Faculty of Dentistry
Short notes on Physics II
Code: SCG 122
Prof. Ahmed Mohamed El-Lawindy
E Mail: ellawindy@su.edu.eg
www.http//dentistry.su.edu.eg
References
1-Lecture notes
2-Internet websites
3-Physics for biology and premedical students, Burns
4-College Physics ,Serway, 7th edition
Chapter 1 Electricity
1-1 Electrostatics
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Chapter 1:
ELECTRICITY
1-1 Electrostatics
1-1-1 Coulomb force
So far, most of the forces, pressures and stresses we have studied have been communicated by direct
contact. The sole exception has been gravity, which we have unquestioningly accepted as a given part of our
experimental environment. In fact, there is a more causal way to deal with gravity, and it is intimately
related to the way which we treat electricity.
The basic problem which electricity presents to our intuition is: how can the electric force act at a
distance, without direct contact? When a boy gets closer to a charged sphere, his hair will rise, defying
gravity. What actually happens is a separation of “electric charges" from the boy's hair such that the
charges on hair nearer to the charged sphere are of opposite type of the charge on it. This is an attractive
force between what is called positive and negative charges.
Figure 1-1-1: The attractive force between charges on a charged sphere and charges on a boy's hair.
At the heart of this simple phenomenon is one of the most universally applicable ideas in physics: action at
a distance is caused by charges, which are the sources of forces. In the case of the electric "Coulomb"
force, the magnitude is proportional to the product of the charges, and inversely proportional to the square
of the distance between them. Like all forces, it is a vector quantity, in scalar form, the force is given by:
F
q1q2
qq
 K 1 22 ,
2
4o r
r
1
K
1
4o .
(1-1-1)
Here, q1 and q2 are the magnitudes of the charges, F is a vector which points from one charge to the other,
r is the distance between them and K=1.9x1019 Nm2/C2, is called Coulomb’s constant. Equation (1-1-1) was
first postulated by Coulomb, and is known after his early experiment. He started his experiment by an
unkown charge Q, and by conduction he obtianed different sizes of charges relative to it, as shown in
figure 1-2.
Q
Q
Q/2
Q/2
Q/4
Q/2
Q/4
Q/4
Q/8
Q/8
Q/1
6
Q/8
Q/1
6
Fig 1-1-2 Coulombs apparatus
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Chapter 1 Electricity
1-1 Electrostatics
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The magnitudes of the charges are measured in "Coulombs" (abbreviated C), which is a new fundamental
unit for us. The proportionality constant  is called the "electrical permittivity" of the medium through
which the force acts. It is a measure of the "effectiveness" with which the electrical force is felt across
the medium. It is often written as the product of  and , where  = 8.85 x 10 - 12 C2/ N m 2 is the electrical
permittivity of the vacuum and is called the “relative permittivity". In case of vacuum,  Note that
the force can be either attractive, when the charges have opposite sign, or repulsive, when the charges
have like sign, figure 1-1-3. When the product of the charges is positive, the force which one charge exerts
on the other is directed away from the first charge, and so the other charge is repulsed. If the product
of the charges is negative, the force one charge exerts on the other is directed toward the first charge,
and they are attracted.
+q1--------r--------+q2
+q1---------r--------q2
Figure 1-1-3 Forces between electric charge
Figure 1-1-4
Physicists have identified the two objects which account for the vast majority of freely moving electrical
charge in the universe as microscopic "electrons" and "protons", Figure 1-1-4. For the present, we will
simply consider them as VERY small "particles": the electron is negative (by convention) and the proton is
positive. The mass of the electron is 9.109 x 10 -31 kg, the mass of the proton is 1.673 x 10 -27 kg, and the
magnitude of their electrical charge, determined by Millikan’s experiment is
e  1.6 x 10 - 19 C.
(1-1-2)
Free charges less than e have never been observed, and all charges are integer multiples of e. Often we will
express the macroscopic charges observed experimentally using a "number”,n, variable to count the number
of particles or ions involved as
q= ne
n=0,+1,+2,+ 3, …..
(1-1-3)
+ and – signs indicate positive and negative charges, respectively. The
elementary charge is one of the fundamental constants of nature.
Figure 1-1-5 Milliken’s apparatus to measure the electric charge
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Chapter 1 Electricity
1-1 Electrostatics
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In addition to electrons and protons, charge is also carried by "ionized" atoms and molecules, especially in
biological systems. These "ions" are formed by the addition or removal of electrons; their protons are
tightly bound in their nuclei.
An object is electrically neutral; even it contains a very large quantity of charges. Objects are said to be
electrically charged if extra charges are added or subtracted from it. Neutral objects are charged by
different ways. The most common ways are charging by conduction, induction, and rubbing.
Sample Problem 1-1-1 A penny, being electrically neutral, contains equal amounts of positive and negative
charge. What is the magnitude of these equal charges?
Solution The charge q is given by NZe, in which N is the number of atoms in a penny and Ze is the
magnitude of the positive and the negative charges carried by each atom.
The number N of atoms in a penny, assumed for simplicity to be made of copper, is NAm/M, in which NA is
the Avogadro constant. The mass m of the coin is 3.11 g, and the mass M of 1 mol of copper (called its molar
mass) is 63.5 g. We find
N= NAm =
M
(6.02 x 1023 atoms/mol x (3.11 g)
63.5g/mol
= 2.95 x 1022 atoms.
q= NZe=2.95 x 1022 atomsx26x2 atoms-1x1.6x10-19 C= 245440 C
It is a huge quantity of charge, but remembers that these charges neutralize each other so that we don’t
feel it.
The force of gravity is very similar to the Coulomb force; the charges are now the masses of the objects
which are gravitationally attracted and the constant of proportionality ("Newton's Constant", equal to
6.673 x 10 - 11 N m 2 / kg 2) is now in the numerator:
F = - G m 1 m 2 / r 2.
(1-1-4)
Figure 1-1-6 The forces between masses and charges.
Note that gravitational forces are always attractive, since mass is always positive. Note also that both the
Coulomb and gravitational forces are vectors: therefore the superposition of two or more forces is equal to
their vector sum.
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Chapter 1 Electricity
1-1 Electrostatics
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Sample Problem 1-1-2 The average distance r between the electron and the proton in the hydrogen atom
is 5.3x10-11 m. (a) What is the magnitude of the average electrostatic force that acts between these two
particles? (b) What is the magnitude of the average gravitational force that acts between these particles?
Solution (a) From Eq. (1-1) we have, for the electrostatic force,
q1q 2
r2
(1.6 x10 19 ) 2
F  9 x10 9
 8.2 x10 8 N
11 2
(5.3x10 )
FK
While this force may seem small (it is about equal to the weight of a speck of dust), it produces an immense
effect, namely, the acceleration of the electron within the atom.
(b) For the gravitational force, Eq. (1-3), we have
Fg= G m1m2/ r2
= (6.67 x 10-11 N.m/kg2)(9.ll x l0-31 kg) ( 1.67 X 10-27kg)
(5.3 X 10-11 m)2
-47
= 3.6 X l0 N.
It is worthy to mention that the gravitational force can safely be ignored compared to the Coulomb force.
Sample Problem 1-1-3 The nucleus of an iron atom has a radius of about 4x10 -15 m and contains 26
protons. What repulsive electrostatic force acts between two protons in such a nucleus if a distance of one
radius separates them?
Solution From Eq. (1-1), we have
FK
qpqp
r2
(8.99 x l09N.m2/C2)(l.60x l0-19 C)2
(4 X l0-15)2
=14 N.
This enormous force, more than 1 kg and acting on a single proton, must be more than balanced by
the attractive nuclear force that binds the nucleus together. This force, whose range is so short that its
effects cannot be felt very far outside the nucleus, is known as the "strong nuclear force" and is very well
named.
=
PROBLEMS
1.A point charge of +3.12X 10-6C is 12.3 cm distant from a second point charge of -1.48 X 10-6 C. Calculate
the magnitude of the force on each charge.
2. What must be the distance between point charge q 1 = 26.3 C and point charge q2 = - 47.1 C in order
that the attractive electrical force between them has a magnitude of 5.66 N?
3. In the return stroke of a typical lightning bolt a current of 2.5 X l0 4 A flows for 20s. How much
charge is transferred in this event?
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Chapter 1 Electricity
1-1 Electrostatics
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4. Two equally charged particles, held 3.20 mm apart, are released from rest The initial acceleration of
the first particle is observed to be 722 m/s2 and that of the second to be 9.16 m/s2 The mass of the first
particle is 6.31x l0-7 kg. Find (a) the mass of the second particle and (b) the magnitude of the common
charge.
5. Two identical conducting spheres, 1 and 2, carry equal amounts of charge and are fixed a distance apart
large compared with their diameters. They repel each other with an electrical force of 88 mN. Suppose now
that a third identical sphere 3, having an insulating handle and initially uncharged, is touched first to
sphere 1, then to sphere 2, and finally removed. Find the force between spheres 1 and 2 now.
6. Two positive charges, each 4.18 C, and a negative charge, - 6.36 C, are fixed at the vertices of an
equilateral triangle of side 13.0 cm. Find the electrical force on the negative charge.
7. Each of two small spheres is charged positively, the total charge being 52.6C. Each sphere is repelled
from the other with a force of 1.19 N when the spheres are 1.94 m apart. Calculate the charge on each
sphere.
8. Two identical conducting spheres, having charges of opposite sign, attract each other with a force of
0.108 N when separated by 50.0 cm. The spheres are suddenly connected by a thin conducting wire, which is
then removed, and there-after the spheres repel each other with a force of 0.0360 N. What were the
initial charges on the spheres?
9. Two fixed charges, +l.07C and -3.28C, are 61.8cm apart. Where may a third charge be located so that
no net force acts on it?
10.Two free point charges +q and +4q are a distance L apart. A third charge is so placed that the entire
system is in equilibrium. (a) Find the sign, magnitude, and location of the third charge. (b) Show that the
equilibrium is unstable.
11.A certain charge Q is to be divided into two parts (Q - q) and q. What is the relation of Q to q if the
two parts, placed a given distance apart, are to have a maximum Coulomb repulsion?
12.Three small balls, each of mass 13.3 g, are suspended separately from a common point by silk threads,
each 1.17 m long. The balls are identically charged and hang at the corners of an equilateral triangle 15.3cm
on a side. Find the charge on each ball.
13.Find the total charge in coulombs of 75.0 kg of electrons.
14.In a crystal of salt, an atom of sodium transfers one of its electrons to a neighboring atom of chlorine,
forming an ionic bond. The resulting positive sodium ion and negative chlorine ion attract each other by the
electrostatic force. Calculate the force of attraction if the ions are 282 pm apart.
15.The electrostatic force between two identical ions that are separated by a distance of 5.0xl0 -10 m is
3.7x 10-9 N. (a) Find the charge on each ion. (b) How many electrons are missing from each ion?
16(a) How many electrons would have to be removed from a penny to leave it with a charge of +l.15xl0 -7C?
(b) To what fraction of the electrons in the penny does this correspond? See Sample Problem 2.
17. An electron is in a vacuum near the surface of the Earth. Where a second electron should be placed so
that the net force on the first electron, owing to the other electron and to gravity, is zero?
18.Protons in cosmic rays strike the Earth's atmosphere at a rate averaged over the Earth's surface, of
1500 protons/m2 s. What total current does the Earth receive from beyond its atmosphere in the form of
incident cosmic ray protons?
19.Calculate the number of coulombs of positive charge in a glass of water. Assume the volume of the
water to be 250 cm3.
20.(a) What equal amounts of positive charge would have to be placed on the Earth and on the Moon to
neutralize their gravitational attraction? Do you need to know the Moon's distance to solve this problem?
Why or why not? (b) How many metric tons of hydrogen would be needed to provide the positive charge
calculated in part (a)? The molar mass of hydrogen is 1.008 g/mol.
6
Chapter 1 Electricity
1-1 Electrostatics
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1-1-2 Electric Field
In order to model the electrical influence of an isolated charge, we resort to a rather fictional scenario.
We first "fix" a charge (or charges) at specific coordinates in space. This charge is called the
"configuration charge". We think of the configuration charge as the source of an "electrical field", which
has a different magnitude and direction at every point in space according to the following equation:
E (r ) 
qc
4o r 2
1
(1-1-5)
where the subscript "c" on the charge indicates that it is a configuration charge. Here, r as the argument
of the function E specifies a location (the "field point") where we want to compute the field; r in the
denominator specifies the distance from the configuration charge to the field point, and E is a vector
pointing from the configuration charge to the field point. This electric field has dimensions of force over
charge, and the force felt by a "test" particle in this field is
F x = q t E x,
(1-1-6)
("t" indicating a test charge) which is equivalent to the Coulomb force. The notion here is that a charged
particle has an influence independent of whatever forces are felt by other charges. Since the electric field
is a vector, vector superposition again applies: the electric field at any given point in space due to a sum of
configuration charges is equal to the sum of the electric fields at that point. We pretend that the test
charge DOES NOT INFLUENCE the electrical field created by the configuration charges. Since the
configuration charges are "static" (do not move), they are the sources of the "electrostatic field"; the test
charge is a probe which does not affect that field.
If we want a picture of a field, we usually think of many lines of "flux" emerging radially away from the
charge:
Figure 1-1-6 Electric field lines of a
Figure 1-1-7 Superposition of electric fields
positive and a negative charges
For negative charges, the lines point inward. These flux lines indicate the direction a positively charged
test particle would move due to the electric force between it and the configuration charge. The "number"
of lines is proportional to the quantity of the configuration charge. Note that the "density per unit area" of
the flux decreases as the distance from the center increases. That is, the lines are much closer together
when nearer to their origin. Since we live in three spatial dimensions, and the surface area of a sphere
7
Chapter 1 Electricity
1-1 Electrostatics
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centered on the charge increases with the square of the distance, we have a model for the inverse square
force law: the magnitude of the force is proportional to the density of the field lines ("flux density"),
which decreases with the square of the distance. We can also see why a spherical source is equivalent to a
point source at its center: the flux lines emanating from the surface can be extrapolated inwards to the
center with no change in the physical forces felt outside the surface.
Sample Problem 1-1-6 A proton is placed in a uniform electric field E. What must be the magnitude and
direction of this field if the electrostatic force acting on the proton is just to balance its weight?
Solution From Eq. 2, replacing q0 by e and F by mg, we have
27
E=
mg (1.67 x10 kg) X (9.8m / s 2 )
F
=
=
e
1.60 X 1019 C
q0
= 1.0 x l0-17 N/C, directed up
this is a very weak field indeed. E must point vertically upward to the float the (positively charged) proton,
because F = q0E and q0 >0.
Sample Problem 1-1-7 In an ionized helium atom (a helium atom in which one of the two electrons has
been removed), the electron and the nucleus are separated by a distance of 26.5 pm. What is the electric
field due to the nucleus at the location of the electron?
Solution We use Eq.1-5, with q (the charge of the nucleus) equal to +2e:
E = =
1
40
q
r2
2(1.60 x1019 C )
= ( 8.99 x 10 N.m /C ) X
(26.5 x1012 m) 2
9
2
2
= 4.13x 10-12 N/C.
Sample Problem 1-1-8 Figure 1-1-8 shows a charge q1 of +1.5 C and a charge q2 of+2.3C.The first charge
is at the origin of an x-axis, and the second is at a position x = L, where L = 13cm. At what point P along the
x-axis is the electric field zero?
+ 1.5 C
2.3C
L = 13cm
Figure 1-1-8 Sample problem 1-8.
Solution The point must lie between the charges because only in this region do the forces exerted by q1 and
by q2 on a test charge oppose each other. If E1 is the electric field due to q1 and E2 is that due to q2, the
magnitudes of these vectors must be equal, or
E1=E2.
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Chapter 1 Electricity
1-1 Electrostatics
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From Eq. (1-1) we then have
1
q1
1
q2
=
2
40 x
40 ( L  x ) 2
where x is the coordinate of point P. Solving for x, we obtain
x=
13cm
L
=
= 5.8 cm
1  q 2 / q1 1  2.3C / 1.5C
This result is positive and is less than L, confirming that the zero-field point lies between the two charges,
as we know it must.
The bodies that can be charged may be of different forms, e.g. a wire, a thin sheet, and a volumetric shape
such as a cube, a rectangle, etc. In such cases one may speak about, linear charge density, , surface
charge density, , and volume charge density,. If the charge is uniformly distributed over these bodies,
one has the following relations;
= Q/L C/cm
= Q/A
C/cm2
= Q/V
C/cm3 (1-1-7)
Sample Problem 1-1-9 The magnitude of the average electric field normally present in the Earth's
atmosphere just above the surface of the Earth is about 150 N/C, directed downward. What is the total
net surface charge carried by the Earth? Assume the Earth to be a conductor.
Solution Lines of force terminate on negative charges so that, if the Earth's electric field points
downward, its average surface charge density  must be negative. From Eq. (1-5) one finds;
E= (8.85 x 1012C2/N.m2)(-150 N/C) =-l.33 x l0-9 C/m2.
The Earth's total charge q is the surface charge density multiplied by 4R2, the surface area of the
(presumed spherical) Earth. Thus
q = 4R2
= (-1.33 x 10-9 C/m2)( 4)(6.37 x 106 m)2
= -6.8 x 105C = -680 kC.
1-1-2-a The Electric Dipole
Figure 1-1-9 shows a positive and a negative charge of equal magnitude q placed a
distance d apart, a configuration called an electric dipole. We seek to calculate the
electric field E at point P, a distance x along the perpendicular bisector of the line joining
the charges.
The positive and negative charges set up electric fields E+ and E-, respectively. The
magnitudes of these two fields at P are equal, because P is equidistant from the positive
and negative charges. Figure 1-12 also shows the directions of E1 and E2 determined by
the directions of the force due to each charge alone that would act on a. positive test
charge at p. The total electric field at p is determined, according to the vector sum
E=E1+E2
And from Eq. 1-5, the magnitudes of the fields from each charge is given by
Figure 1-1-9 Positive and negative charges of equal magnitude form an electric dipole.
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Chapter 1 Electricity
1-1 Electrostatics
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Because the fields E1 and E2 have equal magnitudes and lie at equal angles with respect to the y
direction as shown, the y component of the total field is (E1 sin - E2 sin = 0. The total field E therefore
has only a x component, of magnitude
(1-1-8)
Equation (1-8) gives the magnitude of the electric field at P due to a dipole. A dipole moment quantity, p, is
a vector quantity whose value p=qa and directed in the direction from negative to positive charge of the
dipole.
Often we observe the field of an electric dipole at points P whose distance y from the dipole is very large
compared with the separation a. In this case If y>>a, one can neglect a2 compared to y2 and write
E  ke
2p
y3
An expression of a similar form is obtained for the field along the dipole axis (the x-axis of Fig. 1-1-9);
E  ke
p
2x 3
In either case, the field at distant points varies with the distance r from the dipole as l/r3. This is a
characteristic result for the electric dipole field. The field varies more rapidly with distance than the l/r2
dependence characteristic of a point charge.
There are also more complicated charge distributions that give electric fields that vary as higher
inverse powers of r. For examples, the field of an electric quadruple, a variation of 1/r4 is obtained.
Figure 1-10 Electric quadruples
Molecules are said to be polarized when a separation exists between the average position of the negative
charges and the average position of the positive charges in the molecule. In some molecules, such as water,
this condition is always present—such molecules are called polar molecules. Molecules that do not possess a
permanent polarization are called nonpolar molecules.
Figure 1-11 Representation of water molecule
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Chapter 1 Electricity
1-1 Electrostatics
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We can understand the permanent polarization of water by inspecting the geometry of the water molecule.
In the water molecule, the oxygen atom is bonded to the hydrogen atoms such that an angle of 105° is
formed between the two bonds, Fig. 1-1-11. The center of the negative charge distribution is near the
oxygen atom, and the center of the positive charge distribution lies at a point midway along the line joining
the hydrogen atoms (the point labeled x in Fig. 1-1-11). We can model the water molecule and other polar
molecules as dipoles because the average positions of the positive and negative charges act as point
charges. As a result, we can apply our discussion of dipoles to the behavior of polar molecules.
Microwave ovens take advantage of the polar nature of the water molecule. When in operation, microwave
ovens generate a rapidly changing electric field that causes the polar molecules to swing back and forth,
absorbing energy from the field in the process. Because the jostling molecules collide with each other, the
energy they absorb from the field is converted to internal energy, which corresponds to an increase in
temperature of the food.
Another household scenario in which the dipole structure of water is exploited is washing with soap and
water. Grease and oil are made up of nonpolar molecules, which are generally not attracted to water. Plain
water is not very useful for removing this type of grime. Soap contains long molecules called surfactants. In
a long molecule, the polarity characteristics of one end of the molecule can be different from those at the
other end. In a surfactant molecule, one end acts like a nonpolar molecule and the other acts like a polar
molecule. The nonpolar end can attach to a grease or oil molecule, and the polar end can attach to a water
molecule. Thus, the soap serves as a chain, linking the dirt and water molecules together. When the water is
rinsed away, the grease and oil go with it.
Sample Problem 1-1-9
A molecule of water vapor (H2O) has an electric dipole moment of magnitude p = 6.2 x l0-30 C.m. (This
large dipole moment is responsible for many of the properties that make water such an important
substance, such as its ability to act as an almost universal solvent.).
E
Figure 1-1-12 Representation of dipole moment of water molecule in an electric field, E
A representation, figure 1-1-12, of this molecule shows three nuclei and a surrounding electron clouds. A
vector on the axis of symmetry represents the electric dipole moment p. (a) How far apart are the
effective centers of positive and negative charge in a molecule of H 2O? (b) What is the maximum torque on
a molecule of H2O in a typical laboratory electric field of magnitude 1.5x104 N/C?
Solution (a) There are 10 electrons and, correspondingly, 10 positive charges in this molecule. We can
write, for the magnitude of the dipole moment,
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Chapter 1 Electricity
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p= qd = (l0e)(d),
in which d is the separation we are seeking and e is the elementary charge. Thus
d
p
6.2 x1030 C.m

10e (10)(1.60 x1019 C )
=3.9x10-12m= 3.9 pm
This is about 4% of the OH bond distance in this molecule.
(b) The torque, pE sinis a maximum when  = 900. Substituting this value in that equation
yields
pE sin= (6.2x10-30C.m)(1.5x104N/C)(sin 900)
= 9.3x10-26 N.m
1-1-2-b A Point Charge in an Electric Field
In this section we try to answer the question:
What happens when we put a charged particle in a known electric field?
From Eq. 1-6, one know that a particle of charge q in an electric field E experiences a force F given by
F = qE.
To study the motion of the particle in the electric field, all we need do is use Newton's second law,
 F  ma , where the resultant force on the particle includes electric force and any other forces that
may act.
We can achieve a simplification if we consider the case in which the electric force is constant. Such
a situation can be achieved in practice by connecting the terminals of a battery to a pair of parallel metal
plates that are insulated from each other. If the distance between the plates is small compared with
their dimensions, the field in the region between the plates will be very nearly uniform, except near the
edges. In the following sample problems, we assume that the field exists only in the region between the
plates and drops suddenly to zero when the particle leaves that region. In reality the field decreases
rapidly over a distance that is of the order of the spacing between the plates; when this distance is small,
we don't make too large an error in calculating the motion of the particle if we ignore the edge effect.
Sample Problem 1-1-10 A charged drop of oil of radius R = 2.76 m and density  = 920 kg/m3 is
maintained in equilibrium under the combined influence of its weight and a down-ward uniform electric
field of magnitude E = 1.65 x 106 N/C (Fig. 1-1-13). (a) Calculate the magnitude and sign of the charge on
the drop. Express the result in terms of the elementary charge e. (b) The drop is exposed to a
radioactive source that emits electrons. Two electrons strike the drop and are captured by it, changing
its charge by two units. If the electric field remains at its constant value, calculate the resulting
acceleration of the drop.
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Chapter 1 Electricity
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Figure 1-1-13 Sample problems 10. A negatively charged drop is placed in a uniform electric field E.
Solution (a) To keep the drop in equilibrium, its weight mg must be balanced by an equal electric force of
magnitude qE acting upward. Because the electric field is given as being in the downward direction, the
charge q on the drop must be negative for the electric force to point in a direction opposite the field.
The equilibrium condition is
 F  mg  qE  0
Taking y components, we obtain
-mg +q(-E) =0
or, solving for the unknown q,
mg
4 / 3R3 g
q

E
E
q
4 / 3 (2.76)(106 m)3 (920kg / m3 )(9.8m / s 2 )
= -4.8x10-19C.
6
1.65 x10 N / C
If we write q in terms of the electronic charge - e as q = n(- e), where n is the number of
electronic charges on the drop, then
n
q
 4.8 x10 19 C

3
 e  1.6 x10 19 N / C
(b) If we add two additional electrons to the drop, its charge will become
q' = (n + 2)(-e) = 5(- 1.6 x l0-19 C) = -8.0 x 10-19 C.
Newton's second law can be written
 F  mg  q E  ma
'
and, taking y components, we obtain
-mg+ q'(-E)= ma
We can now solve for the acceleration:
a  g 
q'E
m
13
Chapter 1 Electricity
1-1 Electrostatics
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=9.80 m/s2 - (-8.0x l0-19 C)(1.65 X 106 N/C)
-6
m)3(920 kg/m3)
= -9.80 m/s2 + 16.3 m/s2 = +6.5 m/s2.
The drop accelerates in the positive y direction.
In this calculation, we have ignored the viscous drag force, which is usually quite important in this
situation. We have, in effect, found the acceleration of the drop at the instant it acquired the extra two
electrons. The drag force, which depends on the velocity of the drop, is initially zero if the drop starts
from rest, but it increases as the drop begins to move, and so the acceleration of the drop will decrease in
magnitude.
This experimental configuration forms the basis of the Millikan oil -drop experiment, which was used
to measure the magnitude of the electronic charge. The experiment is discussed later in this section.
1-1-2-c Electrophoresis
With the concept of an electric field we can now understand the idea behind electrophoresis.
Electrophoresis is a sedimentation technique used in DNA analysis and the determination of protein
molecular weights, but one in which the force of gravity (or the centrifugal force induced by a centrifuge)
is replaced by the electric force.
Problems
1. An object with a net charge of 24 C is placed in a uniform electric field of 610 N/C, directed vertically.
What is the mass of the object if it “floats” in the electric field?
2. (a) Determine the electric field strength at a point 1.00 cm to the left of the
middle charge shown in the Figure (b) If a charge of -2.00 C is placed at this point,
what are the magnitude and direction of the force on it?
3. An electron is accelerated by a constant electric field of magnitude 300 N/C. (a) Find the acceleration
of the electron.
(b) Use the equations of motion with constant acceleration to find the electron’s speed after 1.00 x10-8 s,
assuming it starts from rest.
4. A Styrofoam ball covered with a conducting paint has a mass of 5.0 x 10-3 kg and has a charge of 4.0 C.
What electric field directed upward will produce an electric force on the ball that will balance its weight?
5. Two point charges lie along the y -axis. A charge of q1 = 9.0  C is at y = 6.0 m, and a charge of q2 = -8.0
C is at y = -4.0 m. Locate the point (other than infinity) at which the total electric field is zero.
6. In the Figure, determine the point (other than infinity) at which the total electric field is zero.
14
Chapter 1 Electricity
1-1 Electrostatics
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1-1-3 Electric potential
As the name suggests, the potential is associated with a "potential" ability to perform work. For
gravitational force field on the earth surface, where the vertical acceleration is g, the work performed in
lifting a mass m from ground to a height h is equal to gmh. The potential of the level h above ground is equal
to that work divided by the mass, i.e., gh. This potential is obviously defined with respect to the earth
surface and should be properly labeled as the potential difference, rather than just potential. The zero
potential level is actually quite arbitrary and therefore only the potential difference possesses a good
physical meaning.
:
Mgh
W=Mgh
W=qV
+
+
Drift
velocity
V
1
V
E
h
+
M
g
Drift
velocity
V
qE
-
2
W=W1+W2 = qV= qV1+
qV2V=V1+V2
Figure (1-1-14) Motion of mass M against the force of gravity, and the motion of a +ve charge in an electric field against
the force of charge attraction
V (r )  K
q
1  1  10
r
It has SI units of J / C ("Volts"), and the energy of a test charge in the potential field due to one or more
configuration charges is
U t c = q t V c.
(1-1-11)
The electric (or "electrostatic") potential is analogous to temperature: there is a different temperature at
every point in space, and the temperature gradients tell you where heat flows. Similarly, there is an
electric potential at every point in space, and its gradient (the electric field) tells you where charges move.
As with the cases of the electric field and force, the electrical potential field superposes linearly, so that
the field at a point due to a collection of configuration charges is equal to the sum of the fields due to the
individual charges, at that point.
Sample Problem 1-1-11 Two protons in the nucleus of a 238U atom are 6.0 fm apart. What is the
potential energy associated with the electric force that acts between these two particles.
Solution From Eq. 1-11, with q1 = q2 = +1.6 x 10-19C, we obtain
15
Chapter 1 Electricity
1-1 Electrostatics
___________________________________________________________________________________
U (r )  K
q1q2 (8.99  109 N  m2 / C 2 )(1.60  1019 C ) 2

r
6.0  1015 m
 3.8  1014 J  2.4  105 eV  240keV .
The two protons do not fly apart because they are held together by the attractive strong force that binds
the nucleus together. Unlike the electric force, there is no simple potential energy function that represents
the strong force.
The SI unit of potential that follows from Eq. 9 is the joule/coulomb. This combination occurs so often that
a special unit, the volt (abbreviation V), is used to represent it; that is,
1 volt = 1 joule/coulomb.
The common name of "voltage" is often used for the potential at a point or the potential difference
between points. When you touch the two probes of a voltmeter to two points in an electric circuit, you are
measuring the potential difference (in volts) or voltage between those points.
Charge Acceleration.
In biology research science, the study of cell nuclei may be of scientist interest. One way of such technique
is to insert foreign particles inside a cell nucleus to change its structure in order to observe its new
behavior. These particles must be accelerated and directed towards such cells. Van de Graaff’s
Accelerators are devices which are used in this technique.
This machine is used in nuclear reaction experiment. Ions accelerated in a large potential difference
impinge on a target. The large potential difference is maintained by using a suitable set of battery, which
provides a huge amount of positive charges which are then transferred to a dome through an insulated belt
in a vacuum vessel. The accelerated ions gain kinetic energy, K.E. = potential energy, P.E = charge, Q,
multiplied by the potential difference, V; K.E. = ½ mv2 = P.
Figure 1-17 Van de Graaff`s Accelerator
16
Chapter 1 Electricity
1-1 Electrostatics
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Problems
1. A proton moves 2.00 cm parallel to a uniform electric field of E = 200 N/C. (a) How much work is done
by the field on the proton? (b) What change occurs in the potential energy of the proton? (c) What
potential difference did the proton move through?
2. A uniform electric field of magnitude 250 V/m is directed in the positive x-direction. A 12 C charge
moves from the origin to the point (x) = 20 cm. (a) What was the change in the potential energy of this
charge? (b) Through what potential difference did the charge move?
3-A potential difference of 90 mV exists between the inner and outer surfaces of a cell membrane. The
inner surface is negative relative to the outer surface. How much work is required to eject a positive
sodium ion (Na+) from the interior of the cell?
4. An ion accelerated through a potential difference of 60.0 V has its potential energy decreased by 1.92x
10-17 J. Calculate the charge on the ion.
5. The potential difference between the accelerating plates of a TV set is about 25 kV. If the distance
between the plates is 1.5 cm, find the magnitude of the uniform electric field in the region between the
plates.
6. To recharge a 12-V battery, a battery charger must move 3.6 x 105 C of charge from the negative
terminal to the positive terminal. How much work is done by the charger? Express your answer in joules.
7-Oppositely charged parallel plates are separated by 5.33 mm. A potential difference of 600 V exists
between the plates. (a) What is the magnitude of the electric field between the plates? (b) What is the
magnitude of the force on an electron between the plates? (c) How much work must be done on the
electron to move it to the negative plate if it is initially positioned 2.90 mm from the positive plate?
8. Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V.
(b) Calculate the speed of an electron that is accelerated through the same potential difference.
9. (a) Find the electric potential 1.00 cm from a proton. (b) What is the electric potential difference
between two points that are 1.00 cm and 2.00 cm from a proton?
10. Two point charges are on the y -axis, one of magnitude 3.0 x 10-9 C at the origin and a second of
magnitude 6.0 x 10-9 C at the point y = 30 cm. Calculate the potential at y = 60 cm.
11. A small spherical object carries a charge of 8.00 nC. At what distance from the center of the object is
the potential equal to 100 V? 50.0 V? 25.0 V? Is the spacing of the equipotentials proportional to the
change in voltage?
12. Two point charges Q1 = +5.00 nC and Q2 = -3.00 nC are separated by 35.0 cm.
(a) What is the electric potential at a point midway between the charges? (b) What
is the potential energy of the pair of charges?
16. A point charge of 9.00 = 10-9 C is located at the origin. How much work is
required to bring a positive charge of 3.00 = 10-9 C from infinity to the location x =
30.0 cm?
13. The three charges in the Figure are at the vertices of an isosceles triangle. Let
q = 7.00 nC, and calculate the electric potential at the midpoint of the base.
17