Transcript Electricity PP

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:
F k
F = electrostatic force (N)
q = charge (C)
k = 9x109 N•m2/C2
r = separation between charges (m)
q1q2
r
2
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
6
6
q1q3
9 (8 x10 )(4 x10 )
F1  k 2  9 x10
2
(0.06)
r
= 80 N
6
6
q2 q3
(12
x
10
)(4
x
10
)
9
= 120 N
F2  k 2  9 x10
2
r
(0.06)
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
9
9
q1q3
(4
x
10
)(8
x
10
)
9
-5 N
=
2.88x10
F1  k 2  9 x10
r
(0.1)2
9
9
q2 q3
9 (6 x10 )(8 x10 )
-5 N
F2  k 2  9 x10
=
6.75x10
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
  tan 1
4.45x105
θ = 21˚
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
16-3 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:
E
kq
r
2
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 (6x10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 (6x10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