Transcript Q = Charge

CHAPTER 18 & 19
ELECTRIC FIELD,
ELECTRICAL ENERGY
and CAPACITANCE
Michael Faraday developed
the concept of electric field.
A charge creates an electric
field about it in all
directions. When another
charged object enters this
electric field, it experiences
an electric force.
Suppose I have a +Q
charge. I want to know the
strength of the electric field
(E) at a point which is a
distance r from it.
Electric field intensity (E) at a
point which is a distance r from
+Q is defined as the force
experienced by a small positive
charge +q when it is placed at
that point.
F
+q
Q
r
E
E
q
F
E
q
E  Electric Field Intensity
in Newtons/Co ulomb - (N/C)
F  Force in Newtons - (N)
q  charge in Coulombs - (C)
The direction of the electric field
intensity at a point is the same as
the direction in which a positive
charge +q would move when placed
at that point.
Consider a negative charge –
Q. The electric field intensity
at a point which is at a
distance r away is
F
E
Q
E
r
q
+q
F
E
q
kQq
F 2
r
kQq
2
F
kQ
r
E 
 2
q
q
r
Q
+q E
r
Electric field Lines are
imaginary lines drawn in such a
manner that their direction at
any point is the same as the
direction of the electric field at
that point. The strength of the
electric field is indicated by the
spacing between the lines. The
closer the electric field lines the
stronger is the electric field.
Consider a positive charge.
The electric field lines around
it will point away from the
charge.
Consider a negative charge.
The electric field lines around
it will point towards the
charge.
Field Lines for two positively
charged objects
Field Lines for a negatively and
positively charged object
Electric field lines for objects with
unequal strengths
http://dev.physicslab.org/asp/applets/pointcharges/default.asp
Example
What is the electric field
intensity at a distance 2 m
from a charge -12 x 10-6 C?
kQ
E 2
r
9
6
9 10 12 10

2
2
 2.7 10 4 N / C towards the negative charge
Problem #1
A negative charge of 2 x 10-8 C
experiences a force of 0.060 N
to the right in an electric field.
What is the magnitude of the
electric field?
F
.06N
6
E 

3

10
N
/
C
8
q 2  10 C
Problem #2
A positive test charge of 5 x 10-4
C is in an electric field that
exerts a force of 2.5 x10-4 N on
it. What is the magnitude of the
electric field at the location of
the test charge?
F 2.5  10 N
E 

.
50
N
/
C
4
q
5  10 C
4
Problem #3
What charge exists on a test
charge that experiences a
force of 1.4 x 10-8 N at a point
where the electric field
intensity is 2.0 x 104 N/C?
F 1.4  10
 13
q


7

10
C
4
E
2.0  10
8
Problem #4
Find the electric field intensity
at a point P, located 6 mm to the
left of a point charge of 8C.
What are the magnitude and
direction of the force on a –2nC
charge placed at a point P.
6mm
p
Q
8 x 10-6 C
6mm
Q
p
8 x 10-6 C
E ?
d  r  6mm  .006m
6
Q  8  10 C
kQ 9  10  8  10
E 2 
2
r
(.006)
9
6
 2  10 N / C left (west)
9
6mm
Q
p
8 x 10-6 C
q  2  10 C
9
E  2  10 N / C
9
F  Eq  (2  10 C )( 2  10 N / C )
 4N attractive - towards Q
9
9
Problem #5
Determine the electric field
intensity at the midpoint between
a –60 C charge and a + 40 C
charge. The charges are 70 mm
apart in air.
A
EA
p
-60 x 10-6 C EB
B
40 x 10-6 C
EA
A
p
-60 x 10-6 C EB
kQ 9 10  60 10
EA  2 
2
r
(.035)
9
B
40 x 10-6 C
6
 4.4 10 N / C left
8
kQ 9 10  40  10
EB  2 
2
r
(.035)
9
 3 10 N / C left
8
6
EA
A
p
-60 x 10-6 C EB
B
40 x 10-6 C
E A  4.4  10 N / C left
8
E B  3  10 N /C left
8
NetE  E A  E B  7.4  10 N / C left
8
Problem #6
Find the electric field intensity at
a point 30 mm to the right of a
16 x 10-9 C charge and a 40 mm
to the left of a 9 x 10-9 C charge.
30mm
A
16 x
10-9
C
p
EB
EA
40mm
B
9 x 10-9 C
p
30mm
A
16 x
10-9
C
EB
EA
kQ 9 10 16 10
EA  2 
2
r
(.030)
 160000 N / C right
9
B
40mm
9 x 10-9 C
9
kQ 9 10  9 10
EB  2 
2
r
(.040)
 50625 N / C left
9
9
30mm
A
16 x
10-9
C
p
EB
EA
40mm
B
9 x 10-9 C
E A  160000N /C right
E B  50625N /C left
NetE  E A  E B  160000  50625
 1.1  10 N / C  east  right
5
Problem #7
Two charges of +16 x 10-6 C and
+8 x 10-6 C are 200 mm apart in
air. At what point between the two
charges is the field intensity
equal to zero?
E2
E1
q1=8x10-6 C
q2=16x10-6 C
x
0.2  x
E1  E 2
kq1
kq2

2
2
x
(.2  x )
8  10
16  10

2
2
x
(. 2  x )
1
2

2
x
(. 2  x ) 2
6
6
(. 2  x )  2 x
. 2  x  1. 4 x
.2  1.4 x  x
.2  2.4 x
x  .08m
2
2
- - - - - Take Square Root
of Both sides
CONDUCTORS IN
ELECTROSTATIC EQUILIBRIUM
A good conductor contains
charges that are not bound to
any atom and are free to move
about within the material.
When no net motion of charges
occurs within a conductor, the
conductor is said to be in
electrostatic equillibrium.
AN ISOLATED CONDUCTOR
HAS THE FOLLOWING
PROPERTIES
1.The electric field is zero
everywhere inside the
conductor.
2. Any excess charge on an
isolated conductor resides
entirely on its surface.
AN ISOLATED CONDUCTOR
HAS THE FOLLOWING
PROPERTIES
3. The electric field just outside
a charged conductor is
perpendicular to the
conductor’s surface.
4. On an irregularly shaped
conductor, the charge tends to
accumulate at sharp points.
CHAPTER 19
HOORAY!!!
ENERGY & ELECTRIC
POTENTIAL
Point A
Point B
Consider a fixed positive
charge placed at a point A and
a fixed negative charge at
point B. There is a force of
attraction between the two
charges.
ENERGY & ELECTRIC
POTENTIAL
Point A
Point B
Point C
Therefore to bring the charge at B
to a point C a distance (r) from B,
will require work to be done against
the force of attraction. Thus the
charge at B will undergo a change
in potential energy.
ELECTRICAL POTENTIAL
DIFFERENCE
IS MEASURED IN VOLTS
IT IS THE CHANGE IN POTENTIAL
ENERGY PER UNIT CHARGE.
ΔPE W
V

q
q
V  Electric Potential - Volts(V)
PE  Potential Energy - Joules (J)
W  Work - Joules(J)
q  Charge - Coulombs(C )
W
V
q
F d

q
kqQ
r
2
r

q
kQ

r
This formula is
used when dealing
with charges
W
This formula is
V
used when working
q
with parallel plates
F d

q
F
 d
q
 Ed
This formula is
used when work
on a charge is
given
PE
V
q
W
V
q
Example
The electric field intensity
between the two charged
parallel plates is 8000 N/C. The
plates are 0.05 m apart. What is
the potential difference between
the two plates?
V  E  d  8000  .05  400V
Problem #1
A voltmeter reads 500 V when
placed across two charged,
parallel metal plates. The plates
are 0.02 m apart. What is the
electric field between them?
V  E d
V
E
d
500V

.02m
4
 2.5  10 V / m
Problem #2
What work is done on a 5 C
charge that is raised by a
potential of 1.5 V
W  qV
 5C 1.5V
 7.5J
Problem #3
If 120 J of work is
done to move one
Coulomb of charge
from a positive
plate to a negative
plate, what voltage
difference exists
between the
plates?
W  qV
W
V 
q
120J

1C
 120V
Problem #4
How much work is done to
transfer 0.15 C of charge
through a potential difference
of 9 V?
W  qV
 .15C  9V
 1.4J
Problem #5
An electron is moved through
a potential difference of 500 V.
How much work is done on the
electron?
W  qV
 (1.6  10 )( 500)
 19
 8.0  10 J
 17
Problem #6
A force of 0.053 N is required
to move a charge of 37 x 10-6 C
a distance of 25 cm in an
electric field. What is the size
of the potential difference
between the two points?
V 
V 


W
q
F d
q
.053  .25
6
37  10
2
3.6  10 V
Parallel Plates
When two parallel plates are
connected across a battery, the plates
will become charged and an electric
field will be established between them.
The direction of an electric field is
defined as the direction that a positive
test charge would move. So in this case,
the electric field would point from the
positive plate to the negative plate. The
field lines are parallel to each other and
hence this type of electric field is
uniform and is calculated with the
equation E = V / d.
V  Exd
V = Volts
E = N/C
Therefore 1 N/C = 1 V/m
d=m
Since the field lines are parallel
and the electric field is uniform
between two parallel plates, a
test charge would experience the
same force of attraction or
repulsion no matter where it was
located. That force is calculated
with the equation F = q E.
In the diagram above, the distance between the
plates is 0.14 meters and the voltage across the
plates is 28V.
V
28V
E 
d

.14m
 200V / m
If a positive 2 nC charge were inserted anywhere
between the plates, it would experience a force in the
direction of the negative, bottom plate, no MATTER
where it is placed in the region between the plates.
9
F  Eq  (200V / m)( 2  10 )
7
 4  10 N
Millikan Oil Drop Experiment
•
•
Millikan found the charge of an electron.
Fine oil drops were sprayed from an
atomizer in the air. Gravity acting on the
drops caused them to fall. A potential
difference was placed across the plates.
The resulting electric field between the
plates exerted a force on the charged
drops. The resulting electric field
between the plates was adjusted to
suspend a charged drop between the
plates.

Each electron always carries the same
charge.
 Charges are quantized
 Changes in charges are caused by one or
more electrons being added or removed.
qE  mg
q g

m E
V
E is found out by the formula : E 
d
Example
An oil drop weighs 1.9 x 10-15 N. It
is suspended in an electric field of
6 x 103 N/C. What is the charge of
the drop? How many excess
electrons does it carry?
qE  mg
mg 1.9  10
q

3
E
6  10
 15
 3.1  10 C
 19
3.1  1019
Excess electrons 
 2electrons
19
1.6  10
Problem #7
A positively charged oil drop
weighs 6.4 x 10-13 N. An electric field
of 4 x 106 N/C suspends the drop.
A) What is the charge on the drop?
B) How many electrons are in the
drop?
qE  mg
mg 6.4  10
q

6
E
4  10
 13
 1.6  10 C
 19
Problem #8
A charged particle of mass
2 x 10-9 kg and a charge of 2 x 10-6 C
is placed next to one of the two
parallel plates 10 cm apart with a
2000 N/C electric field intensity.
A) What is the acceleration that the
particle gets because of the force
due to the electric field between the
plates?
B) What is the speed of the particle
at the other plate?
F  ma  qE
qE 2  10  2000
a

9
m
2  10
6
2
 2  10 m / s
6
v  v  2aD
2
f
2
i
v  0  2(2  10 )(. 10)
2
f
6
vf 
2(2  10 )(. 10)
 632m / s
6
Problem #9
Two parallel plates, separated by a
distance of 10 cm have an electric
field of 5000 N/C. If a 4 x 10-6 C
charge with a mass of 5 x 10-15 kg
is placed next to one plate,
A) What will be the acceleration of
the charge?
B) If it starts from rest, what will be
the speed of the charge at the
other plate?
F  ma  qE
qE 4  10  5000
a

 15
m
5  10
12
2
 4  10 m / s
6
v  v  2aD
2
f
2
i
v  0  2(4  10 )(. 10)
2
f
12
v f  2(4  10 )(. 10)
12
 9  10 m / s
5
Capacitor
Capacitor is a device designed to store
electric charge. A typical design of a
capacitor consists of two parallel plates
(metal) separated by a distance d. Plates
are connected to a battery. Electrons leave
one plate giving it a +Q charge, transferred
through the battery and to the other plate
giving it a –Q charge. This charge transfer
stops when the potential difference across
the plates equals the potential difference of
the battery. Thus the charged capacitor acts
as a storehouse of charge and energy that
can be reclaimed when needed for a specific
application.
The capacitance C of a capacitor
is defined as the ratio of the
magnitude of the charge on
either conductor to the
magnitude of the potential
difference between the
Q
conductors.
C
V
C = Capacitance (Farad) (F)
Q = Charge (Coulumb) (C)
V = Potential Difference (Volts)(V)
Problem #10
Both 3.3 x 10-6 F and 6.8 x 10-6 F
capacitor are connected across a
15 V battery. Which capacitor has a
greater charge?
Q
Q
C
C
V
V
6
6
Q  CV  (3.3  10 )(15) Q  CV  (6.8  10 )(15)
 5  10 5 C
 1.02  10 C
4
Thus 6.8 x 10-6 F has a greater
charge.
BATTERIES
Electrons collect on the negative terminal of
the battery. If you connect a wire between
the negative and positive terminals, the
electrons will flow from the negative to the
positive terminal as fast as they can (and
wear out the battery very quickly -- this also
tends to be dangerous, especially with large
batteries, so it is not something you want to
be doing). Normally, you connect some type
of load to the battery using the wire. The
load might be something like a light bulb, a
motor or an electronic circuit like a radio.
Inside the battery itself, a chemical reaction produces
the electrons. The speed of electron production by
this chemical reaction (the battery's internal
resistance) controls how many electrons can flow
between the terminals. Electrons flow from the battery
into a wire, and must travel from the negative to the
positive terminal for the chemical reaction to take
place. That is why a battery can sit on a shelf for a
year and still have plenty of power -- unless electrons
are flowing from the negative to the positive terminal,
the chemical reaction does not take place. Once you
connect a wire, the reaction starts.
The first battery was created by Alessandro Volta in
1800. The voltaic pile was made by alternating layers
of zinc, blotting paper soaked in salt water, and silver.
This arrangement was known as a voltaic pile. The
top and bottom layers of the pile must be different
metals, as shown. If you attach a wire to the top and
bottom of the pile, you can measure a voltage and a
current from the pile. The pile can be stacked as high
as you like, and each layer will increase the voltage
by a fixed amount.