1. Electrostatics Objects become charged due to the movement of electrons
Protons are locked in the nucleus and cannot move
conductor: a material on which
electrons can move easily
example: metals
Metals conduct because outermost electrons are loosely
held; electrons are “bees” and the atoms are “beehives”

2. insulator: a material through which electrons cannot move
Insulators’ electrons are locked in the atom; can’t get out
examples:
glass
plastic

3. Results from Electrostatics Lab
A negatively charged strip is
brought near a neutral ball
+ -
- +
Electrons are repelled to far side;
protons are closer, so it attracts
+ -
After touching, electrons move
from strip to ball
+ -
Strip and ball are no both negative,
so they repel each other
- -
Charged objects attract neutral objects
After neutral objects touch charged objects, charge transfer
occurs; acquire like charge

4. 3 Ways to Charge Objects Friction
( plastic rod in lab, walking across carpet )
Conduction

5. 3 Ways to Charge Objects Friction
( plastic rod in lab, walking across carpet )
Conduction

6. 3 Ways to Charge Objects Friction
( plastic rod in lab, walking across carpet )
Conduction
(contact; acquires same charge as
charging object)

7. electroscope:
plate
stick
stand

8. 1. Bring negatively charged strip near neutral ES 2. Electrons in
- + 1.
Bring negatively
charged strip near
neutral ES
2. Electrons in
plate are repelled
to bottom of ES;
stick is repelled
from stand
3. Touch strip to ES
4. Take strip away; ES
is negatively charged
Electrons transfer
from strip to ES

9. (induces charge movement; acquires charge
Induction
(induces charge movement; acquires charge
opposite that of charging object)
A negatively charged strip is
brought near, but not
touching, a neutral ball
+ -
- +
+ -
Electrons are repelled to far side
Connection to ground is made;
electrons are repelled to ground e-
+ -
Ground connection is broken;
ball is now positive
+ +
+ +
Take strip away; left with positively charged ball

10. 1. Bring a negatively charged strip near two connected metal
spheres that are neutral and insulated from the ground A B
2. While the strip is held near (but not touching), separate the spheres A B
3. Take the strip away; what is the charge on each sphere? A B

11. The force between two point charges is directly
proportional to the product of the charges and
inversely proportional to the square of the distance
between them
Coulomb’s Law:
q1 = first charge
q2 = second charge
r = distance
Mathematically,
k q1 q2
F = r2
k = 9 x 109 N m2/C2
Unit of charge:
coulomb
Charge of an electron: x C
Charles Coulomb
( )

12. ex. Find the force between charges of +1.0 C and -2.0 C located
0.50 m apart in air.
k q1 q2
F = r2
( 9 x 109 Nm2/C2 )( +1.0 C )( -2.0 C ) =
( 0.50 m)2
F = x 1010 N
negative sign indicates attraction

13. Electric Fields
An electric field is said to exist in a region of space if an electric
charge in the region is subject to an electric force
The amount of force on the charge is a measure
of electric field intensity
Electric Field
Intensity F
F = force
q = charge
E = q

14. ex. A charge of 4.0 µC experiences a force of 12 N when placed in an
electric field. Find the electric field intensity.
q = 4.0 μC
= 4.0 x 10-6 C
F = 12 N F
12 N
E = = q
4.0 x 10-6 C
E = 3.0 x 106 N/C

15. Electric Field Lines
To map an electric field, draw the trajectory of a
very small positive test charge
Field around a positive charge: q+ +

16. Electric Field Lines
To map an electric field, draw the trajectory of a
very small positive test charge
Field around a positive charge: +

17. Field around a negative charge:_

18. _ + Field around unlike charges:
Field lines begin on positive charges and end on negative charges
Field lines never intersect

19. Field around like charges:+ +

20. the work done per unit charge as a charge is
Potential Difference:
the work done per unit charge as a charge is
moved between two locations in an electric field; denoted by V W
Potential
Difference
V = q
ex.  24 J of work is done on a 2.0-C charge to move it in an
electric field. Find the potential difference. W
24 J =
= 12 J/C
V = q
2.0 C
= 12 volts
= 12 V
Alessandro Volta
Definition: volt = 1 V = 1 J/C
( )

21. ex. An electron is accelerated through a potential difference of 120 V.
(a) What energy does the electron acquire?
By Work-Energy Theorem,
Energy acquired = Work done on electron W
W = q V
V = q
= ( x C )( 120 V )
Work done = Energy acquired = x J

22. (b) What speed does it acquire?
Energy = x J
KE = 1/2 mv2
me = x kg
2 KE
2 ( 1.92 x J )
v = = m
9.11 x kg
v = x 106 m/s

23. Distribution of Charge
- All static charge on a conductor resides on its surface
- No electric field can exist on the inside of a conductor
- There is no potential difference between any two points
on a conducting surface
If charge is put on a spherical conducting surface, it will
distribute itself evenly on the surface

24. _ _ _ _ _ _ _ _ _ - If charge is placed on a non-spherical conductor,
it will congregate at points of high curvature

25. - If charge is placed on a non-spherical conductor,_ _ _ _ _ _ _ _ _ _ _ _ _ _
it will congregate at points of high curvature
the higher the curvature, the greater the density of charge

26. St. Elmo’s Fire

28. Parallel-plate capacitors
A device that will store electric charge
; made of two
parallel metal plates separated by a small distance +
Battery -
When connected to a battery, charge will flow to the plates and
stay there indefinitely

29. Amount of stored charge depends on: (1) voltage of battery + V -
Amount of stored charge depends on:
(1) voltage of battery
(2) size of capacitor
Q = charge
V = voltage
C = capacitance
Q = C V

30. ex. When hooked to a 12-V battery, 6.0 mC of charge is stored on a
capacitor. Find the capacitance.
Q = C V Q
6.0 x 10-6 C
C = = V
12 V
C = 5.0 x 10-7 C/V
= 5.0 x 10-7 farads
= 5.0 x 10-7 F
Definition: 1 C/V = 1 farad = 1 F
Michael Faraday
( 1791 – 1867 )

31. Electric Field in a Parallel-plate Capacitor + V -
Electric field has the same value anywhere within the plates
(away from the edges)
Easy to control; used to deflect charges

32. Electric Field in a Parallel-plate Capacitor + V d -
Strength of field in capacitor
depends on:
(1) voltage of battery
(2) distance between plates
Electric Field
in a Capacitor V
E = d

33. ex. A 250 mF capacitor, with 1.5 mm separation between the plates,
is hooked to a 12-V battery. Find the electric field intensity
between the plates. V
12 V
E = = d m
E = V/m