Transcript Electric Charges - St. Joseph's Anglo

Electric Charges
• Matter is made up of tiny
particles called atoms. In the
centre of each atom, there is a
nucleus. This is made up of
protons and neutrons. Orbiting
round the nucleus are electrons.
• Protons are positively charged.
Electrons are negatively charged.
The size of charge in an electron is
equal to that in a proton. Neutrons
have no charge.
Unit of charge
• Charge is measured in
coulomb C.
• Charge of a proton
= 1.6 x 10-19 C
• Charge of an electron
= -1.6 x 10-19 C
Electric force
• Like charge repel;
F
Q1
Q2
+
+
F
• Unlike charge attract.
Q1
+
F
F
Q2
–
F
Q1
Q2
–
–
F
Coulomb’s experiment
suspension
head
fibre
Q1
Q2
• A quantity of charge, Q1, on a
fixed sphere repels a like charge,
Q2, on a mobile sphere. This
mobile sphere is attached to the
end of a rod that is free to rotate
because the rod is suspended by a
fiber.
• The repulsion between the two
charges pushes Q2 away a
distance, r, until increasing
tension in the twisting fiber
balances the repulsion.
• The repulsion between the two
charges equals the amount of
force the twisted fiber exerts,
which is known for any given
angle.
Coulomb’s law
• Coulomb’s law states that 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.
Q1
Q2
F12
F21
r
Q1Q2
F 2
r
or
F21  F12  F
Q1Q2
F k 2
r
(Action and reaction)
Permittivity
• The force between two charges also depends on the medium
between them.
• If the charges are situated in an insulating medium instead of a
vacuum, the force between them will be reduced.
• A medium is said to have permittivity; a material with high
permittivity is one which reduced the force between two
charges greatly compared with the vacuum value.
Q1Q2
• The force between two charges is expressed as F 
 2
4e r
1
where e is the absolute permittivity of the medium in which the
charged are placed.
• In free space (vacuum), e = e0 = 8.85 x 10-12 C2 N-1 m-2 where e0
is called the permittivity of free space.
Relative permittivity
• Relative permittivity er is introduced to compare the
permittivity of different media easily.
• The permittivity of a medium is the product of the permittivity
of free space and its relative permittivity. i.e. e = e0 er
Medium
Relative permittivity er
Vacuum (free space)
1
Air
1.0005
Paraffin
2.1
Mica
6
Acetone
27
Methyl alcohol
28
Water
81
• Note:
Since the permittivity of air is very close to
that of free space (relative permittivity of air is 1.0005 ≈
1), the force between the charges in air can be
approximated as
Q1Q2
F
 2
4e 0 r
1
TAS Lab
Cycle
Group
20
(10/4)
21
(22/4)
22
(30/4)
1
B7
C8
A2
2
B7
C8
A2
3
A2
B7
C8
4
A2
B7
C8
Example 1 Coulomb’s law
Find the magnitude of the electric forces in the following cases.
(a)
Two +1C point charges separated by 1 mm in air.
(b)
Two point charges, + 0.1 C and – 0.2 C, are separated by 20
cm in water.
• Solution:
(a) F = [1/(4e0)](1 x 1)/(0.0012) = 8.99 x 1015 N
(b) F = [1/(4e0er)](0.1 x 0.2)/(0.22) = 5.55 x 107 N
Example 2 Coulomb’s law and centripetal force
In a hydrogen atom, an electron moves around a nucleus in a circular
path. Estimate the speed of the electron given the following
information.
Mass of an electron = 9.31 x 10-31 kg
Charge of an electron = 1.6 x 10-19 C
Radius of the orbit = 5 x 10-11 m
Solution:
• Centripetal force = electric force
= [1/(4eo)] (1.6 x 10-19 )(1.6 x 10-19 )/ (5 x 10-11 )2
= 9.2076 x 10-8 N
• By F = mv2/r
9.2076 x 10-8 = (9.31 x 10-31)v2 / (5 x 10-11)
v = 2.22 x 106 ms-1
speed of an orbiting electron is 2.22 x 106 ms-1
Example 3 Addition of electric forces
Three charged particles A, B and C are fixed at the corners of a
square of length a as shown below. What is the resultant force acting
on particle B? [Let k = 1/ (4e0)]
2F
√5 F
A (+ 1 C)
q
a
B (+ 1 C)
1F
a
C (+ 2 C)
• F = k(1 x 1)/(a2) = k/a2
• Magnitude of resultant force
= √5 F = √5 k/a2
• Direction of resultant force:
q = tan -1 (2/1) = 63.4o
The resultant force makes an
angle 63.4o with AB.
Electric field
• An electric field is a region where an electric charge
experiences a force. Electric charges are sources in setting
up electric field which in turn affect other charges placed
inside the field.
• Electric field is represented by field lines and the field lines
cannot cross
• The direction of the field lines shows the direction of the
electric force acting on a positive test charge.
• Where the lines are parallel and uniform spaced, the field is
uniform.
Electric field patterns
The electric field patterns for different arrangement of charges are
shown below.
+
An isolated positive charge
–
An isolated negative charge
Electric field patterns
The electric field patterns for different arrangement of charges are
shown below.
A pair of opposite
charges
A pair of positive
charges
A pair of parallel
plate with opposite
charges
Electric field strength
• The density of field lines is proportional to the electric field
strength.
• Where the field lines are closely spaced, the field is strong.
Where they are widely spaced, the field is weak.
weak field
strong
field
Electric field strength
• If a (positive) test charge Q0 placed at a point experience a
electric force F, then the electric field strength E at that point
is defined to be the force experienced per unit charge.
F
Q0
+
Force
Electricfield strength
Charge
F
or
E
Q0
Force
Gravitatio nal field strength 
Mass
W
or
g
m
Note:
• 1. The direction of E is same as that of the force F.
• 2. Q0 should be extremely small so that it would not disturb
the original electric field.
F
Q0
+
Electric field strength of a positive
point charge
+Q
+Q0
QQ 0
F
 2
4e r
1
+
+
r
• By Coulomb’s law, the electric force acting on the test charge
+Q0 is
1 QQ 0
F
 2
4e r
• The electric field at point P is E  F
Q0
i.e.
E
1
Q
4e r 2

F
QQ 0
4e r 2
1

Electric field strength of a
negative point charge
–Q
+Q0
–
QQ 0
F
 2
4e r
1
+
r
• By Coulomb’s law, the electric force acting on the test
charge – Q0 is F  
QQ0
4e r 2
1

• The electric field at point P is
F
E
Q0
i.e.
1
Q
E
 2
4e r
Note: E depends only on Q which produces the field, and not on the
value of the test charge Q0.
Electric field strength
• If a (positive) test charge Q0 placed at a point experience a
electric force F, then the electric field strength E at that point
is defined to be the force experienced per unit charge.
F
Q0
+
Force
Electricfield strength
Charge
F
or
E
Q0
Force
Gravitatio nal field strength 
Mass
W
or
g
m
Electric field strength of a positive
point charge
+Q
+Q0
QQ 0
F
 2
4e r
1
+
+
r
• By Coulomb’s law, the electric force acting on the test charge
+Q0 is
1 QQ 0
F
 2
4e r
• The electric field at point P is E  F
Q0
i.e.
E
1
Q
4e r 2

Example 4
A negative point charge of -6p C is placed at point A in vacuum.
Calculate the electric field strength at point P, where the distance
between A and P is 2 m. [Take 1/(4e0) = 9 x 109 Nm2C-2]
• Solution:
Electric field strength = [1/(4e0)](Q/r2)
= 9 x 109(6 x 10-12)/22
= 0.0135 NC-1
Example 5
ABC is an equilateral triangle of side a. A positive and a negative
point charge are placed at B and C respectively. The magnitude of
each of the point charge is Q. Find the resultant electric field strength
at A. [Let k = 1/(4e0)]
Electric field strength E due to +Q
Er
= [1/(4e0)](Q/r2)
= kQ/a2
E
E
Electric field strength E due to –Q
A
= kQ/a2
a
Resultant E field strength = kQ/a2
a
B
+Q
C
a
Note: E is a vector. The
–Q resultant field must be obtained
by using vector addition.
Electric field strength of a
charged conducting sphere
+
+
+
+
+
+
+
+
+
A point charge
A charged conducting sphere
• The charges in a charged conductor repel one another. Therefore,
they distributed uniformly over the surface of the sphere.
• Electric field is always perpendicular to the surface of the
conducting sphere.
• A charged sphere behaves like a point charge at its centre.
Inside the sphere, the electric field is zero.
1
Q
Outside the sphere, (r  a ), the electric field is E 
 2
4e 0 r
E
1
4e 0

Q
r2
At the surface of the sphere, (r = a),
E
1
4e 0

Q
a2
or

E
e0
where  
Q
4a 2
E
is known as the charge
per unit area of the
surface of the conductor
or the charge density.

eo
a
r
Comparison between gravitational
field and electric field
Gravitational field Electric field
Symbol
g
Objects affected Objects with mass
Meaning
Unit
W
g
m
N kg-1
Vector
E
Objects with
charge
E
N C-1
Vector
F
Q
Electric potential energy and
gravitational potential energy
The mass gains
gravitational
potential energy
+Q
+Q0
+
++
B
r
x
A
ground
• The gravitational potential energy of a mass is increased if it is moved to a
higher level (against the gravitational field).
• Similarly, the electric potential energy of a positive charge is increased if it is
moved from A to B (against the electric field). When the charge is allowed to
return from B to A, the electrical potential energy gained previously is changed
into kinetic energy.
• In lower forms (F4 – F5), the gravitational potential energy at ground level is
taken as zero.
• Now, theoretically, the electric potential energy of a charge at infinity is taken
as zero. (Remember: the p.e. of gas molecules)
V 
W
Q
Electric potential energy and electric
potential
• The electric potential energy W of a charge Qo at a point in an
electric field is defined as the work done (energy) to move the
charge Qo from infinity to that point.
• The electric potential V at a point in an electric field is defined as
the work done in moving a unit positive charge from infinity to
that point.
W
V 
Q
Potential due to a uniform field E
• The potential difference across two points A and B is defined as
the work done required to move a unit positive charge from point
A to point B.
• Consider two points A and B of distance d apart in the following
uniform field.
Proof:
d
Work done = F x s, F = EQ0
Put Q0 = 1 (unit charge)
B
x
Q0
+
+
A
x
Potential difference V
= Work done
= [E(1)]d = Ed
The electric potential difference between A and B is given by V = Ed.
Example 6
Electric potential is more commonly known as voltage. The potential across the
two parallel plates is 5 V and the distance between the plates is 3 cm.
(a)
Find the magnitude of electric field strength set up by the charged plates.
(b)
If a proton is released from the positive plate, find its speed when it
reaches the negative plate. Assume that the medium between the plates is
vacuum.
Given that the mass of a proton is 1.67 x 10–27 kg and the charge of a
proton is 1.6 x 10–19 C.
Solution:
(a) By V = Ed
(5) = E(0.03)
E = 167 Vm-1
Note: There are two units for E.
(1) E = F/Q (unit : N C-1)
(2) E = V/d (unit: V m-1)
Example 6
Electric potential is more commonly known as voltage. The potential across the
two parallel plates is 5 V and the distance between the plates is 3 cm.
(a)
Find the magnitude of electric field strength set up by the charged plates.
(b)
If a proton is released from the positive plate, find its speed when it
reaches the negative plate. Assume that the medium between the plates is
vacuum.
Given that the mass of a proton is 1.67 x 10–27 kg and the charge of a
proton is 1.6 x 10–19 C.
Solution:
(b) Electric potential energy stored is changed
into kinetic energy
QV = ½ mv2
5(1.6 x 10-19) = ½ (1.67 x 10-27)v2
v = 3.10 x 104 ms-1
Potential due to a point charge
The potential of a point at a distance of r from a positive point charge Q is
V
+Q
+
r
Q
4e r
+Q0
++
x
1

+Q0
++
QQ 0
F
 2
4e x
1
Force
r
Displacement
Potential due to a point charge
Force
r
Displacement
Electric potential V
= Work done in moving
a unit positive charge from infinity to a point
1 QQ
F

4e x – displacement graph
= area under the force
0
2
 F  dx
r
=
(– sign indicates the displacement is in the opposite direction
as the field)

Q
dx
2
4e x
1 Q

=
4e r
r
=  
1

V 
1
Q
4e r

Example 7
Find the potential at points A and B due to two point charges X and Y,
1 m apart in air and carrying charges of 2 x 10-8 C and -2 x 10-8 C
respectively.
• For point A:
B
1m
A
1m
X
1m
Y
1m
Potential due to X
= [1/(4e0)](QX/1)
= 9 x 109 (2 x 10-8) / 1 = 180 V
Potential due to Y
= [1/(4e0)](QY/2)
= 9 x 109 (-2 x 10-8) / 2 = -90 V
• Potential at point A
= 180 + (-90) = 90 V
Note:
Potential is a scalar.
Thus, the resultant can be
obtained by simple addition.
Example 7
Find the potential at points A and B due to two point charges X and Y,
1 m apart in air and carrying charges of 2 x 10-8 C and -2 x 10-8 C
respectively.
B
1m
A
1m
X
1m
Y
1m
• For point B:
Potential due to X
= [1/(4e0)](QX/1)
= 9 x 109 (2 x 10-8) / 1 = 180 V
Potential due to Y
= [1/(4e0)](QY/1)
= 9 x 109 (-2 x 10-8) = -180 V
• Potential at point B
= 180 + (-180) = 0 V
Electrical potential energy stored in a
system of charges
The electrical potential energy stored when two positive charges Q1
1 Q1Q2
and Q2 at a separation of r is W 
4e r
+Q1
+Q2
+
+
Proof:
r
The electrical potential energy stored in the system
= work done to move the charges from infinity to the positions required
= work done to move Q2 from infinity while Q1 is fixed
Q1Q2
dx
2
 4e
x
r
   Fdx   


1
Q1Q2
4e
r

r
1

Example 8
Two positively-charged balls are tied together by a string. One ball
has a mass of 30 g and a charge of 1 mC; the other has a mass of 40 g
and a charge of 2mC. The distance between them is 5 cm. Initially
they are at rest, but when the string is cut they move apart. What are
their speeds v1 and v2 when they are very far apart?
Solution:
30 g
40 g
30 g
40 g
Initial potential energy of the system
1 QQ
=
kQ
Q /r
4e
r 1 2
1
2
0
= 9 x 109 (1 x 10-6)(2 x 106) / 0.05
= 0.36 J
When the balls are very far apart, the potential energy is zero and final kinetic energy
= ½ mv12 + ½ mv22
Energy is conserved, so the final kinetic energy = initial potential energy
½ (0.03)v12 + ½ (0.04)v22 = 0.36
⇒3v12 + 4v22 = 72 --- (1)
By conservation of momentum,
m1v1 = m2v2
0.03v1 = 0.04 v2
⇒ v1 = 4v2/3 --- (2)
Solving (1) and (2): v1 = 3.70 ms-1 and v2 = 2.78 ms-1.
Potential due to a charged sphere
• Inside the sphere, electric field is zero.
Thus, the electric potential is constant,
but not zero.
• The field at any point outside the
sphere is exactly the same as if the
whole charge were concentrated at the
1 Q
centre of the sphere. i.e. V 

4e r
• The potential at the surface is
1
Q
V

where a is the radius of
4e a
the sphere.
Equipotential lines
100 V
150 V
200 V
equipotential
lines
electric field lines
Field and equipotential lines
for a positive point charge
400 V
300 V
200 V
100 V
0 V (arbitrary zero)
Field and equipotential
lines for a uniform field
• Equipotential lines are connected lines of the same potential.
• The equipotentials are always perpendicular to the field lines.
• The density of the equipotentials represents the strength of the
electric field.
• The equipotentials never cross each other.
• If a charge moves along an equipotential line, no work is done; if a
charge moves between quipotential lines, work is done.
Equipotential surfaces and field lines
• The equipotentials are always
perpendicular to the field lines.
• The density of the
equipotentials represents the
strength of the electric field.
• The equipotentials never cross
each other.
A conducting Material in an Electric
Field
• Consider a pair of oppositely charged plates which established a
uniform field between them.
conductor
V/V
+
-
0
E/V m-1
x/m 0
x/m
Electrostatic Shielding
+
-
• The field inside the hollow metal box is zero.
• A conducting box used in this way is an effective device for
shielding delicate instruments and electronic circuit from
unwanted external electric field.
• The inside of a car or an airplane is relatively safe from
lightning.
Influence on a charged conductor
an earthed
object
a neutral
object
a negatively
charged object
a positively
charged object
Consider an isolated positively charged conductor.
The potential is decreased by an earthed object, an uncharged
object or a negatively charged object.
The potential is increased by a positively charged object.
Flame probe investigation of electric
potential
• A flame probe is used to
measure the potential in
space.
• The ions in the fame
neutralize the charge on the
needle. Therefore, there is
no excessive charge on the
needle. i.e. the probe will
not influence the potential
to be measured.
• the potential can be read
from the deflection of the
leaf