge qB, separated a distance d, is proportional to the magnitude of the charge and inversely proportional to the square of the distance:
F = K qAqB d2 qA qB d

14. Coulomb’s Law: Key Facts Need to know
The charge of an electron is:
-1.60 x coulombs (C)
The charge of a proton is:
1.60 x coulombs (C)
The charge, q, is measured in coulombs. The distance, d, is measured in meters. The force, F, is measured in newtons.
The constant, K = 9.0 x 109 Nm2/C2

15. Problem Solving Strategy
Sketch the system showing all distances
Diagram the vectors
Use Coulomb’s law to find the magnitude of the force. Note: it is unnecessary to include the sign of the charges or the distance. The answer is always positive.
Use your diagram along with trigonometric relations to find the direction of the force

16. Example Problem 1
Two charges are separated by 3.0 cm. Object A has a charge of +6.0 C, while object B has a charge of +3.0 C. What is the force on object A?
Known: Unknown:
qA = +6.0 x 10-6 C FB on A = ?
qB = +3.0 x 10-6 C
d = m

17. Example 1 Solution F = K qAqB d2
= (9.0 x 109 Nm2/C2)(6.0 x 10-6C)(3.0 x 10-6C)
(3.0 x 10-2 m)2
FB on A = 1.8 x 102 N

18. Example 2: Three Charges
Given:
Find the net force on the -2 µC charge
Known:
-2 µC
+6 µC
2 µC
6 cm
2 cm

19. FNet = FA on B + FC on B = - 30 N + 90 N = 60 N +6 µC
FA on B= KqAqB d2
= (9x109 Nm2C2)(6x10-6 C)(2x10-6 C)
(0.06 m)2
= - 30 N
FC on B= KqCqB
= (9x109 Nm2C2)(2x10-6 C)(2x10-6 C)
(0.02 m)2
= + 90
FNet = FA on B + FC on B = - 30 N + 90 N = 60 N
-2 µC
+6 µC
2 µC
6 cm
2 cm

20. Example Problem 3
A sphere with a charge 6.0 C is located near two other charged spheres. A -3.0 C is located 4.00 cm to the right and a 1.5 C sphere is located 3.00 cm directly underneath. Determine the net force on the 6.0 C sphere.
FB on A
FC on A
Fnet  A
qA= 6 C
qc = 1.5 C
qB = -3 C C B
dAB
dAC

21. Example 3 Solution FB on A = FC on A = Fnet =  = B A dAB Fnet
qA= 6 C
qc = 1.5 C
qB = -3 C C B
dAB
dAC
FB on A
FC on A
Fnet 
FB on A =
FC on A =
Fnet =
 =

22. Static Charge Generator

23. Electric Field Need to know
An electric field extends outward from every charge and permeates all of space

24. Investigating the Electric Field
We can quantify the strength of an electric field by measuring the force on a small positive test charge
So small that the force it exerts does not significantly alter the distribution of the charges that create the field
qA+
+qB a

25. Electric Field
An electric field, E, at any point is defined as the force, F, exerted on a tiny positive test charge at that point divided by the magnitude of the test charge:
E = F/qB
qA+
+qB

26. Electric Field Equation
E = F/qB
E = K qB qA/r2 qB
E = KqA/r2
qA+
+qB

27. Example
Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge qA = -3.0 x 10-6 C
E = KqA/r2
Answer: 3.0 x 105 N/C

28. Electric Field Lines
Drawn so that they indicate the direction of the force due to the given field on a positive charge
qA+
+qB

29. Electric Field Lines Need to Know Lines indicate direction of the force due to the given field on a positive test charge

30. Properties of Field Lines Need to know
The field lines indicate the direction of the electric field
The lines are drawn so that the magnitude of the electric field, E, is proportional to the number of lines crossing unit area perpendicular to the lines. The closer the lines, the stronger the field.
Electric field lines start on positive charges and end on negative charges

31. Electric Potential Difference Need to know
V = Won q’ = PE: Potential difference often q’ q’ referred to as Voltage
Electric Potential Difference Units: Volt =J/C + g E
displacement
displacement
W = Fd = mgd
W = Vq +
Big
Negative
Charge

32. Typical Voltages Source Voltage Thundercloud to ground
High voltage power line
Power supply for TV tube
Auto ignition
Household outlet
Auto battery
Resting potential across nerve membrane
Potential changes on skin
(EKG)
Voltage
108 V
106 V
104 V
102 V
12 V
10-1 V
10-4 V

33. Example
Two parallel plates are charged to a voltage of 50V. If the separation between the plates is m, calculate the electric field between them.
E = V/d = 50V/0.050m =1000V/m + -
E = 1000 V/m
d = 5 cm

34. Practice
Two parallel plates are charged to a voltage of 500V. If the separation between the plates is m, calculate the electric field between them.
Two parallel plates are charged to a voltage of 50V. If the electric field between them is 10,000 V/m, what is distance between the plates?
Answers:
100,000 V/m
.0050 m

35. Capacitors Need to Know
A capacitor is a device that can store electric charge
Consists of two conducting objects placed near each other but not touching
They store charge for later use
Usage: camera flash, energy back-up for computers and as surge protectors

36. Capacitors
Consists of a pair of parallel plates of area, A, and separated by a small distance d.
In a diagram, they are represented by the symbol:
If a voltage is applied to a capacitor, one plate acquires a negative charge and the other an equal amount of positive charge.

37. Capacitors Need to Know
The amount of charge acquired by each plate is proportional to the potential difference
Q = CV
Where C is constant and is called the capacitance of the capacitor
Unit: Coulombs/Volt = Farad
Typical capacitor range is 1pF (10-12) to 1F (10-6)

38. Determining Capacitance
Constant for a given a capacitor
Depends on structure and dimensions of he capacitor itself:
C = o A/d
A = area
d = separation distance between plates
o = 8.85 x C2/Nm2
= permittivity of free space