Download Charge

W. Sautter 2007
Electrostatics is the study of the effects of stationary
charges on each other in their surroundings.
Charges are created by the transfer of electrons to or from
one body to another. (Protons are NEVER transferred.)
Objects with equal numbers of protons and
electrons are neutral. They have no net charge.
Objects with more electrons than
protons are charged negatively
Objects with less electrons than
protons are charged positively.
-
+
-
-
-
-
+
+ Likes +
+
-
Unlikes
attract
Likes
repel
repel
+
Unlikes
attract
Charge on 1 electron = - 1.6 x 10 –19 coulombs
Charge on 1 proton = + 1.6 x 10 –19 coulombs
6.25 x 10 18electrons
1 coulomb
Fg = gravity force between
m1 and m2 separated by
a distance r
G is the Universal
Gravitational Constant
The weight of an object
is its mass times g’, the
gravity value at location r
Fe = electrostatic force
Between m1 and m2
separated by a distance r
k is Coulomb’s
Constant
Coulomb’s Law is an inverse square law
similar to the Law of Gravitation
It is dissimilar in that electrostatic forces can be attraction or
repulsion. Gravity is attraction only.
Electrostatic force is strong, gravity is very weak.
Gravity is classified as a weak force because huge
amounts of mass are required to create a reasonably
large force.
Note the very small value of the force constant.
Electric forces is classified as a strong force because
small charge quantities can to create large forces.
Note the very large value of the force constant
Electric charges are detected by the presence of an electric field (E).
Recall that a gravity field can be detected by its influence
(attractive force) on a mass (often called a test mass).
Electric fields are detected by their influence (attractive or
repulsive forces) on a charge (often called a test charge).
Electrically charged bodies can be created by physically rubbing
electrons off one object on to another. For example, rubbing a
rubber rod with fur will transfer electrons from the fur to the rod
which becomes negatively charged (it has extra electrons).
Rubbing a glass rod with silk will transfer electrons from the rod to
the silk. The rod is charged positively (it has missing electrons).
- - - - -
Moveable
leaves
-
Charged rod
An Electroscope is an
instrument used to measure
the presence of an Electric field
(presence of charged bodies).
Leaves of electroscope
Diverge (like charges repel)
CHARGING BY
CONDUCTION
OBJECT
IS NOW
- NEUTRAL
NEGATIVELY
OBJECT
CHARGED
- -
- - - -- - -- -
WHEN OBJECT IS TOUCHED
BY THE CHARGED ROD
ELECTRONS MOVE FROM
THE ROD TO THE SPHERE
UNTIL ELECTROSTATIC
EQUILIBRIUM IS REACHED
INDUCTIVE CHARGING
CHARGED RUBBER ROD
(EXCESS ELECTRONS)
Electrons on sphere move
to the opposite side due to
repulsion of electrons on
the charged rod
Electric Field
Charged Plates
E is always out of plus (+) into minus (-)
Recall that a gravity field (g) is measured by dividing the force
acting on it (its weight(w) in Newtons ) by the mass quantity (m)
in kilograms
g in N / Kg
An electric field (E) is measured by dividing the force
acting on it (in Newtons ) by the charge quantity (q) in
coulombs (C ).
E in N / C
-
+
E
Scale
Test charge
-
Scale
+ Test charge
E
As Electric
field strength
increases, the
Force on a test
Charge increases.
This is similar to
the weight of a
Mass increasing
in a stronger
Gravity field
Electric Field
+
Force
Charged Plates
E = Force (N) / Charge (Coul)
+
E = k q / r2
Lines of Flux spread
over a greater area
as distance from
charge increases and
field strength
weakens
(1) E = F/q (by definition)
(2) F = kq1 q2 / r2 (Coulomb’s Law)
(3) E = kq1 q2 / r2 q2 ( by substitution)
E = kq / r2
q = point charge in coulombs
k = coulomb’s constant
r= distance from charge in
meters
E = electric field strength
at that point (N / C)
Electrical Potential is defined as the work required to move a
charge over a distance in an electric field.
Electrical potential is measured in volts.
One volt equals one joule (work) divided by charge (coulombs)
In a uniform electric field (a field between two parallel charged
metal plates) , potential (V) equals work (W) divided by charge (q).
Therefore, since work equal force times distance and force equals
the electric field strength (N/C) times charge in coulombs (C) ,
potential equals electric field times distance the charge moves.
V = W / q = (F x d) / q = (F / q) x d = E x d
V (volts) = E ( N/C) x d (m)
Electric Field
+
work
x Point B
+
work
x Point A
Charged Plates
V = Work (Joules) / Charge (coul)
In an electric field issued from a point charge the field varies
inversely with the square of the distance from the charge,
therefore the work required to move the charge is continuously
changing.
This is similar to the varying gravity field surrounding a mass.
The work required to move a mass in that field also
continuously changing.
To find the work required to move the charge in a varying
field we must integrate the force times distance equation
using calculus.
The following frames show the use of calculus to find work
form force versus distance relationships. If your course
does not require calculus, skip these frames
BOX METHOD
WORK = AREA UNDER THE CURVE
W =  F  X (SUM OF THE BOXES)
F
O
R
C
E
AREA
MISSED
- INCREASING
AS THE
NUMBER
OF BOXES
THE
NUMBERTHE
BOXES
WILL
INCREASES,
ERROR
REDUCE
THIS ERROR!
DECREASES!
(N)
WIDTH OF EACH BOX =  X
X1
X2
DISPLACEMENT (M)
Finding Area Under Curves Mathematically
• Areas under force versus distance graphs (work)
can be found mathematically. The process
requires that the equation for the graph be
known and integral calculus be used.
• Recall that integration is also referred to as
finding the antiderivative of a function.
• The next slide reviews the steps in finding the
integral of the basic function, y = kxn.
INTEGRATION – THE ANTIDERIVATIVE
INTEGRATION IS THE REVERSE PROCESS OF
FINDING THE DERIVATIVE. IT CAN ALSO BE USED
TO FIND THE AREA UNDER A CURVE.
THE GENERAL FORMAT FOR FINDING THE INTEGRAL
OF A SIMPLE POWER RELATIONSHIP, Y = KXn
ADD ONE
TO THE
POWER
 is the symbol
for integration
 
DIVIDE THE
EQUATION
BY THE N + 1
ADD A
CONSTANT
APPLYING THE INTEGRAL FORMULA
GIVEN THE
EQUATION
FORMAT TO
FIND THE
INTEGRAL
 

Integration can be used to find area under a curve between
two points. Also, if the original equation is a derivate, then
the equation from which the derivate came can be determined.
APPLYING THE INTEGRAL FORMULA
Find the area between x = 2 and x = 5 for the equation y = 5X3.
First find the integral of the equation as shown on the previous
frame. The integral was found to be 5/4 X4 + C.
The values 5 and 2 are
called the limits.
each of the limits is
placed in the integrated
equation and the results
of each calculation are
subtracted (lower limit
from upper limit)
From Infinity ( a great distance) to a point in the field

r

r
Coulomb’s Law
W= Fdr = ( k q1 q2 / r2 )dr
00
00
r
W = k q1 q2 r-1 = - k q1 q2 / r |
00
( 1/ infinity ~ 0 ) therefore
W = - k q1 q2 / r
Recall: Work to
move a charge in
a gravity field:
By Analogy: Work
to move a charge in
an electric field:
Work = - qG m1 m2
r
Work = - k q1 q2
r
Dividing the work equation obtained in the previous slides
by the charge quantity of the point charge we get a
relationship which describes absolute potential
at point in the field
Absolute potential is measured by using the
Work to move a charge from infinity to a point
+
V=kq/r
Point A
x+ work
+
work
Work to move a charge
from infinity to a point
in an electric field is
called Absolute
Potential
1 volt = 1 joule/ 1coul
Potential = Work / Charge
Volts = Joules / Coulomb
Recall from previous slide:
W = - k q1 q2 / r
Absolute potential is
measured from infinity
to a point r in the field
VAbsolute = ( - k q1 q2 / r ) / q2
q1 = point charge
VAbsolute = - k q1 / r
r= distance from the
point charge
V = W = - k q\ 1 q2
=kq/r
/
q
r
q
\1
V=kq/r
Where V is the absolute potential at a point (r) in the electric
field caused by charge (q) , we can now find the potential
difference between two points in the field.
Potential Difference
From point A to B
Absolute
at point B
Absolute
at point A
Capacitors are electrical devices used to store electrical energy.
They are not to be confused with batteries which create
electrical energy via chemical reaction.
The structure of a capacitor is shown on the next slide.
Essentially, electrons are pumped onto one of the metal
plates shown and pushed off the other plate by the electric
field developed on the first plate. The electrons are pumped
be a battery, power supply or other voltage source.
Once the charge is isolated on the metal plates it represents
stored energy since when the capacitor is connected to an
electrical circuit, electrons will flow from the plate of high
electron concentration through the connecting device,
thus releasing energy and back to the plate of low electron
concentration until charge equilibrium is reached.
Electrons
in
+
+
+
+
+
+
+
+
+
+
-
Capacitors
Store electrical
energy
Although it may seem illogical, in
conventional current the + plate
has the excess electrons !
C (farads) = q (coul) / V (volts)
A Farad is a unit of capacitance
1 F = 1 coul / 1 volt
Large capacitor can hold more
charge at lower voltages and
can store more energy
CAPACITOR TRENDS
(1) Large capacitors can hold more charge and
larger amounts of energy than smaller ones.
(2) The closer the plates are together, the more
energy that can be stored since the electric field
between the plates becomes stronger.
(3) The larger the dielectric constant the less
the tendency of electrons to “jump the gap”
(short out) between the plates at higher
voltages and more energy can be stored.
Metal plates (A) in meters2
+
+
+
+
+
+
+
+
+
+
-
Dielectric (non conductive material)
C = e0k A / d
e0 = permittivity constant
8.85 x 10-12 Coul2 / N m2
K = dielectric constant (air = 1.0)
It depends on the nature of the dielectric
As plate area increases, C increases
As plate separation increases, C decreases
Plate separation (d) in meters
Area under the graph
gives the work to charge
the capacitor
Charging a Capacitor
The slope of the graph
line equals the capacitance (C )
C=q/V
C C
H
A
R
G
E
(C)
V
VOLTAGE (V)
W = ½ qV,
Also since q = CV, by
Substitution
W = ½ CV2, and
since V = q / C
W = ½ q 2/ V
+
+
+
-
+
+
+
-
+
+
+
-
Effectively, the capacitors in parallel
become one larger capacitor
C(total)
= C (1) + C (2) + C (3)
Since each individual capacitor is directly
connected to the voltage source
(battery) each bears the same voltage
The charge on each capicator is dependent
on its individual capacitance. (Plate
size, separation, dielectric)
+
+
+
1/C(total)
-
+
+
+
-
+
+
+
-
= 1/ C (1) + 1/ C (2) + 1/C (3)
Capacitors in series have different characteristics than
Those connnected in parallel
(1) Since they are not all directly connnected to the
voltage source, the voltage on each varies according
to its individual capacitance. Larger ones have
lower voltage than smaller ones.
(2) Each capacitor, despite its size, has the same charge.
1/C(total)
= 1/ C (1) + 1/ C (2) + 1/C (3)