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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”
insulator: a material through which electrons cannot move
• Insulators’ electrons are locked in the atom; can’t get out
examples:
glass
plastic
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
3 Ways to Charge Objects
• Friction ( plastic rod in lab, walking across carpet )
• Conduction
3 Ways to Charge Objects
• Friction ( plastic rod in lab, walking across carpet )
• Conduction
3 Ways to Charge Objects
• Friction ( plastic rod in lab, walking across carpet )
• Conduction
(contact; acquires same charge as
charging object)
electroscope:
plate
stick
stand
- - - - - - - - - - - - -
1. Bring negatively
charged strip near
neutral ES
- - - - - - - - - - - - + - + -
+
+
-
3. Touch strip to ES
-
-
-
-
-
4. Take strip away; ES
is negatively charged
- - - - - - - - - - - - - ++++
Electrons transfer
from strip to ES
2. Electrons in
plate are repelled
to bottom of ES;
stick is repelled
from stand
+ + + +
- - - -
-
-
• 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
+ + -
Ground connection is broken;
ball is now positive
++
++
Take strip away; left with positively charged ball
++
++
e-
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
Coulomb’s Law: 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
Mathematically,
F =
Unit of charge:
k q1 q2
r2
q1
q2
r
k
=
=
=
=
first charge
second charge
distance
9 x 109 N m2/C2
coulomb
Charge of an electron: 1.6021 x 10-19 C
Charles Coulomb
( 1736 - 1806 )
ex.
Find the force between charges of +1.0 C and -2.0 C located
0.50 m apart in air.
F =
=
k q1 q2
r2
( 9 x 109 Nm2/C2 )( +1.0 C )( -2.0 C )
( 0.50 m)2
F = - 7.2 x 1010 N
negative sign indicates attraction
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
E=
F
q
F = force
q = charge
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
E=
F
q
=
12 N
4.0 x 10-6 C
E = 3.0 x 106 N/C
F = 12 N
Electric Field Lines
To map an electric field, draw the trajectory of a
very small positive test charge
Field around a positive charge:
+
q+
Electric Field Lines
To map an electric field, draw the trajectory of a
very small positive test charge
Field around a positive charge:
+
Field around a negative charge:
_
Field around unlike charges:
+
_
•Field lines begin on positive charges and end on negative charges
•Field lines never intersect
Field around like charges:
+
+
Potential Difference: the work done per unit charge as a charge is
moved between two locations in an electric field; denoted by V
Potential
Difference
V=
W
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.
V=
W
q
=
24 J
2.0 C
= 12 J/C
= 12 volts
= 12 V
Definition:
1 volt = 1 V = 1 J/C
Alessandro Volta
( 1745 - 1827 )
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
V=
W
q
W = qV
= ( 1.6021 x 10-19 C )( 120 V )
Work done = Energy acquired = 1.92 x 10-17 J
(b) What speed does it acquire?
Energy = 1.92 x 10-17 J
KE = 1/2 mv2
v =
2 KE
me = 9.11 x 10-31 kg
=
m
v = 6.50 x 106 m/s
2 ( 1.92 x 10-17 J )
9.11 x 10-31 kg
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
- If charge is placed on a non-spherical conductor,
_
_
_
_
_
__
__
it will congregate at points of high curvature
- 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
St. Elmo’s Fire
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
C
V
+ + +
+ + +
-
-
+ + +
+ +
+
-
-
-
Amount of stored charge depends on:
-
- -
-
-
-
(1) voltage of battery
(2) size of capacitor
Q=CV
-
Q = charge
V = voltage
C = capacitance
ex. When hooked to a 12-V battery, 6.0 mC of charge is stored on a
capacitor. Find the capacitance.
Q=CV
C=
Q
V
=
6.0 x 10-6 C
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 )
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
Electric Field in a Parallel-plate Capacitor
+ + +
V
+ + +
+ + +
+ +
+
d
-
-
-
Strength of field in capacitor
depends on:
Electric Field
in a Capacitor
E=
-
-
-
-
-
-
-
-
(1) voltage of battery
(2) distance between plates
V
d
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.
E =
V
d
=
E = 8000 V/m
12 V
0.0015 m