Transcript Document

ELECTROSTATICS:
The study of the behavior of stationary charges
ELECTRIC CHARGE
There are two types of electric charge, arbitrarily called
positive and negative. Rubbing certain electrically
neutral objects together (e.g., a glass rod and a silk
cloth) tends to cause the electric charges to separate. In
the case of the glass and silk, the glass rod loses
negative charge and becomes positively charged while
the silk cloth gains negative charge and therefore
becomes negatively charged. After separation, the
negative charges and positive charges are found to
attract one another.
When a rubber rod is rubbed against fur, electrons
are removed from the fur and deposited on the rod.
Electrons
negative
move from
- positive
fur to the
- rubber rod. + + + +
The rod is said to be negatively charged because of
an excess of electrons. The fur is said to be positively
charged because of a deficiency of electrons.
When a glass rod is rubbed against silk, electrons are
removed from the glass and deposited on the silk.
Electrons
move from
glass to the
silk cloth.
glass
silk
positive
negative
- - - -
+ +
+ +
The glass is said to be positively charged because
of a deficiency of electrons. The silk is said to be
negatively charged because of an excess of
electrons.
Laboratory devices used to study the
existence of two kinds of electric charge.
Pith-ball
Electroscope
Gold-leaf
Electroscope
1. Charge the rubber rod by rubbing against fur.
2. Transfer electrons from rod to each pith ball.
The two negative charges repel each
other.
1. Charge the glass rod by rubbing against silk.
2. Touch balls with rod. Free electrons on the balls
move to fill vacancies on the cloth, leaving each of
the balls with a deficiency. (Positively charged.)
The two positive charges repel each
other.
Rubber
glass
Attraction
fur
silk
Note that the negatively charged (green) ball is
attracted to the positively charged (red) ball.
Opposite Charges Attract!
Like charges repel; unlike charges attract.
Neg
Pos
Neg
Pos
Neg Pos
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.
The law of conservation of electric charge: "The net
amount of electric charge produced in any process is
zero." Another way of saying this is that in any process
electric charge cannot be created or destroyed, however,
it can be transferred from one object to another.
Charged comb attracts
neutral bits of paper.
Charged comb attracts neutral
water molecules.
Applications of Electrostatic
Charging
Fine mist of negatively charged
gold particles adhere to
Negatively charged paint
positively charged protein on
adheres to positively
fingerprint.
charged metal.
The quantity of charge (q) can be defined in terms
of the number of electrons, but the Coulomb (C) is
a better unit for later work.
The Coulomb: 1 C = 6.25 x 1018 electrons
The charge on a single electron is:
1 electron: e- = -1.6 x 10-19 C
The coulomb (selected for use with electric
currents) is actually a very large unit for static
electricity. It is common to use the metric prefixes.
1 mC = 1 x 10-6 C
1 nC = 1 x 10-9 C
1 pC = 1 x 10-12 C
COULOMB’S LAW
Coulomb’s Law states that two point
charges exert a force (F) on one another
that is directly proportional to the product
of the magnitudes of the charges (q) and
inversely proportional to the square of the
distance (r) between their centers. The
equation is:
q1q2
Fk 2
r
F = electrostatic force (N)
q = charge (C)
k = 9x109 N•m2/C2
r = separation between charges (m)
The value of k can also be expressed in terms of the
permittivity of free space (εo):
k
1
4 o
 9x109 N. m2/C2
The proportionality constant (k) can only be used if the
medium that separates the charges is a vacuum. If the
region between the point charges is not a vacuum then
the value of the proportionality constant to be used is
determined by dividing k by the dielectric constant (K).
For a vacuum K = 1, for distilled water K = 80, and for
wax paper K = 2.25
Problem-Solving Strategies
1. Draw and label a figure indicating positive and
negative charges along with the given distances.
2. Draw the force of attraction or repulsion on the
given charge on a neat, labeled FBD.
3. Find the resultant force.
Important: Do not use the signs of the charges
when applying Coulomb's law!
11.1 Two charges q1 = - 8 μC and q2= +12 μC are placed 120 mm
apart in the air. What is the resultant force on a third charge
q3 = - 4 μC placed midway between the other charges?
FR
F2
q1 = - 8x10-6 C
F1
q2= +12x10-6 C
q3 = - 4x10-6 C
+
0.06 m
0.06 m
q
q
q
r = 0. 120 m
1
3
2
q1q3 9 x109 (8x106 )(4 x106 )
F1  k 2 
= 80 N
2
(0.06)
r
q2 q3 9 x109 (12 x106 )(4 x106 )
= 120 N
F2  k 2 
2
(0.06)
r
FR = 80 + 120
= 200 N, to the right
11.2 Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are
arranged as shown. Find the resultant force on q3 due to the other
two charges.
F1
FR
-9
q1 = +4x10 C
q2= -6x10-9 C
F2
37˚
θ
q3 = -8x10-9 C
q1q3 9 x109 (4 x109 )(8x109 )
-5 N
=
2.88x10
F1  k 2 
r
(0.1)2
q2 q3 9 x109 (6 x109 )(8x109 )
-5 N
=
6.75x10
F2  k 2 
r
(8x102 )2
FR
F1
37˚
θ
F2
From the FBD:
Σ Fx = F2 - F1 cos 37˚
= (6.75x10-5) - (2.88x10-5)(cos 37˚)
= 4.45x10-5 N
F  (4.45x105 ) 2  (173
. x105 ) 2
5
173
.
x
10
= 21˚
  tan 1
5
4.45x10
Σ Fy = F1 sin 37˚
= (2.88x10-5)(sin 37˚)
= 1.73x10-5 N
= 4.8x10-5 N
FR (4.8x10-5 N, 21˚)
ELECTRIC FIELD
An electric field is said to exit in a region of space in
which an electric charge will experience an electric
force. The magnitude of the electric field intensity is
given by:
F
E
q
Units: N/C
The direction of the electric field intensity at a point in
space is the same as the direction in which a positive
charge would move if it were placed at that point. The
electric field lines or lines of force indicate the direction.
The electric field is strongest in regions where the lines
are close together and weak when the lines are further
apart.
11.3 The electric field intensity between two plates is constant and
directed downward. The magnitude of the electric field intensity is
6x104 N/C. What are the magnitude and direction of the electric
force exerted on an electron projected horizontally between the
two plates?
E = 6x104 N/C
qe = 1.6x10-19 C
F = qE
= 1.6x10-19 (6x104)
= 9.6x10-15 N, upward
11.4 Show that the gravitational force on the electron of example
11-3 may be neglected.
me = 9.11x10-31 kg
FG = mg
= 9.11x10-31 (9.8)
= 8.92x10-30 N
The electric force is larger than the gravitational force by
a factor of 1.08x1015!
The electric field intensity E at a distance r from a single
charge q can be found as follows:
kq
E 2
r
Units: N/C
11.5 What is the electric field intensity at a distance of 2 m from a
charge of -12 μC?
r=2m
q = -12 μC
kq 9 x109 (12 x106 )
E 2 
2
2
r
= 27x103 N/C, towards q
When more than one charge contributes to the field, the
resultant field is the vector sum of the contributions
from each charge.
kq
E 2
r
Units: N/C
11.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart,
as shown in the figure. Determine the electric field a. At point A
q1 = -6 x10-9 C
q2 = +6 x10-9 C
E1
ER
E2
kq
E 2
r
9 x109 (6 x109 )
4 N/C, left
E1 
=
3.38x10
2 2
(4 x10 )
9 x109 (6 x109 )
3 N/C, left
E2 
=
8.44x10
(8 x102 ) 2
E1
ER
E2
ER = E 1 + E2
= 3.38x104 + 8.44x103
= 4.22x104 N/C, left
11.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart,
as shown in the figure. Determine the electric field b. At point B
q1 = -6x10-9 C
q2 = +6x10-9 C
E2
37º
θ
ER
E1
9 x109 (6 x109 )
E1 
(9 x102 ) 2
= 6.67x103 N/C
9 x109 (6 x109 )
3 N/C
E2 
=
2.4x10
2 2
(15x10 )
From vector diagram:
E2
37º
θ
Σ Ex = - E2cos 37˚
= - (2.4x103)(cos 37˚)
= -1916.7 N/C
ER
E1
Σ Ey = E2 sin 37˚- E1
= (2.4x103)(sin 37˚) - (6.67x103)
= - 5225.6 N/C
E R  (1916.7) 2  (5225.6) 2 = 5566 N/C
5225.6
  tan
= 70˚
1916.7
1
E2
37º
θ
180˚ + 70˚ = 250˚
ER (5566 N/C, 250˚)
ER
E1
GRAVITATIONAL POTENTIAL ENERGY
Consider that a mass m is moved from level A to level B.
An external force F equal to the weight mg must be
applied to move the mass against gravity. The work done
by this force is W = mgh
When the mass m reaches level B, it has a potential for
doing work relative to level A. The system has a
potential energy (U) that is equal to the work done
against gravity. The potential energy coverts into kinetic
energy as the mass falls from level B.
ELECTRIC POTENTIAL ENERGY
When a charge q is moved against a constant electric
force for a distance d, the potential energy of the system
is equal to the work done to move the charge:
PE = W = F d = qE d
If the charge is released, it will acquire a kinetic energy
as it returns the same distance.
An important difference between gravitational potential
energy and electric potential energy is that in the case of
gravity, there is only one kind of mass, and the forces
involved are always forces of attraction.
Therefore, a mass at higher elevations always has the
greater potential energy relative to the earth.
This is not true in the electrical case because of the
existence of the negative charge. In the figure below, a
positive charge has a greater potential energy at point
B than at point A. Work has been done against the
electric field.
On the other hand, if a negative charge were moved
from point A to B, work will be done by the field.
A negative charge would have a lower potential energy
at B, exactly the opposite situation of a positive charge.
Whenever a positive charge
is moved against an electric
field, the potential energy
increases.
Whenever a negative charge
is moved against an electric
field, the potential energy
decreases.
When a positive charge is
placed in an electric field:
- It moves in the direction of
the field
- It moves from a point of
higher potential to a point of
lower potential
- Its electrical potential energy
decreases and its kinetic
energy increases
+
-
When a negative charge is
placed in an electric field
- It moves opposite to the
direction of the field
- It moves from a point of
lower potential to a point of
higher potential
- Its electrical potential energy
increases and its kinetic
energy increases
- Work has to be done on the
charge for it to move from
point A to point B
+
-
ELECTRIC POTENTIAL ENERGY
The electric potential energy of a system composed of a
charge q and another charge Q separated by a distance
r is equal to the work done against the electric forces in
moving a charge +q from infinity to that point.
qQ
PE  k
r
Joules (J)
11.7 A charge of + 2 nC is 20 cm away from another charge of
+ 4 μC. a. What is the potential energy of the system?
+2x10-9C
q=
Q = + 4x10-6 C
r = 0.20 m
kQq
9 x109 (4 x106 )(2 x109 )
PE 

r
0.2
= 3.6x10-4 J
b. What is the change in potential energy if the 2 nC charge is
moved to a distance of 8 cm from the 4 μC charge?
r = 0.08 m
9 x109 (4 x106 )(2 x109 )
= 9.0x10-4 J
PE 
0.08
ΔPE = 9.0x10-4 - 3.6x10-4
= 5.4x10-4 J
• If the charges have the same sign, PE is positive
– Positive work must be done to force the two
charges near one another
– The like charges would repel
• If the charges have opposite signs, PE is negative
– The force would be attractive
– Work must be done to hold back the unlike charges
from accelerating as they are brought close
together
ELECTRIC POTENTIAL
The advantage of the concept of an electric field as force
per unit charge was to assign an electrical property to
space. If the electric field intensity is known at some
point, the force on a charge placed at that point can be
predicted.
It is equally convenient to assign another property to the
space surrounding a charge. This will allow us to predict
the potential energy due to another charged placed at
any point. This property of space is called electric
potential.
The electric potential V is defined in terms of the work to
be done on a charge to move it against an electric field.
The electric potential V is a scalar quantity defined as
the potential energy per unit charge.
PE
VA 
q
Units: J/C = Volt (V)
The potential at a point is defined in terms of a positive
charge. The potential due to a positive charge is
positive, and the potential due to a negative charge is
negative.
Substitution of the equation for potential energy yields:
PE kQq / r
VA 

q
q
kQ
VA 
r
Units: Volt (V)
ELECTRIC POTENTIAL due to POINT CHARGES
The potential V at a point a distance r from a charge Q is
equal to the work per unit charge done against electric
forces in bringing a positive charge +q from infinity to
that point.
The point of zero electric potential is taken to be at an
infinite distance from the charge
In other words, the potential at some point A as shown
in the figure, is equal to the potential energy per unit
charge.
A
kQ
VA 
r
The potential is the same at equal distances from a
spherical charge. The dashed lines in the figure below
are called equipotential lines. Note that the lines of equal
potential are always perpendicular to the electric field
lines.
Equipotential Lines
An equipotential surface
is a surface on which all
points are at the same
potential
No work is required to
move a charge at a
constant speed on an
equipotential surface
Electric Potential Due to Point Charges
These plots show the
potential due to
(a) positive and
b) negative charge.
11.8 a. Calculate the potential at a point A that is 30 cm distant from
a charge of- 2 μC.
Q = - 2x10-6 C
r = 0.3 m
kQ 9 x109 ( 2 x106 )

VA 
= - 6x104 V
0.3
r
b. Find the potential energy if a + 4 nC is placed at point A.
q = +4x10-9 C
PE = qVA
= 4x10-9 (-6x104)
= -24x10-5 J
The potential in the vicinity of a number of charges is
equal to the algebraic sum of the potentials due to each
charge.
11.9 Two charges + 6 μC and - 6 μC are separated by 12 cm, as
shown. Calculate the potential a) at point A
Q1 = + 6x10-6 C
Q2 = - 6x10-6 C
r1 = 4x10-2 m
r2 = 8x10-2 m
kQ
V
r
6
6
9 x10 (6 x10 ) 9 x10 ( 6 x10 )
5 V
VA 

=
6.75x10
4 x102
8 x102
9
9
positive = the field does the work
b) at point B.
kQ
V
r
9 x109 (6 x106 ) 9 x109 ( 6 x106 )
6 V
=
1.01x10
VB 

16 x102
4 x102
negative = an external source of energy must perform
the work
11.10 What is the electric potential at the center of a square of
sides equal 1 m if the charges placed at the corners are +1 nC, -2
nC, +3 nC and -4 nC (read in a CW direction)
+1
-2
The center of the square is equidistant
from all four charges, a distance r of
d
= 0.707
2
-4
+3
V = V 1 + V2 + V 3 + V 4 
k (q1  q2  q3  q4 )
V
r
kq1 kq2 kq3 kq4



r
r
r
r

9 x109 (1  2  3  4) 109
0.707
= - 25 V
The potential difference between two points A and B to
be equal to:
W
V  VA  VB 
q
In other words the
potential difference
between two points is the
work per unit positive
charge done by electric
forces in moving a small
test charge from the point
of higher potential to the
point of lower potential.
11.11 Consider the two charges on problem 11.8
a. What is the potential difference between points A and B?
VA = 6.75x105 V
VB = -10.1x105 V
ΔV = VA-VB
= 6.75x105 - (-1.01x106)
= 1.69x106 V
b. How much work is done by the electric field in moving a -2 nC
charge from A to B?
q = -2x10-9 C
W = q ΔV
= (-2x10-9)(1.69x106)
= - 3.37x10-3 J
POTENTIAL AND ELECTRIC FIELD
The potential difference between two oppositely charged
plates is equal to the product of the field intensity and
the plate separation.
W = q ΔV
=Fd
=qEd
ΔV = E d
Units:
ΔV = Volts (V)
E = V/m or N/C
11.12 The potential difference between two plates 5 mm apart is
10 kV. Determine the electric field intensity between the two plates.
d = 5x10-3 m
V = 10x103 V
V 10 x103
E 
3
5
x
10
d
= 2x106 V/m
CAPACITORS
A capacitor is a device that can store electric charge and
consists of two conducting objects placed near one
another but not touching. A typical capacitor consists of
a pair of parallel plates of area A separated by a distance
d. Often the two plates are rolled into the form of a
cylinder with paper or other insulator.
If a potential difference or voltage is applied to a
capacitor it becomes charged. The amount of charge
acquired by each plate is proportional to the voltage.
Q
C
V
C is the capacitance in farads (F),
Q is the charge in (C)
V is the voltage (V)
The capacitance of a parallel-plate capacitor depends on
the geometry of the capacitor according to this equation:
A
C  o
d
εo = 8.85x10-12 C2/Nm2
Where C is the capacitance (F),
A is the area of the plates (m2)
and d is the separation between
the plates (m)
Applications of Capacitors:
Keyboards and Defibrillators
• Computers use capacitors in the keyboard:
• The keyboards use capacitors at the bases of the keys
• When the key is pressed, the capacitor spacing decreases
and the capacitance increases
• The key is recognized by the change in capacitance
11.13 a. Find the capacitance of a capacitor whose plates are
12 cm x 12 cm separated 1 mm by an air gap.
A = 0.12 x 0.12 m2
d = 0.001 m
V = 12 V
. )(012
. )
A  8.85x1012 (012
= 1.27x10-10 F
C  o
(0.001)
d
b. Find the charge on each plate if the capacitor is connected to a
12 V battery.
Q
C
V
Q = CV
= 1.27x10-10 (12)
= 1.53x10-9 C
ENERGY OF A CHARGED CAPACITOR
Consider a capacitor initially uncharged. When a source
of voltage is connected to it, the voltage between the
plates increases as charge is transferred.
The energy of a charged capacitor can be found with
these equations:
2
1
1
Q
2
PE  QV  CV 
2
2
2C
Units: Joules (J)
CAPACITORS IN SERIES AND PARALLEL
These are the symbols used in different
arrangements of capacitors:
CAPACITORS IN SERIES
The concept of equivalent capacitance
refers to the capacitance of a single
capacitor that could replace all the
capacitors in a circuit.
CAPACITORS IN
SERIES
 Series capacitors always
have the same charge.
Q Q1 Q2 Q3
 
  The voltage across the
C C1 C2 C
equivalent
capacitor Ceq is
3
the sum of the voltage
across both capacitors.
Q1
+
+
C1
Q2
- +
- +
Q3
- +
- +
C2
Battery
-
C3
Charge is same:
series connection of
capacitors.
Q = Q1 = Q2 =Q3
V1
+
+
C1
•A
V2
- +
- +
V3
- +
- +
C2
Battery
Total voltage (V)
-
C3
•
B
series connection =
Sum of voltages
V = V1 + V2 + V3
Equivalent Capacitance in Series:
Where does the equation come from?
V1
+
+
V2
- +
- +
C1
C2
V3
-+
-+
-
Q
Q
C ; V 
V
C
C3
V = V1 + V2 + V3
Q1= Q2 = Q3
Q Q1 Q2 Q3
 

C C1 C2 C3
1 1 1 1
  
Ce C1 C2 C3
Equivalent Ce
for capacitors
in series:
n
1
1

Ce i 1 Ci
CAPACITORS IN PARALLEL
CAPACITORS IN PARALLEL
Parallel Capacitors
always have the same
voltage drop across
each of them.
The charge on the
equivalent capacitor
CEQ is the sum of the
charges on both
capacitors.
Equivalent Capacitance in Parallel:
Where does the equation come from?
Q
C  ; Q  CV
V
Parallel capacitors
in Parallel:
- -
Ce = C1 + C2 + C3
C3
+
+
- -
C2
+
+
+
+
C1
- -
Q = Q1 + Q2 + Q3
Equal Voltages:
CV = C1V1 + C2V2 + C3V3
Equivalent Ce
for capacitors
in parallel:
n
Ce   Ci
i 1
11.14 a. Find the equivalent capacitance of the circuit.
C2 = 2 μF, C3 = 3 μF, C4 = 4 μF
V = 120 V
C2 and C4 are in series
C2 ,4
C2 C4
2( 4)


= 1.33 μF
C2  C4 2  4
C3 is now in parallel with C2,4
Ceq= C3 + C2,4
= 3 +1.33
= 4.33 μF
1
1
1


C2 ,4 C2 C4
b. Determine the charge on each capacitor.
First, we need to find the total charge of the
system
QT = CeqV
Q3 = C3V
= 4.33 (120)
= 3(120)
= 520 μC
= 360 μC
Q2 and Q4 have the same charge since they are
in series:
Q 2 = Q4 = Q T - Q 3
= 520 - 360
= 160 μC
Q3 = 360 μC, Q2 = Q4 =160 μC
c. What is the voltage across the 4 μF capacitor?
Q4 160
V4 
= 40 V

C4
4
The remaining voltage (120 - 40 = 80 V) goes through the
C2 capacitor.