Download by electric field

Week 1 - S2
Feb. 15 – 20, 2016
Fields
4.1.Electric Fields
4.2. Magnetic Fields
Teacher: Mohammad Mezaael
4.1.Electric Fields
4.1.1 state that electric fields are created by electric charges;
4.1.2 define electric field strength as force per unit positive charge;
4.1.3 select and use the equation E = F/q
4.1.4 select and use the equation for Coulomb's law: F= Qq/( 4 πε0 r2)
4.1.5 select and use the equation for the electric field strength of a point
charge: E=Q/(4πε0 r2)
4.1.6 state that a uniform electric field exists between oppositely charged
parallel plates:
4.1.7 select and use the equation for field strength between charged parallel
plates: E = V/d ;
4.1.8 sketch the electric field lines and equi-potentials for a point charge, a
spherically symmetric charge distribution, the region between two parallel
plates and an electric dipole;
4.1.9 explain the effect of a uniform electric field on the motion of charged
Vocabulary Terms










charge
electrically neutral
static electricity
positive charge
negative charge
electric forces
charge by friction
electroscope
protons
neutrons







electrons
gravitational field
charged
induction
Coulomb’s law
capacitor
parallel plate
capacitor
 microfarad
 coulomb
 electric field










capacitance
charge
polarization
shielding test
charge
farad
field inverse
square law
discharged field
lines
Electric Charge
Fundamental Charge: The charge on one electron.
e = 1.6 x 10 -19 C
Unit of charge is a Coulomb (C)
Two types of charge:
 Positive Charge: A shortage of electrons.
 Negative Charge: An excess of electrons.
Conservation of charge – The net charge of a closed system remains
constant.
A quantity of charge should always be identified with a
positive or a negative sign.
Nucleus
-
-
n + n
+ +
n
+
n
n
+
+ n
-
-
-
-
Negative
NeutralAtom
Atom
Positive
Atom
Number
Numberof
ofelectrons
electrons><=Number
Numberof
ofprotons
protons
Number
of
electrons
Number
of
protons
-19
-2e
=
-3.2
x
10
CC
+2e = +3.2 x 10-19
Electric Forces
• Electric forces are created between all electric charges.
• Because there are two kinds of charge (positive and negative)
the electrical force between charges can attract or repel.
Like Charges - Repel
F
+
+
Unlike Charges - Attract
-
F
F
+
F
Electric Field
Space around a charge
What is the difference?
The electric field is the space
around an electrical charge
just like
a gravitational field is the space
around a mass.
The electric field strength, E , is the force, Felec,
per unit charge, q, at a point. The equation for
EFS is (arrows show the units for each symbol)
E = Felec
q
• E = F/ qo
•
•
•
•
qo , positive test charge
E is a vector quantity
Unit: N/C
E is analogous to the gravitational field, g,
where g=F/m
Example 1:
If the electric field strength due to some charged
object is 1.5E3 N/C, find the force on a positive
charge due to 2E-4 C placed in the field.
• Known: E = 1.5E3 N/C & q = 2E-4 C
• Unknown = Felec
• Equation: E = Felec / q (solve for Felec)
Felec = (1.5E3 N/C) x (2E-4 C)
= 0.30 N
• If the charge on the object in example #1 was
doubled, does that change the electric field
strength? Hint: what change would increasing
q have on Felec?
• Increasing q will increase Felec so ratio between
the two is constant.
• Conclusion = size of test charge doesn’t effect
“E”
Coulomb’s Law
• Coulomb’s Law: the electric force is directly
proportional to the product of their charges
and inversely proportional to the square of
the distance between their charges.
Felec = K x q1x q2
d2
 The force between two
charges gets stronger as
the charges move closer
together.
 The force also gets
stronger if the amount of
charge becomes larger.
Electric Field Strength for a Point Charge
• Consider the force felt by a charge q in the field of another charge Q,
where the charges are separated by a distance r.
F = kQq / r2
+Q
r
+q
F
• But E = F/q, so
E = k Q / r2
• The electric field strength for a point charge Q is inversely proportional to
the square of the distance from the point charge.
Example 1: Work out the field strengths at the points labelled A and B in
the diagram below. What do you notice about the values, and why is this?
Add arrows at A and B to indicate the electric field strengths there.
-5C
20 cm
10 cm
B
A
EA = kQ/r2 = -1.124 x 1012 N C-1
EB = kQ/r2 = -4.5 x 1012 N C-1 (attractive force).
Electric Field Strength Between Charged Parallel
Plates
•
•
Work done on the charge = energy transformed by the charge
Fd = Vq
•
Rearranging this equation, we have
F/q =V/d
•
By definition, F/q = E, so
E=V/d
• E is directly proportional to the p.d (or V) between the two plates
and inversely proportional to the separation of the plates (d).
•
Units of electric field strength are V m-1 or N C-1
+ + + + + + ++ + + + + + + +
•
For a given separation and p.d, the electric field strength between
two parallel plates is constant (uniform field), except at the edges.
A point charge q is moved between the plates and experiences a
constant force F due to the uniform electric field.
E = V/d
+q
d
-
Two parallel conductors
Example 2:
A charged dust particle is stationary between two horizontal charged metal
plates. The metal plates have a separation of 3.6 cm and the p.d between
the plates is 720 V. The dust particle has a charge of +7e, where e = 1.6 x
10-19 C. Calculate:
(a) The electric field strength between the plates
b) The weight of the dust particle
Solution:
a) E = V / d = 720 / 0.036 = 2.0 x 104 V m-1
b) The particle is stationary, therefore the net force must be zero.
Weight = Electric force = Eq = 2.0 x 104 x (7 x 1.6 x 10-19) = 2.2 x 10-14 N
1.
2.
3.
4.
Facts
Increasing the amount of charge will increase the
electric force. (Why ?)
Increasing the distance between the charges will
decrease the electric force (Why ?)
Felec is + if the 2 charges are alike (similar) indicating
that they will repel each other (Why ?)
Felec is – if the 2 charges are different indicating that
they will attract each other (Why ?)
Example 3:
Two 40 gram masses each with a charge of 3μC are placed
50cm apart. Compare the gravitational force between the two
masses to the electric force between the two masses. (Ignore
the force of the earth on the two masses).
3μC
40g
3μC
40g
50cm
m1m2
Fg  G 2
r
 6.67 10
11
(.04)(.04)
2
(0.5)
 4.27 10
q1q2
FE  k 2
r
6
6
(
3

10
)(
3

10
)
9
 9.0 10
(0.5) 2
13
N
 0.324 N
The electric force is much greater than the gravitational
force
To summarise
Gravitational Fields
Electrical Fields
A gravitational field exists in a region of
space in which a stationary mass
experiences a force.
An electric field exists in a region of space
in which a stationary charge experiences a
force.
Gravitational field strength is the force per
unit mass on a point mass placed at that
point.
g=F/m
Electric field strength is the force per unit
positive charge on a point charge placed at
that point.
E=F/Q
F = Gm1m2/r2
F = kQ1Q2/r2
Inverse Square Law
Inverse Square Law
F is attractive only
F is attractive or repulsive
g = Gm/r2
E = kQ/r2
For the field strength due to a point mass
For the field strength due to a point charge
Example 4
Three charged objects are placed as shown. Find the net force
on the object with the charge of -4μC.
F k
- 5μC
45º
202  202  28cm
20cm
q1q2
r2
(5 106 )(4 106 )
F1  9 10
 4.5N
2
(0.20)
9
(5 106 )(4 106 )
F2  9 10
 2.30 N
2
(0.28)
9
5μC
F1 45º
- 4μC
20cm
F2
F1 and F2 must be added together as vectors.
F1
2.3cos45≈1.6
45º
F2
F1 = < - 4.5 , 0.0 >
+ F2 = < 1.6 , - 1.6 >
Fnet = < - 2.9 , - 1.6 >
2.3sin45≈1.6
- 1.6
- 2.9
29º
θ
3.31
Fnet  2.9 2  1.6 2  3.31N
  1.6 

  tan 

29

  2.9 
1
3.31N at 209º
Example 5
Two 8 grams, equally charged balls are suspended on earth as
shown in the diagram below. Find the charge on each ball.
20º
L = 30cm
FE
q
10º 10º
L = 30cm
30sin10º
q
r
r =2(30sin10º)=10.4cm
2
q1q2
q
FE  k 2  k 2
r
r
FE
Draw a force diagram for one charge and treat as an
equilibrium problem.

T
FE
q
80º
Tcos80º
Fg = .08N
T sin 80  .08
.08
T
 .081N

sin 80
Tsin80º
FE  T cos 80
q2

k

(.
081
)
cos
80
.104 2
.014
2
q 
(.104) 2
k
q  1.3 10 7 C
Electric Potential Energy
charges also have electrical potential energy
W  Fd
+Q
E
 QEd
F  QE
d
U e  QEd
+Q
v
Electric Potential Energy
• Work done (by electric field) on
charged particle is QEd
• Particle has gained Kinetic Energy
(QEd)
• Particle must therefore have lost
Potential Energy U=-QEd
Electric Potential
The electric potential energy depends on the charge
present
Change in potential is change in
potential energy for a test charge
per unit charge
We can define the electric potential V which
does not depend on charge by using a “test”
charge
U
V 
Q0
for uniform field
U  Q0 Ed
U
V 
  Ed
Q0
Electric Potential
compare with the Electric Field and Coulomb Force
U
V 
Q0
F
E
Q0
U  QV
F  QE
If we know the potential field this allows us to calculate changes
in potential energy for any charge introduced
Electric Potential
Electric Potential is a scalar field
it is defined everywhere
it doesn’t depend on a charge being there
but it does not have any direction
Electric Potential, units
SI Units of Electric Potential
U
V 
Q0
Units are J/C
Alternatively called Volts (V)
We have seen
E  V / d
V  Ed
Thus E also has units of V/m
Potential in Uniform field
WBC  F||d  0
E
C
+Q
WAB  F||d  QEd||
WAC  WAB  WBC
 QEd ||
d||
+Q
+Q
A
B
U AC  QEd||
V AC   Ed||
A
Electric Potential of a single charge
E
B
r
+
Capacitors
A system of two conductors, each
carrying equal charge is known as a
capacitor
Capacitance of charged sphere
Q
C
V
r=
40 r
V 
1 Q
R
+Q
-
definition
potential due to
isolated charge
Capacitors
e.g. 1: two metal spheres
e.g. 2: two parallel sheets
-
+
Each conductor is called a plate
+Q
-Q
Equi-potential Lines
Like elevation, potential can be displayed as contours
Like elevation, potential requires a zero point, potential V=0
at r=
Like slope & elevation we can obtain
the Electric Field from the potential
field
V
E
r
A contour diagram
Electric field lines around a positive point charge are directed radially
outward. The reason is the fact that test charge is defined as a small
positive charge. When test charges are placed around a positive charge,
they will be repelled and move radially outward. On the contrary, when test
charges are placed around a negative charge, they will be pulled radially
inward. The field lines for a single positive and a single negative point
charge when each acts alone are shown in the following figure:
Equipotential surfaces are necessarily spheres centered at the point
charge. At any point on an equipotential surface (or sphere around a
single point charge), the potential is the same. That is the reason for the
choice of word “equipotential.” field lines are always perpendicular to
equipotential lines. This concept may be used to draw field lines around
any type of charge distribution if equipotential lines are first determined.
The field lines and equipotential lines around an electric dipole are shown below:
The field lines and equipotential lines around a charge distribution similar
to that of a parallel-plate capacitor are shown below: