Download DC CIRCUITS

AP Physics Summer Institute
2007
ELECTROSTATICS
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.
If the glass rod is suspended from a
string and a second positively
charged glass rod is brought near,
a force of electrical repulsion
results. Negatively charged objects
also exert a repulsive force on one
another.
These results can be
summarized as follows:
unlike charges attract
and like charges repel.
CONSERVATION OF ELECTRIC CHARGE
In the process of rubbing two solid objects together,
electrical charges are not created. Instead, both objects
contain both positive and negative charges.
During the rubbing process, the negative charge is
transferred from one object to the other leaving one
object with an excess of positive charge and the other
with an excess of negative charge.
The quantity of excess charge on each object is exactly
the same.
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.
The SI unit of charge is the coulomb (C).
1 C = 6.25 x 1018 electrons or protons
The charge carried by the electron is represented by the
symbol -e, and the charge carried by the proton is +e. A
third particle, which carries no electrical charge, is the
neutron.
e = 1.6 x 10-19 C
melectron = 9.11 x 10-31 kg
mproton = 1.672 x 10-27 kg
mneutron = 1.675 x 10-27 kg.
Experiments performed early in this century have led to
the conclusion that protons and neutrons are confined
to the nucleus of the atom while the electrons exist
outside of the nucleus. When solids are rubbed together,
it is the electrons that are transferred from one object to
the other. The positive charges, which are located in the
nucleus, do not move.
Rubber scrapes electrons
from fur atoms.
INSULATORS AND CONDUCTORS
An insulator is a material in which the electrons are
tightly held by the nucleus and are not free to move
through the material. There is no such thing as a perfect
insulator, however examples of good insulators are:
glass, rubber, plastic and dry wood.
A conductor is a material through which electrons are
free to move through the material. Just as in the case of
the insulators, there is no perfect conductor. Examples
of good conductors include metals, such as silver,
copper, gold and mercury.
A few materials, such as silicon, germanium and carbon,
are called semiconductors. At ordinary temperature,
there are a few free electrons and the material is a poor
conductor of electricity. As the temperature rises,
electrons break free and move through the material. As a
result, the ability of a semiconductor to conduct
improves with temperature.
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
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 = - 8 μC
F1
q2= +12 μC
q1q2
q3 = - 4 μC
+
Fk 2
q
q
q
r = 0. 120m
1
3
6
2
r
6
(8 x10 )(4 x10 )
F1  9 x10
= 80 N
2
(0.06)
6
6
9 (12 x10 )(4 x10 ) = 120 N
F2  9 x10
(0.06) 2
9
FR = 80 + 120
= 200 N, to the right
2. Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are arranged
as shown. Find the resultant force on q3due to the other two
charges.
F1
FR
q1 = +4 nC
q2= -6 nC
F2
37˚
θ
q3 = -8 nC
q1q2
Fk 2
r
9
9
(
4
x
10
)(
8
x
10
)
9
-5 N
=
2.88x10
F1  9 x10
(01
. )2
9
9
(
6
x
10
)(
8
x
10
)
9
-5 N
=
6.75x10
F2  9 x10
(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.
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
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
3. 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 nC
q2 = +6 nC
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
3. 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 = -6 nC
q2 = +6 nC
E2
37º
θ
ER
E1
9 x109 (6x109 )
E1 
(9 x102 ) 2
= 6.67x103 N/C, down
9 x109 (6 x109 )
3 N/C at 37˚
E2 
=
2.4x10
2 2
(15x10 )
E2
From FBD
37º
θ
Σ Ex = - E2cos 37˚
= - (2.4x103)(cos 37˚)
= -1916.7 N/C
ER
E1
Σ Fy = E2 sin 37˚- E1
= (2.4x103)(sin 37˚) - (6.67x103)
= - 5225.6 N/C
E R  (1916.7)  (5225.6)
2
2
= 5566 N/C
5225.6
  tan
= 70˚
1916.7
1
E2
37º
θ
180˚ + 70˚ = 250˚
ER (5566 N/C, 250˚)
ER
E1
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 A
VA 
q
Units: J/C = Volt (V)
As discussed in the energy chapter, only differences in
potential are measurable.
We find the potential difference between two points A
and B to be equal to:
VAB
WBA
 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.
Since the electric potential is defined as the potential
energy per unit charge, then the change in potential
energy of a charge q when moved between two points A
and B is:
ΔPE = PEB - PEA = qVBA
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.
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.
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 VBA
W = F d = q Ed
(d is distance)
VBA = E d
Units: VBA = Volts (V)
E = V/m or N/C
4. The potential difference between two plates 5 mm apart is 10 kV.
Determine the electric field intensity between the two plates.
r = 5x10-3 m
V = 10x103 V
V 10 x103
E 
3
5
x
10
d
= 2x106 V/m
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.
In other words, the potential at some point A as shown
in the figure, is equal to the potential energy per unit
charge.
A
Q
VA  k
r
Units: J/C = Volt (V)
The potential in the vicinity of a number of charges is
equal to the algebraic sum of the potentials due to each
charge.
V = V1 + V2 + V3 + ........... + VN
= (Qi/ri)
kQ
V 
r
Units: Volts (V)
5. 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
2
-4
+3
kq1 kq2 kq3 kq4



V = V 1 + V2 + V 3 + V 4 
r
r
r
r
k
9 x109 (1  2  3  4) x109
V  (q1  q2  q3  q4 ) 
= - 25 V
1
r
2
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)
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.
6. A charge of + 2 nC is 20 cm away from another charge of
μC. a. What is the potential energy of the system?
q = +2 nC
Q = + 4 μC
r = 20 cm
9
6
9
9
x
10
(
4
x
10
)(
2
x
10
)
kQq

PE 
0.2
r
= 36x10-5 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 = 8 cm
9 x109 (4 x106 )(2 x109 )
= 90x10-5 J
PE 
0.08
ΔPE = 90x10-5 - 36x10-5
= 54x10-5 J
+4