Download AP Electrostatics

Do Now (11/11/13):

What do you know about electric charges?
What do you think the word “electrostatics”
means?

Pass your HW in please!

Electrostatics
Bad Hair Day
Static Charges
Rub a balloon on a wool sweater and it will stick
to the wall. Why?
Rubbing a balloon
on a wool
sweater
creates charges
on the surfaces.
Electrons are
added or
subtracted from
the atoms.
Charges That Things Accumulate
•Neutral
• Very positive
• Human hands (usually too moist,
though)
• Rabbit Fur
• Glass
• Human hair
• Nylon
• Wool
• Fur
• Lead
• Silk
• Aluminum
• Paper
• Cotton
• Steel
• Neutral
• Steel
• Wood
• Amber
• Hard rubber
• Nickel, Copper
• Brass, Silver
• Gold, Platinum
• Polyester
• Styrene (Styrofoam)
• Saran Wrap
• Polyurethane
• Polyethylene (like Scotch Tape)
• Polypropylene
• Vinyl (PVC)
• Silicon
• Teflon
• Very negative
Charging an Object by Touching
+ +
+
+
+
+
+
+
+
Two Objects—one
is charged
+
+
Objects touch—charge
is transferred
+
+
+
+
Objects separate—
both are charged
Behavior of Electric Charges
Charging an Electroscope
An electroscope is a device that
permits us to explore the concepts of
induction and conduction charging.
Charging by Contact
Some electrons leave rod and spread over sphere.
Charging by 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.
Charge Distributions
Charge on Metals Charge on Insulators Charge on Metal Points
Excess charge on
the surface of a
metal of uniform
curvature spreads
out.
Charge on insulating
materials doesn't
move easily.
Excess charge on a metal
accumulates at points.
Lightning, lightning rods.
Charges on a Conductor
Attracting Uncharged Metallic
Objects
Electrons are free to
move in metals.
Nuclei remain in place;
electrons move to bottom.
Charges on an Insulator
Attracting Uncharged Nonmetallic
Objects
Charges Accumulate on Points
A Shocking Experience
How Lightning Occurs
Electrostatics Is Not Friction


Electrostatic charges are not
caused by friction.
The materials involved and the
pressure and speed of contact
and separation affects the
magnitude of the charge. This
contact and separation
process is known as
"triboelectrification," or
"tribocharging.“
The
suffix tribo means to rub in Greek, thus triboelectrification
simply means to electrify (or charge) by rubbing, or by contact.
Applications of Electrostatic
Charging
Fine mist of negatively charged gold
particles adhere to positively charged
protein on fingerprint.
Negatively charged paint
adheres to positively
charged metal.
Electrostatic Air Cleaner
Electric Forces

The strength of the electric force varies with
the square of the distance between the
charges
k q1q2
F=
r2


Where k = 8.988 x 109 Nm2/C2 (but approximate 9x109)
and a coulomb is the charge which results in a force of
9 x 109 N if placed on two objects 1.0 m apart
Important Numbers
Charge of the electron: -1.6 x 10-19 C = -e
Charge of the proton: 1.6 x 10-19 C = +e
Mass of the electron:
9.11 x 10-31 kg
Mass of the proton: 2000 times electron
(1.67 x 10-27kg)
Charges

A coulomb is an extremely large charge



Charges produced by rubbing objects are typically
about a microcoulomb
The charge of an electron is 1.602 x 10-19 C
Sometimes the force between charges is written
as:
F = (1/4πε0) (Q1Q2/r2) where ε0 is the permittivity
of free space = 1/4πk =
= 8.85 x 10-12 C2/Nm2
Forces Between Charges


The force field between charges depends
on their sign and their magnitude
Electric forces are vectors like all other
forces
0.30 m
Q1 = -8.0 μC
0.20 m
Q2 = +3.0 μC
Q3 = -4.0 μC
Net force on charge 3 will be the sum of F31 and F32
Simple Force Calculation
F = k Q1Q2/r2
-------------------------------------k = 9 x 109 N-m2/C2
F = (9 x 109) (5)(8)/22
= 9 x 1010 N
This is an enormous force,
because a Coulomb is
a huge charge:
What is the force between the
charges?
If the two charges are of opposite
sign, what is the direction of the
force?
One Coulomb is the charge
on 6.25 x 1018 electrons.
Do Now (11/12/13): Three
Charges on a Line
Where may any test
charge q be placed
between the charges
if it is to experience
zero electric force?
Three Charges on a Line: Part I
Force between any two charges:
F = kq1q2/r2
----------------------------------------------------------------
Forces by the two charges must be equal
but opposite:
Force by red charge
= k(5)q / x2
Force by yellow charge = k(8)q / (4-x)2
Where may any test charge q be
placed between the charges if it
is to experience zero electric
force?
Forces are equal:
k(5)q / x2 = k(8)q / (4-x)2
Solve for x:
x = 1.77 m
Three Charges on a Line: Part II
On the line in which region,
A, B, or C, may a point be
found at which the net force
on a positive test charge q
would be zero?
How Lightning Occurs
Electric Force Vectors
Consider the forces exerted on the charge
in the top right by the other three:
Electric Fields Produce Forces
The Electric Field Due to a Point
Charge
F = kQq0/r2
Define: E = F/q0 = kQ/r2
Electric Fields


An electric field extends outward from every
charge and permeates all of space
The electric field is given by the force on a very
small test charge q, such that:
E = F/q

The field at a distance r from a charge Q is:
E =F/q = kQq = kQ/r2
r2
q
Electric Fields
Electric field due to a positive point charge.
Arrows point in the direction along which a
positive test charge would accelerate.
--------------------------------------------------------F = kQq0/r2
E = F/q0 = kQ/r2
Electric field due to
a negative point
charge.
-----------------------------------Arrows point toward
negative charge.
Field is spherically
symmetric.
Field Lines



The field lines indicate the direction of the
electric field; the field points in the direction
tangent to the field line at any point
The lines are drawn so that the magnitude of the
field, E, is proportional to the number of lines
crossing a 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 and the number
starting or ending is proportional to the
magnitude of the charge
E-Field of Spherical Charge
Distributions
Radius of the ball is r = 0.5 m.
What is the electric field E
2 m from the center of the ball?
(Assume uniform distribution)
E = kQ/r2 = (9x109)(5)/22
= 1.125 x 1010 N/C
Electric Field Calculation
E2 = (3.0)2 + (2.0)2
= 13.0
E = 3.61 N/C
q = tan-1(2/3)
= 33.7 degrees
Symmetry In Electric
Field Calculations
Electric Field of Dipoles
Electric Fields Under the Sea
Elephant Gnathonemus detects
nearby objects by their effects on
the electric field.
Cells in shark detect weak electric
fields caused by the operation of
the muscles of its prey. Fields as
weak as 10-6 N/C are
detectable
The Electric Field of a Lightning
Strike
The
direction of
the electric field is
from positive to
negative despite
the fact that the
current flow is
from negative to
positive
This is consistent
with the force on a
POSITIVE test
charge
Examples of Electric Field
Strengths
Source
E
(N/C)
Source
E
(N/C)
House wires
0.01
Thunderstorm
10,000
Near stereo
100
Breakdown of air
3 x 106
Atmosphere
150
Cell membrane
Shower
800
Laser
1011
Sunlight
1000
Pulsar
1014
Compare to the field detectable by sharks, 10-6 N/C
107
Practice:

Complete Problem #10 and #11 in your
textbook in Chapter 15
Do Now (11/13/13):


Pick up a green/yellow half sheet from the
back of the room on your way in
Review yesterday’s Do Now (the solution
is on the back board)
A Parallel Plate Capacitor
Example:
A = 0.15 m2
q = 6 x 10-6 C
s = q/A
= 6 x 10-6 C/
0.15 m2
= 40 x 10-6 C/m2
s = q/A = charge density
E = s/e0 e0 = 8.85 x 10-12 N-m2/C2
e0 is called the "permittivity of vacuum"
E = s/e0
= 40 x 10-6/
8.85 x 10-12
= 4.52 x 106 N/C
Do Now (11/14/13):



Find a place in the room where you are as
far away from as many people as
possible.
Write it down.
Go stand there.
Electric Field Inside a Conductor
If E weren't zero inside, the
Excess charge inside a
free electrons (not shown)
metal moves to the surface.
would accelerate.
At equilibrium, all excess charge on a metal resides on the
surface of the metal.
Electric Fields and Conductors

In a static situation (charges not moving) the
electric field inside a conductor is zero


If there were a field, there would be a force on the
free electrons, since F=qE. They would move until
they reached positions where the force on them
would be zero
Therefore, any net charge on a conductor
distributes itself on the surface

The charges get as far away from each other as
possible
Electric Fields and Conductors (cont’d)

A charge
placed inside
a conducting
sphere
results in
charges as
shown in the
figure
Electric Fields and Conductors (cont’d)

The electric field of static charges
is always perpendicular to the
surface outside of a conductor

If there were a parallel component
of the field, the electrons would
move along the surface until they
reached positions at which no force
was exerted on them.
E-Field is Perpendicular to
Conductors in Equilibrium
Uncharged Metal Plate in an
Electric Field
Metal plate is polarized by the
external electric field.
Sheets of charges on plate
set up electric field (not shown)
which cancels the external
electric field.
If the electric field E weren't
zero inside the metal, what
would happen?
What is the field inside a hollow box
placed between two charged plates?



If the box was a solid block
of conducting material the
field inside would be zero
For a hollow box the external
field does not change, since
the electrons can still move
in the same ways
A hollow box is a useful way
to protect sensitive
electronics from external
electric fields, such as
produced by lightning
Recognizing Incorrect Electric
Field Patterns
This field configuration can't exist
because the bottom of the ball will be
positively charged, so a field should
exist between the plate and the
bottom of the ball.
On the left and right sides in
this view, the electric field E is
tangent to the metal ball, so a
tangential force on the
electrons would exist,
contradicting the fact of
equilibrium.
Using Metal to Shield Electronic
Components
Electric Flux Through a Plane
Surface
Electric Flux = F = EA cos q
Electric Flux Through a Closed
Surface
Electric Flux = F = E DA cos q
(Some texts use DS for the area)
---------------------------------------------------If there is no net charge inside this
closed surface, the net flux is zero:
every arrow that enters must exit.
E-field vectors which enter a surface
provide negative flux, while vectors
which exit give positive flux.
Electric Flux
Visually we can try to understand that the flux is simply
the # of electric field lines passing through any given
area.
In the left figure, the
flux is zero.
In the right figure,
the flux is 2.
• When E lines pass outward through a closed surface,
the FLUX is positive
• When E lines go into a closed surface, the FLUX is
negative
Gauss's Law
Friedrich Gauss (1777-1855)
Gauss's Law:
S AE cosq = q/e0
q = net charge inside Gaussian surface
This is useful if q = 0 and E = constant.
Gauss’ Law
Where does a fluid come from? A spring! The spring is the
SOURCE of the flow. Suppose you enclose the spring with a
closed surface such as a sphere. If your water accumulates
within the sphere, you can see that the total flow out of the
sphere is equal to the rate at which the source is producing
water.
In the case of electric fields the source of the field is the CHARGE!
So we can now say that the SUM OF THE SOURCES WITHIN A
CLOSED SURFACE IS EQUAL TO THE TOTAL FLUX
THROUGH THE SURFACE. This has become known as Gauss'
Law
Gauss’ Law
The electric flux (flow)
is in direct proportion to
the charge that is
enclosed within some
type of surface, which
we call Gaussian.
The vacuum permittivity
constant is the constant of
proportionality in this case as
the flow can be interrupted
should some type of material
come between the flux and
the surface area. Gauss’
Law then is derived
mathematically using 2
known expressions for flux.
 E  dA =
qenc
eo
Gauss & Michael Faraday
Faraday was interested in
how charges move when
placed inside of a
conductor. He placed a
charge inside, but as a
result the charges moved
to the outside surface.
Then he choose his
Gaussian surface to be
just inside the box.
E  da =
qenc = 0
qenc
eo
 0( A) =
qenc
eo
He verified all of this
because he DID NOT
get shocked while
INSIDE the box. This
is called Faraday’s
cage.
Gauss’s Law

For Physics B: E-field inside a conductor
is zero
ESA cos q =
qenc = 0
qenc
eo
 0( A) =
qenc
eo
For Closed Surfaces:
Calculus:
Gauss's Law Gives Field Due to a
Point Charge
Gauss's Law:
SAE cosq = q/e0
A = area of sphere
= 4pr2
E is the same at all points
on the surface
q=0
cos q = 1
(4pr2)E = q/e0
E = q/(4pe0r2)
Gauss's Law Application
SAE cosq = q/e0
q = sA where s = charge density
This is a sheet of charge--not a metal
A1E + A2 (0) + A3E = sA/e0
plate. Sheet is very large (edges are
2AE = sA/e0
not shown); near center of sheet, the
E vector is perpendicular to the sheet. E = s/2e0
Gauss’ Law – How does it work?
Consider a POSITIVE POINT CHARGE, Q.
Step 1 – Is there a source of symmetry?
Yes, it is spherical symmetry!
You then draw a shape in such a way
as to obey the symmetry and
ENCLOSE the charge. In this case,
we enclose the charge within a
sphere. This surface is called a
GAUSSIAN SURFACE.
Step 2 – What do you know about the electric
field at all points on this surface?
It is constant.
The “E” is then brought out of the
integral.
E  da =
qenc
eo
Gauss’ Law – How does it work?
Step 3 – Identify the area
of the Gaussian surface?
In this case, summing each
and every dA gives us the
surface area of a sphere.
E (4pr ) =
2
qenc
eo
Step 4 – Identify the charge
enclosed?
The charge enclosed is Q!
E (4pr ) =
2
Q
eo

Q
E=
4pr 2e o
This is the
equation for
a POINT
CHARGE!
Cylinder with Charge distribution

Charge distribution:
Q
=
L
Q = L = qenc
Gauss’ Law and cylindrical
dersymmetry
a line( or rod) of charge that is very long (infinite)
+
+
+
+
+
+
+
+
+
+
+
+
We can ENCLOSE it within a
CYLINDER. Thus our Gaussian
surface is a cylinder.
E  da =
qenc
eo
L
E (2prL) =
eo

E=
2pre o
E (2prL ) =
qenc
eo
RECALL : Macro   =
Q = L = qenc
Acylinder = 2prL
This is the same equation we got doing
extended charge distributions.
Q
L
Gauss’ Law for insulating sheets and
A charge is distributed with a uniform charge density over an infinite
disks
plane INSULATING thin sheet. Determine E outside the sheet.
For an insulating sheet the charge resides
INSIDE the sheet. Thus there is an electric
field on BOTH sides of the plane.
+
 E  dA =
qenc
EA  EA =
Q
s=
eo
eo
 2 EA =
Q
eo
Q
sA
, 2 EA =
A
eo
s
E=
2e o
This is the same
equation we got doing
extended charge
distributions.
Gauss’ Law for conducting sheets and disks
A charge is distributed with a uniform charge density over an infinite
thick conducting sheet. Determine E outside the sheet.
+
For a thick conducting sheet, the
charge exists on the surface only
+
E =0
+
+
+
+
+
 E  dA =
EA =
qenc
eo
Q
eo
Q
sA
s = , EA =
A
eo
s
E=
eo
In summary
Whether you use electric charge distributions or Gauss’ Law you get the
SAME electric field functions for symmetrical situations.
E=
Q
4peo r 2
 E  dA =
Function
Equation
 dE =
qenc
eo
Point, hoop,
or Sphere
(Volume)
E=
dq
4peo r 2
Q
4peo r 2
Disk or Sheet
(AREA)
“insulating
and thin”
s
E=
2e o
Line, rod, or
cylinder
(LINEAR)
E=

2peo r
Practice:

Complete the multiple choice questions in
Chapter 15
Gauss's Law Applied to Parallel
Plate Capacitor
Large plates close together; ignore
E is zero at the left end and E is
fringing at edges. Electric field inside
parallel to the side.
the metal is zero. E is perpendicular to
the plates (far from the edges).
q = sA
EA = sA/e0
We assume a charge density s
E = s/e0
Wimshurst Machine

Invented by James
Wimshurst in 1882


The first studies of sparks and
oscillating electrical discharge
were made using this type of
machine.
Electrostatic machines were
fundamental in the early
studies of electricity, starting
in the XVII century, in the
form of "friction machines",
and their development
culminated at the end of the
XIX century with the
development of powerful
"influence machines".
Theory Of Operation Of A
Wimhurst Machine







The disks can be made of plastic, glass, or hard rubber
The counter-rotating disks cause air molecules to become electrically
activated by the frictional movement between the disks.
This rotating action causes the disks to become continually charged and
an electrostatic charge builds up, which will cause a flash over if not bled
off.
To prevent flash over, a series of foil sections are attached to the center
portion of each disk and equally spaced and back to back with foil
sections on the outer sides.
To remove the charge, collection arms are arranged to collect the charge
and transfer the charge to a storage capacitor.
At 45 degrees to these collection points is a neutralizing bar that extends
the full length of the disk and has brushes at both ends.
A neutralizing brush equals the charges on the metal foil position at both
positions on both sides. The neutralizing bar on opposite side disk is at
ninety degrees to the one for the other side.
Van de Graaff Generator
Van de Graaff Generator
How It Works


When the motor is turned on, the lower roller
(charger) begins turning the belt.
Belt is made of rubber and the lower roller is
covered in silicon tape,

Lower roller begins to build a negative charge and the
belt builds a positive charge.


Silicon is more negative than rubber; therefore, the lower
roller is capturing electrons from the belt as it passes over
the roller
Positive charges from belt are deposited on sphere
Cereal Storm
Van de Graaff Generator
A. Output terminal—an aluminum or
steel sphere
B. Upper Brush—A piece of fine
metal wire
C. Upper Roller—A piece of nylon
D. Belt--A piece of tubing
E. Power supply
F. Lower Brush
G. Lower roller—nylon covered with
silicon tape
Do Now (11/18/13):
Define the following in your own words. If
you do not know, hypothesize:
 Capacitance
 Voltage
 Potential
Definitions


Electric Field = force per unit charge
Electric Potential = potential energy per unit charge
electric potential = electric potential energy
charge
Vab = Va – Vb = -Wab/q

The change in electric potential is the work done on a unit
charge
1 volt = 1 joule/coulomb
Brainstorm:
The charges that flow through the wires in your
home ____.
a. are stored in the outlets at your home
b. are created when an appliance is turned on
c. originate at the power (energy) company
d. originate in the wires between your home
and the power company
e. already exist in the wires at your home
Voltage Sources

To do useful work voltage sources
capable of maintaining a steady current
flow are required



Generators
Batteries
Fuel cells
Voltage provides the force to “push” electrons
through a circuit
Electric Potential


Just as with gravitational
potential energy, the zero
point of electric potential
is an arbitrary location
The larger rock has the
greater potential energy;
the larger charge has the
greater electric potential
energy
Relationship Between Electric
Potential and Electric Field

The effects of a charge distribution can be described
using either the electric field or the electric potential





Electric potential is a scalar which makes it sometimes easier to
use
Work done by the electric field to move a positive charge
q from b to a is:
W = qVba
If there is a uniform field between two plates, the work
can be written as:
W = Fd = qEd
Therefore, Vba = Ed or E = Vba/d
The units of electric field are either V/m or N/C,
1 N/C = 1 V/m
Example


Two parallel plates are
charged to 50 V. If the
separation between the
plates is 0.050 m,
calculate the electric field
between them
E = V/d = 50 V/ 0.050 m
= 1000 V/m
Equipotential Lines and
Surfaces

Along equipotential
lines and surfaces, all
points are at the
same potential

An equipotential
surface must be
perpendicular to the
electric field at any
point
Equipotential Examples #1
The potential along an
equipotential curve is the
same at any point
Equipotential lines are
perpendicular to the electric
field lines
Equipotential Examples #2
V = W/qmoved
As we move a charge from
one equipotential line to
another we change its
electric potential
It takes the same amount of work to pull a
charge to one spot on the curve as it
does to pull it out to a different spot on
the curve. That means that the work
done per unit of charge (electric
potential) is also the same.
The work done was 10J on 1C so the
potential difference is 10J/C or 10
volts.
Electron Volts


A joule is a large unit of measure when charges
of the size of electrons are considered
An electron volt (eV) is defined as the energy
acquired by a particle carrying a charge equal to
that of an electron when it is moved through a
potential difference of one volt
1 eV = 1.6 x 10-19 J
Electric Potential of a Point Charge


The electric potential at a distance r from a
point charge Q is given by:
V = (1/4πε0) (Q/r)
= k (Q/r)
V goes to zero as r → ∞
Work to Force Two + Charges
Together
What is the minimum work required to move a
charge q = 3.0 μC from a great distance (r = ∞)
to a point 0.5 m from a charge Q = 20.0 μC?
 The work required is the change in potential
energy:
W = qVab = q (kQ/rb – kQ/ra)
= (3 x 10-6 C) (9 x 109 Nm2) (2.0 x 10-5 C) = 1.08J
(0.5m)

Which Has the Most Potential
Energy?
Largest negative energy
Hardest to separate
Positive energy
Capacitors

A capacitor is a device for storing electric
charge


The simplest capacitor consists of two parallel
conducting surfaces
If a voltage is applied to a capacitor it
becomes charged


The amount of charge is given by Q = CV where
C is called the capacitance of the capacitor
Capacitance is measured as coulombs per volt
and this unit is called a farad
Capacitance

The capacitance C is constant
for a given capacitor


It does not depend on Q or V; it
depends only on the structure of
the capacitor
For parallel plates of area A
separated by a distance d in air
the capacitance is given by:
C = ε0A/d
Dielectrics



In most capacitors the conducting layers are
separated by an insulating material that is called
a dielectric
The dielectric increases the voltage that can be
applied to the plates before they short out and
they can be placed closer together
The dielectric increases the capacitance of the
capacitor by a factor K which is called the
dielectric constant
C = Kε0A/d or C = εA/d where ε = Kε0
How a Dielectric Works

Consider a capacitor
with charges +Q and
–Q on its plates

The voltage between
the plates is Q = CAVA
where the subscript A
refers to having air
between the plates
How a Dielectric Works #2

Now place a dielectric
between the plates



The electric field between
the plates will induce
charges in the dielectric
even though the charges
can’t flow
The net effect is as if there
were a net charges on the outer
surfaces of the dielectric
The force on a test charge
q within the dielectric is
reduced by the factor K
because some of the field
lines no longer go through
the dielectric
How a Dielectric Works #3

Because the field is reduced within the dielectric
the force on the test charge is reduced by a factor
of K



The voltage is now given by V = VA/K
But the charge on the plates has not changed so
Q = CV where C is the capacitance with the
dielectric present
We can write:


C = Q/V = Q/(VA/K) = QK/VA = KCA
Therefore the capacitance is increased by the factor K
Common Dielectric Constants
Example



A capacitor consists of
two plates of area A
separated by a distance d
connected to a battery of
voltage V from which it
acquires a charge Q
While connected to the
battery a dielectric is
inserted
Will Q increase,
decrease, or stay the
same?



Since the capacitor remains
connected to the battery, the
voltage V must remain the
same
But inserting a dielectric
increases the capacitance C
and Q = CV
Therefore, if C increases, Q
must also
Storage of Electric Energy

A charged capacitor stores electric energy


The net effect of charging a capacitor is to move
a charge from one plate to another


The energy in a capacitor is equal to the work done to
charge it
As more and more charge accumulate on a plate, the
harder it becomes to put more charge on it
The energy in a capacitor is
U = ½QV = ½CV2 = ½Q2/C since Q = CV
Example

A camera flash unit stores energy in a
150 μF capacitor at 200 V



How much electric energy is stored?
U = ½CV2 = ½(150 x 10-6 F)(200 V)2
= 3.0 J
Notice that
FV2 = (C/V)(V2) = CV = C(J/C) = J
Cathode Ray Tubes (CRTs)

In a cathode ray tube, electrons are boiled off a hot electrode and
are accelerated by a potential of 5-50 kV
 The electrons are steered onto the screen by pairs of parallel
deflection plates
 Changing the voltage on the deflection plates will change the
position of the electrons on the screen
Do Now (11/19/13):



Draw a parallel circuit
Draw a series circuit
What is the difference between the two?
Multiple Capacitors






When used in circuits capacitors can be
either in series or parallel
When connected in parallel, the voltage is
the same across all capacitors
Q = Q1 + Q2 + Q 3 = C 1 V + C 2 V + C 3 V
A single capacitor with the equivalent
capacitance can be written as Ceq
Therefore,
CeqV = C1V + C2V + C3V or
= C1 + C2 + C 3
Capacitors in series just add
The effect is as if the surface area of the
plates was increased
How Lightning Occurs
When Charges Move Against
Forces, Work Is Done
•
•
In order to bring two like charges near each other work
must be done. In order to separate two opposite
charges, work must be done.
As the monkey does work on the positive charge, he
increases the energy of that charge. The closer he brings
it, the more electrical potential energy it has. When he
releases the charge, work gets done on the charge which
changes its energy from electrical potential energy to
kinetic energy.
Practice:

Complete the multiple choice questions in
Ch. 16