CHAPTER 1:

ELECTRIC CHARGE AND ELECTRIC FIELDS

  i)     Electrostatic
  ii)    conductors and insulators
  iii)   Coulomb’s Law
  iv)    Electric Fields
  v)     Electric Fields Calculation
  vi)    Electric Field Lines
  vii)   Electric Dipole


1.1 Electrostatic

Electrostatic is a study on the electric charges in the
static or steady state condition. In this chapter, we will
discuss the basic and the fundamental concept of electric
charges, electric fields and their characteristics.

Plastic rods and fur are good for demonstrating
electrostatics.

Benjamin Franklin (1706-1790) suggested charges
negative and positive.
“Two positive charges or two negative charges repel
  each other. A positive charge and a negative charge
                  attract each other”.

Caution: Electric attraction and repulsion
The attraction and repulsion of two charged object are
sometimes summarized as “like charges repel, and
opposite charges attract”

But keep in mind that the phase “like charge” does not
mean that the two charges are exactly identical, only that
both charges have the same algebraic sign (both positive
or both negative). “Opposite charges” mean that both
objects have an electric charge, and those charges have
different sign (one positive and the other negative).
1.1.1 Electric charge and the structure of matter




The structure of atoms can be described in terms of
three particles: the negatively charged electron, the
positively charged proton, and uncharged neutron (Fig
above).

Proton and neutron in an atom make up a small, very
dense core called the nucleus. (10-15 m)

Surrounding the nucleus are the electrons
(10 -10 m a far from nucleus).
The negative charge of the electron has exactly the
magnitude as the positive charge of proton.

1.1.2 Electric Charge is Conserved

Principle of conservation charge:
  i)     The algebraic sum of all electric charges in any
         closed system is constant
  ii)    The magnitude of charge of the electron or
         proton is a natural unit of charge.

In any charging process, charge is not created or
destroyed: it is merely transferred from one body to
another.

“The electric charge is quantized. (1, 2, 3, 4…)”
1.2   Conductors and Insulators


                           Materials that allow easy
                           passage of charge are called
                           conductors. Materials that
                           resist electronic flow are
                           called insulators.

                           The motion of electrons
                           through conducts and
                           about insulators allows us
                           to observe “opposite
                           charges attract” and “like
                           charges repel.”



Charging by induction
1.3 Coulomb’s Law
Charles Augustin de Coulomb (1736-1806) studied the
interaction forces of charged particles in detail in 1784.
Point charges
Coulomb found that
  i) The electric force is proportional to   ଶ


                                ଶ
  ii)   The electric force between two point charges
        depends on the quantity of charge on each body,
        which we will denote by q or Q. ( positive or
        negative)

                     +              +

                      -             -

                     +              -

                     -              +
                            r

  iii) The forces that two point charges ଵ and ଶ exert
       on each other are proportional to each charge and
       therefore are proportional to the product ଵ ଶ of
       the two charges
                                 ଵ ଶ
Coulomb’s law state that;

The magnitude of the electric 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.

In mathematical term, the magnitude F of the force that
each of two point charge ଵ and ଶ a distance r apart
exerts on the other can be expressed as;

                                 ૚ ૛
                                  ૛


where k is a constant.

1.3.1 Electric Constants, k

In SI units the constant, k is               where ( ଴ - epsilon
                                         ଴
nought or opsilon zero)


                                     ଽ           ଶ       ଶ


By approximation
                                 ଽ           ଶ       ଶ
Magnitude of the charge of an electron or proton, e

                                     ିଵଽ


One Coulomb represents the negative of the total charge
              ଵ଼
of about         electron.

So that, the electric force is given as
                                   ૚ ૛
                                    ૛
                               ૙

Superposition of Forces: holds for any number of
charges. We can apply Coulomb’s law to any collection of
charges.
Example

Two point charges ଵ               and ଶ              , are
separated by a distance of 3.0 cm. Find the magnitude
and direction of
  (i)    the electric force that ଵ exerts on ଶ , and
  (ii)   the electric force that ଶ exerts on ଵ .




Solution
  a) This problem asks for the electric forces that two
    charges exert on each other, so we will need to use
    Coulomb’s law. After we convert charge to coulombs
    and distance to meters, the magnitude of force that
      ଵ exerts on ଶ is


                   ଵ ଶ
 ଵ௢௡ଶ             ଶ
               ଴
                                  ିଽ             ିଽ
           ଽ       ଶ ିଶ
                                          ଶ
Since the two charges have opposite signs, the force is
attractive; that is the force that acts on ଶ is directed
toward ଵ along the line joining the two charges.

  b) Newton’s third law applies to the electric force. Even
     though the charges have different magnitude, the
     magnitude of the force that ଶ exerts on ଵ is the
     same the magnitude of the force that ଵ exerts
     on ଶ . So that
                         ଶ௢௡ଵ


Example

Two point charges are located on the positive x-axis of a
coordinate system. Charge q1 = 1.0 nC is 2.0 cm from the
origin, and charge q2 = -3.0 nC is 4.0 cm from the origin.
What is the total force exerted by these two charges on a
charge q3 = 5.0 nC located at the origin?




Solution
Find the magnitude of   ଵ௢௡ଷ
ଵ ଶ
 ଵ௢௡ଷ              ଶ
                ଴
                                 ିଽ                ିଽ
            ଽ           ଶ ିଶ
                                           ଶ
             ିସ

*(this force has a negative x-component because q3 is
repelled by q1)


Then Find the magnitude of       ଶ௢௡ଷ
                    ଵ ଶ
 ଶ௢௡ଷ              ଶ
                ଴
                                   ିଽ                   ିଽ
            ଽ           ଶ ିଶ
                                               ଶ
            ିହ

*(this force has a positive x-component because q3 is
attracted by q2)

So the sum of x-component is
                    ௫
There are no y or z- components. Thus the total force on q3 is
directed to the left, with magnitude     .
1.4 Electric Fields

We defined the electric field at point as the electric
force experienced by a test charge q0 at the point,
divided by the charge q0.



                               ଴


The direction of      and   is the same.
Electric field of a point charge

   Consider we have a charge q as a point source. If we
   place a small test charge q0 at the field point, P at a
   distance r from the point source, the magnitude F0 of
   the force is given by Coulomb’s law
                                     ଵ   ȁ௤௤బ ȁ
                            ଴       ସగఢబ ௥ మ


so that, the magnitude of electric field, E is

                                               ଶ
                                          ଴


But the direction of and is the same. Then the
electric field vector is given as,

                                               ଶ
                                         ଴
 is a vector unit in r direction.
Example 1: Electric-field magnitude for a point charge
What is the magnitude of electric field at a field point 2.0
m from a point charge q = 4.0 nC ?

Solution
We are given the magnitude of charge and the distance
                                                      ͳ   ȁ‫ݍ‬ȁ
from the object to the field point, so by using
                                                    Ͷߨ߳Ͳ ‫ʹݎ‬
we could calculate the magnitude of


                                      ଶ
                                 ଴
ିଽ
            ଽ       ଶ ିଶ
                                     ଶ




Example 2: Electric Field Vector for a point charge.

A point charge q = -8.0 nC is located at the origin. Find
the electric-field vector at the field point x = 1.2 m,
y = -1.6 m?

Solution




The vector of field point P is

The distance from the charge at point source, S to the
field point, P is
                ଶ     ଶ          ଶ        ଶ

The vector unit,
Ԧ
      ௥       ଵǤଶపƸିଵǤ଺ఫƸ
     ȁ௥ ȁ          ଶǤ଴
Hence the electric-field vector is
      ͳ ‫ݍ‬
ሬԦ
‫ܧ‬ൌ            ‫ݎ‬Ƹ
     Ͷߨ߳଴ ‫ ݎ‬ଶ
                   ଽ     ଶ ିଶ ሻ
                                  െͺǤͲ ൈ ͳͲିଽ ‫ܥ‬
ൌ  ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬                          ሺͲǤ͸ଓƸ െ ͲǤͺଔƸሻ
                                     ሺʹǤͲሻଶ
            ܰ        ܰ
ൌ െͳͳ     ଓƸ ൅ ͳͶ ଔƸ
            ‫ܥ‬        ‫ܥ‬

Example 3: Electron in a uniform field

When the terminals of a battery are connected to two
large parallel conducting plates, the resulting charges on
                                  ሬԦ
the plate cause an electric field ‫ ܧ‬in the region between
the plates that is very uniform.




If the plate are horizontal and separated by 1.0 cm and
the plate are connected to 100 V battery, the magnitude
of the field is E = 1.00 x 104 N/C. Suppose the is
vertically upward,
a) If an electron released from rest at the upper plate,
     what is its acceleration?
  b) What speed does the electron acquire while
     traveling 1.0 cm to lower plate?
                                             ିଵଽ
Given electron charge is                          and
                        ିଷଵ
mass ௘

Solution:
  a) Noted that         is upward but is downward because the
       charge of electron is negative. Thus Fy is negative. Because
       Fy is constant, the electron moves with constant
       acceleration ay given by,
                                                               ସ
                                       ିଵଽ
            ௬
   ௬                                              ିଷଵ
             ௘       ௘
                               ଵହ     ଶ

  b) The electron starts from rest, so its motion is in the y
     –direction only. We can find the electron’s speed at
     any position using constant-acceleration
               ଶ     ଶ                               ଶ
     formula ௬       ଴௬      ௬        ଴ . We have ଴௬
     and y0 = 0 so speed ௬ when y = -1.0 cm.

                                            ݉
       ห‫ݒ‬௬ ห ൌ ටʹܽ௬ ‫ ݕ‬ൌ ටʹ ቀെͳǤ͹͸ ൈ ͳͲଵହ      ቁ ሺെͳǤͲ ൈ ͳͲିଶ ݉ሻ
                                            ‫ݏ‬ଶ
                 ൌ ͷǤͻ ൈ ͳͲ଺ ݉Ȁ‫ݏ‬
Example 4: An electron trajectory




If we launch an electron into the electric field of Example
3 with an initial horizontal velocity v0, what is the
equation of its trajectory?

Solution
The acceleration is constant and in the y-direction. Hence
we can use the kinematic equation for 2-dimensional
motion with constant acceleration.

                  ଵ       ଶ                             ଵ       ଶ
      ଴     ଴௫        ௫       and          ଴       ଴௬       ௬
                  ଶ                                     ଶ

We have ax=0 and ay = (-e)E /m . at t =0 , x0 =y0=0, v0x = v0
and v0y=0, hence at time t,

                  ଵ       ଶ         ଵ ௘ா ଶ
      ଴   and         ௬
                  ଶ                 ଶ௠


Eliminating t between these equations, we get
                                               ଶ
                                       ଶ
                                       ଴
1.5 Electric Fields Calculation

In real situations, we encounter charge that is distributed
over space. To find the field caused by a distribution, we
imagine the distribution to be made up of many point
charges, q1,q2,q3…….qn. At any given point P, each point
charge produces its own electric field ଵ ଶ ଷ             ௡,
so a test charge q0 placed at P experiences a force ଵ   ଴ ଵ
from charge q1 and a force ଶ        ଴ ଶ from charge q2 and so
on.
From the principle of superposition of forces, the total
forces ଴ that the charge distribution exerts on the q0 is the
vector sum of these individual forces,
       ଴     ଵ     ଶ     ଷ          ଴ ଵ         ଴ ଶ   ଴ ଷ
Then the total electric field at point P,
                         ଴
                               ଵ     ଶ      ଷ
                         ଴
Example 1:
Point charge q1 and q2 of +12nC and -12nC respectively,
are placed 0.10 m apart. This combination of two charges
with equal magnitude and opposite sign is called an
electric dipole. Compute the electric field caused by q1,
the field caused by q2, and total field (a) point a, (b) at
point b, and (c) at point c.




Solution
    a) At point a: the electric field, ଵ caused by the
       positive charge q1 and the field ଶ caused by the
       negative charge q2 are both directed toward the
       right. The magnitude of ଵ and ଶ are;

            ͳ ȁ‫ݍ‬ଵ ȁ               ଽ   ଶ ିଶ ሻ
                                             ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ
    ሬሬሬሬԦ
   ห‫ܧ‬ଵ ห ൌ          ൌൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬
           Ͷߨ߳଴ ‫ ݎ‬ଶ                            ሺͲǤͲ͸݉ሻଶ
ൌ ͵ǤͲ ൈ ͳͲସ ܰȀ‫ܥ‬

        ͳ ȁ‫ݍ‬ଶ ȁ                ଽ   ଶ ିଶ ሻ
                                          ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ
 ሬሬሬሬԦ
ห‫ܧ‬ଶ ห ൌ          ൌൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬
      Ͷߨ߳଴ ‫ ݎ‬ଶ                              ሺͲǤͲͶ݉ሻଶ
  ൌ ͸Ǥͺ ൈ ͳͲସ ܰȀ‫ܥ‬


The component of        ଵ   and        ଶ   are;
                                   ே
               ‫ܧ‬ଵ௫ ൌ ͵ǤͲ ൈ ͳͲସ           and ‫ܧ‬ଵ௬ ൌ Ͳ
                                   ஼
                                  ସே
               ‫ܧ‬ଶ௫ ൌ ͸Ǥͺ ൈ ͳͲ            and ‫ܧ‬ଶ௬ ൌ Ͳ
                                   ஼
Hence at point a the total electric field
 ௔     ଵ     ଶ has components.

                                                   Ͷܰ
  ௔௫      ଵ௫      ଶ௫                                ‫ܥ‬

  ௔௬      ଵ௬      ଶ௬
                                                            ସே
At point a the total field has magnitude 9.8                 ஼
and is directed toward the right.
                 Ͷܰ
  ௔               ‫ܥ‬


b) At point a: the electric field, ଵ caused by the
   positive charge q1 is directed toward left and the
   field ଶ caused by q2 is directed toward the right.
   The magnitude of ଵ and ଶ are;
           ͳ ȁ‫ݍ‬ଵ ȁ               ଽ   ଶ ିଶ ሻ
                                            ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ
   ሬሬሬሬԦ
  ห‫ܧ‬ଵ ห ൌ          ൌൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬
          Ͷߨ߳଴ ‫ ݎ‬ଶ                            ሺͲǤͲͶ݉ሻଶ
  ൌ ͸Ǥͺ ൈ ͳͲସ ܰȀ‫ܥ‬
ͳ ȁ‫ݍ‬ଶ ȁ               ଽ   ଶ ିଶ ሻ
                                            ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ
   ሬሬሬሬԦ
  ห‫ܧ‬ଶ ห ൌ          ൌൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬
          Ͷߨ߳଴ ‫ ݎ‬ଶ                            ሺͲǤͳͶ݉ሻଶ
  ൌ ͲǤͷͷ ൈ ͳͲସ ܰȀ‫ܥ‬


  The component of              ଵ   and    ଶ   are
                                       ே
               ‫ܧ‬ଵ௫ ൌ  െ͸Ǥͺ ൈ ͳͲସ           and ‫ܧ‬ଵ௬ ൌ Ͳ
                                       ஼
                                     ସ ே
                ‫ܧ‬ଶ௫ ൌ ͲǤͷͷ ൈ ͳͲ           and ‫ܧ‬ଶ௬ ൌ Ͳ
                                      ஼
Hence at point a the total electric field
‫ܧ‬௕ ൌ ሬሬሬሬԦ  ൅ ‫ܧ‬ଶ has components
ሬሬሬሬԦ ‫ܧ‬ଵ ሬሬሬሬԦ

                                          ே
‫ܧ‬௕௫ ൌ ‫ܧ‬ଵ௫ ൅ ‫ܧ‬ଶ௫ ൌ ሺെ͸Ǥͺ ൅ ͲǤͷͷሻ  ൈ ͳͲସ
ሬԦ
                                          ஼
ሬԦ௕௬ ൌ ‫ܧ‬ଵ௬ ൅ ‫ܧ‬ଶ௬ ൌ Ͳ ൅ Ͳ
‫ܧ‬

                                                               ସே
At point b the total field has magnitude 6.2                    ஼
and is directed toward the right.
                     Ͷܰ
 ௕                    ‫ܥ‬


c) At point c, both ଵ and ଶ have same magnitude,
   since this point is equidistant from both charges
   and charge magnitude are the same;
                     ͳ ȁ‫ݍ‬ଵ ȁ
   ሬሬሬሬԦ   ሬሬሬሬԦ
  ห‫ܧ‬ଵ ห ൌ ห‫ܧ‬ଶ ห ൌ            ൌ
                    Ͷߨ߳଴ ‫ ݎ‬ଶ
                            ଽ       ଶ ିଶ ሻ
                                             ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ
              ൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬
                                               ሺͲǤͳ͵݉ሻଶ
  ൌ ͸Ǥ͵ͻ ൈ ͳͲସ ܰȀ‫ܥ‬
The direction of ଵ and ଶ are shown in Figure.
      The x-component of the both vector as the same

                                   ே                 ே   ହ
‫ܧ‬ଵ௫ ൌ ‫ܧ‬ଶ௫ ൌ ‫ܧ‬ଵ …‘• ߙ ൌ ͸Ǥ͵Ͳ ൈ ͳͲଷ …‘• ߙ ൌ ͸Ǥ͵Ͳ ൈ ͳͲଷ ቀ ቁ ൌ
                                   ஼                 ஼   ଵଷ
              ே
ʹǤͶ͸ ൈ   ͳͲସ
              ஼
    From symmetry the y-component are equal and
    opposite direction so add to zero.
                                    ே            ே
    ‫ܧ‬௖௫ ൌ ‫ܧ‬ଵ௫ ൅ ‫ܧ‬ଶ௫ ൌ ʹሺʹǤͶ͸ሻ ൈ ͳͲଷ ൌ ͶǤͻ ൈ ͳͲଷ
    ሬԦ
                                    ஼            ஼
    ሬԦ௖௬ ൌ ‫ܧ‬ଵ௬ ൅ ‫ܧ‬ଶ௬ ൌ Ͳ
    ‫ܧ‬


    So at point c the total electric field has magnitude
                  ே
    ͶǤͻ ൈ ͳͲଷ and its direction toward the right.
                  ஼
                      ͵ܰ
      ௖                ‫ܥ‬


1.6 Electric Field Lines

Electric field lines can be a big help for visualizing electric
fields and making them seem more real. An electric field
line is an imaginary line or curve drawn through a region
of space so that its tangent at any point is in the
direction of the electric field vector at that point.
Figure below shows some of the electric field lines in a
plane (a) a single positive charge, (b) two equal-
magnitude charges, one positive and one negative
(dipole), (c) two equal positive charges.

Diagram called Field Map.

Chapter1: Coulomb's Law

  • 1.
    CHAPTER 1: ELECTRIC CHARGEAND ELECTRIC FIELDS i) Electrostatic ii) conductors and insulators iii) Coulomb’s Law iv) Electric Fields v) Electric Fields Calculation vi) Electric Field Lines vii) Electric Dipole 1.1 Electrostatic Electrostatic is a study on the electric charges in the static or steady state condition. In this chapter, we will discuss the basic and the fundamental concept of electric charges, electric fields and their characteristics. Plastic rods and fur are good for demonstrating electrostatics. Benjamin Franklin (1706-1790) suggested charges negative and positive.
  • 2.
    “Two positive chargesor two negative charges repel each other. A positive charge and a negative charge attract each other”. Caution: Electric attraction and repulsion The attraction and repulsion of two charged object are sometimes summarized as “like charges repel, and opposite charges attract” But keep in mind that the phase “like charge” does not mean that the two charges are exactly identical, only that both charges have the same algebraic sign (both positive or both negative). “Opposite charges” mean that both objects have an electric charge, and those charges have different sign (one positive and the other negative).
  • 3.
    1.1.1 Electric chargeand the structure of matter The structure of atoms can be described in terms of three particles: the negatively charged electron, the positively charged proton, and uncharged neutron (Fig above). Proton and neutron in an atom make up a small, very dense core called the nucleus. (10-15 m) Surrounding the nucleus are the electrons (10 -10 m a far from nucleus).
  • 4.
    The negative chargeof the electron has exactly the magnitude as the positive charge of proton. 1.1.2 Electric Charge is Conserved Principle of conservation charge: i) The algebraic sum of all electric charges in any closed system is constant ii) The magnitude of charge of the electron or proton is a natural unit of charge. In any charging process, charge is not created or destroyed: it is merely transferred from one body to another. “The electric charge is quantized. (1, 2, 3, 4…)”
  • 5.
    1.2 Conductors and Insulators Materials that allow easy passage of charge are called conductors. Materials that resist electronic flow are called insulators. The motion of electrons through conducts and about insulators allows us to observe “opposite charges attract” and “like charges repel.” Charging by induction
  • 6.
    1.3 Coulomb’s Law CharlesAugustin de Coulomb (1736-1806) studied the interaction forces of charged particles in detail in 1784.
  • 7.
    Point charges Coulomb foundthat i) The electric force is proportional to ଶ ଶ ii) The electric force between two point charges depends on the quantity of charge on each body, which we will denote by q or Q. ( positive or negative) + + - - + - - + r iii) The forces that two point charges ଵ and ଶ exert on each other are proportional to each charge and therefore are proportional to the product ଵ ଶ of the two charges ଵ ଶ
  • 8.
    Coulomb’s law statethat; The magnitude of the electric 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. In mathematical term, the magnitude F of the force that each of two point charge ଵ and ଶ a distance r apart exerts on the other can be expressed as; ૚ ૛ ૛ where k is a constant. 1.3.1 Electric Constants, k In SI units the constant, k is where ( ଴ - epsilon ଴ nought or opsilon zero) ଽ ଶ ଶ By approximation ଽ ଶ ଶ
  • 9.
    Magnitude of thecharge of an electron or proton, e ିଵଽ One Coulomb represents the negative of the total charge ଵ଼ of about electron. So that, the electric force is given as ૚ ૛ ૛ ૙ Superposition of Forces: holds for any number of charges. We can apply Coulomb’s law to any collection of charges.
  • 10.
    Example Two point chargesଵ and ଶ , are separated by a distance of 3.0 cm. Find the magnitude and direction of (i) the electric force that ଵ exerts on ଶ , and (ii) the electric force that ଶ exerts on ଵ . Solution a) This problem asks for the electric forces that two charges exert on each other, so we will need to use Coulomb’s law. After we convert charge to coulombs and distance to meters, the magnitude of force that ଵ exerts on ଶ is ଵ ଶ ଵ௢௡ଶ ଶ ଴ ିଽ ିଽ ଽ ଶ ିଶ ଶ
  • 11.
    Since the twocharges have opposite signs, the force is attractive; that is the force that acts on ଶ is directed toward ଵ along the line joining the two charges. b) Newton’s third law applies to the electric force. Even though the charges have different magnitude, the magnitude of the force that ଶ exerts on ଵ is the same the magnitude of the force that ଵ exerts on ଶ . So that ଶ௢௡ଵ Example Two point charges are located on the positive x-axis of a coordinate system. Charge q1 = 1.0 nC is 2.0 cm from the origin, and charge q2 = -3.0 nC is 4.0 cm from the origin. What is the total force exerted by these two charges on a charge q3 = 5.0 nC located at the origin? Solution Find the magnitude of ଵ௢௡ଷ
  • 12.
    ଵ ଶ ଵ௢௡ଷ ଶ ଴ ିଽ ିଽ ଽ ଶ ିଶ ଶ ିସ *(this force has a negative x-component because q3 is repelled by q1) Then Find the magnitude of ଶ௢௡ଷ ଵ ଶ ଶ௢௡ଷ ଶ ଴ ିଽ ିଽ ଽ ଶ ିଶ ଶ ିହ *(this force has a positive x-component because q3 is attracted by q2) So the sum of x-component is ௫ There are no y or z- components. Thus the total force on q3 is directed to the left, with magnitude .
  • 13.
    1.4 Electric Fields Wedefined the electric field at point as the electric force experienced by a test charge q0 at the point, divided by the charge q0. ଴ The direction of and is the same.
  • 14.
    Electric field ofa point charge Consider we have a charge q as a point source. If we place a small test charge q0 at the field point, P at a distance r from the point source, the magnitude F0 of the force is given by Coulomb’s law ଵ ȁ௤௤బ ȁ ଴ ସగఢబ ௥ మ so that, the magnitude of electric field, E is ଶ ଴ But the direction of and is the same. Then the electric field vector is given as, ଶ ଴ is a vector unit in r direction.
  • 15.
    Example 1: Electric-fieldmagnitude for a point charge What is the magnitude of electric field at a field point 2.0 m from a point charge q = 4.0 nC ? Solution We are given the magnitude of charge and the distance ͳ ȁ‫ݍ‬ȁ from the object to the field point, so by using Ͷߨ߳Ͳ ‫ʹݎ‬ we could calculate the magnitude of ଶ ଴
  • 16.
    ିଽ ଽ ଶ ିଶ ଶ Example 2: Electric Field Vector for a point charge. A point charge q = -8.0 nC is located at the origin. Find the electric-field vector at the field point x = 1.2 m, y = -1.6 m? Solution The vector of field point P is The distance from the charge at point source, S to the field point, P is ଶ ଶ ଶ ଶ The vector unit,
  • 17.
    Ԧ ௥ ଵǤଶపƸିଵǤ଺ఫƸ ȁ௥ ȁ ଶǤ଴ Hence the electric-field vector is ͳ ‫ݍ‬ ሬԦ ‫ܧ‬ൌ ‫ݎ‬Ƹ Ͷߨ߳଴ ‫ ݎ‬ଶ ଽ ଶ ିଶ ሻ െͺǤͲ ൈ ͳͲିଽ ‫ܥ‬ ൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬ ሺͲǤ͸ଓƸ െ ͲǤͺଔƸሻ ሺʹǤͲሻଶ ܰ ܰ ൌ െͳͳ ଓƸ ൅ ͳͶ ଔƸ ‫ܥ‬ ‫ܥ‬ Example 3: Electron in a uniform field When the terminals of a battery are connected to two large parallel conducting plates, the resulting charges on ሬԦ the plate cause an electric field ‫ ܧ‬in the region between the plates that is very uniform. If the plate are horizontal and separated by 1.0 cm and the plate are connected to 100 V battery, the magnitude of the field is E = 1.00 x 104 N/C. Suppose the is vertically upward,
  • 18.
    a) If anelectron released from rest at the upper plate, what is its acceleration? b) What speed does the electron acquire while traveling 1.0 cm to lower plate? ିଵଽ Given electron charge is and ିଷଵ mass ௘ Solution: a) Noted that is upward but is downward because the charge of electron is negative. Thus Fy is negative. Because Fy is constant, the electron moves with constant acceleration ay given by, ସ ିଵଽ ௬ ௬ ିଷଵ ௘ ௘ ଵହ ଶ b) The electron starts from rest, so its motion is in the y –direction only. We can find the electron’s speed at any position using constant-acceleration ଶ ଶ ଶ formula ௬ ଴௬ ௬ ଴ . We have ଴௬ and y0 = 0 so speed ௬ when y = -1.0 cm. ݉ ห‫ݒ‬௬ ห ൌ ටʹܽ௬ ‫ ݕ‬ൌ ටʹ ቀെͳǤ͹͸ ൈ ͳͲଵହ ቁ ሺെͳǤͲ ൈ ͳͲିଶ ݉ሻ ‫ݏ‬ଶ ൌ ͷǤͻ ൈ ͳͲ଺ ݉Ȁ‫ݏ‬
  • 19.
    Example 4: Anelectron trajectory If we launch an electron into the electric field of Example 3 with an initial horizontal velocity v0, what is the equation of its trajectory? Solution The acceleration is constant and in the y-direction. Hence we can use the kinematic equation for 2-dimensional motion with constant acceleration. ଵ ଶ ଵ ଶ ଴ ଴௫ ௫ and ଴ ଴௬ ௬ ଶ ଶ We have ax=0 and ay = (-e)E /m . at t =0 , x0 =y0=0, v0x = v0 and v0y=0, hence at time t, ଵ ଶ ଵ ௘ா ଶ ଴ and ௬ ଶ ଶ௠ Eliminating t between these equations, we get ଶ ଶ ଴
  • 20.
    1.5 Electric FieldsCalculation In real situations, we encounter charge that is distributed over space. To find the field caused by a distribution, we imagine the distribution to be made up of many point charges, q1,q2,q3…….qn. At any given point P, each point charge produces its own electric field ଵ ଶ ଷ ௡, so a test charge q0 placed at P experiences a force ଵ ଴ ଵ from charge q1 and a force ଶ ଴ ଶ from charge q2 and so on. From the principle of superposition of forces, the total forces ଴ that the charge distribution exerts on the q0 is the vector sum of these individual forces, ଴ ଵ ଶ ଷ ଴ ଵ ଴ ଶ ଴ ଷ Then the total electric field at point P, ଴ ଵ ଶ ଷ ଴
  • 21.
    Example 1: Point chargeq1 and q2 of +12nC and -12nC respectively, are placed 0.10 m apart. This combination of two charges with equal magnitude and opposite sign is called an electric dipole. Compute the electric field caused by q1, the field caused by q2, and total field (a) point a, (b) at point b, and (c) at point c. Solution a) At point a: the electric field, ଵ caused by the positive charge q1 and the field ଶ caused by the negative charge q2 are both directed toward the right. The magnitude of ଵ and ଶ are; ͳ ȁ‫ݍ‬ଵ ȁ ଽ ଶ ିଶ ሻ ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ ሬሬሬሬԦ ห‫ܧ‬ଵ ห ൌ ൌൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬ Ͷߨ߳଴ ‫ ݎ‬ଶ ሺͲǤͲ͸݉ሻଶ
  • 22.
    ൌ ͵ǤͲ ൈͳͲସ ܰȀ‫ܥ‬ ͳ ȁ‫ݍ‬ଶ ȁ ଽ ଶ ିଶ ሻ ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ ሬሬሬሬԦ ห‫ܧ‬ଶ ห ൌ ൌൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬ Ͷߨ߳଴ ‫ ݎ‬ଶ ሺͲǤͲͶ݉ሻଶ ൌ ͸Ǥͺ ൈ ͳͲସ ܰȀ‫ܥ‬ The component of ଵ and ଶ are; ே ‫ܧ‬ଵ௫ ൌ ͵ǤͲ ൈ ͳͲସ and ‫ܧ‬ଵ௬ ൌ Ͳ ஼ ସே ‫ܧ‬ଶ௫ ൌ ͸Ǥͺ ൈ ͳͲ and ‫ܧ‬ଶ௬ ൌ Ͳ ஼ Hence at point a the total electric field ௔ ଵ ଶ has components. Ͷܰ ௔௫ ଵ௫ ଶ௫ ‫ܥ‬ ௔௬ ଵ௬ ଶ௬ ସே At point a the total field has magnitude 9.8 ஼ and is directed toward the right. Ͷܰ ௔ ‫ܥ‬ b) At point a: the electric field, ଵ caused by the positive charge q1 is directed toward left and the field ଶ caused by q2 is directed toward the right. The magnitude of ଵ and ଶ are; ͳ ȁ‫ݍ‬ଵ ȁ ଽ ଶ ିଶ ሻ ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ ሬሬሬሬԦ ห‫ܧ‬ଵ ห ൌ ൌൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬ Ͷߨ߳଴ ‫ ݎ‬ଶ ሺͲǤͲͶ݉ሻଶ ൌ ͸Ǥͺ ൈ ͳͲସ ܰȀ‫ܥ‬
  • 23.
    ͳ ȁ‫ݍ‬ଶ ȁ ଽ ଶ ିଶ ሻ ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ ሬሬሬሬԦ ห‫ܧ‬ଶ ห ൌ ൌൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬ Ͷߨ߳଴ ‫ ݎ‬ଶ ሺͲǤͳͶ݉ሻଶ ൌ ͲǤͷͷ ൈ ͳͲସ ܰȀ‫ܥ‬ The component of ଵ and ଶ are ே ‫ܧ‬ଵ௫ ൌ െ͸Ǥͺ ൈ ͳͲସ and ‫ܧ‬ଵ௬ ൌ Ͳ ஼ ସ ே ‫ܧ‬ଶ௫ ൌ ͲǤͷͷ ൈ ͳͲ and ‫ܧ‬ଶ௬ ൌ Ͳ ஼ Hence at point a the total electric field ‫ܧ‬௕ ൌ ሬሬሬሬԦ ൅ ‫ܧ‬ଶ has components ሬሬሬሬԦ ‫ܧ‬ଵ ሬሬሬሬԦ ே ‫ܧ‬௕௫ ൌ ‫ܧ‬ଵ௫ ൅ ‫ܧ‬ଶ௫ ൌ ሺെ͸Ǥͺ ൅ ͲǤͷͷሻ ൈ ͳͲସ ሬԦ ஼ ሬԦ௕௬ ൌ ‫ܧ‬ଵ௬ ൅ ‫ܧ‬ଶ௬ ൌ Ͳ ൅ Ͳ ‫ܧ‬ ସே At point b the total field has magnitude 6.2 ஼ and is directed toward the right. Ͷܰ ௕ ‫ܥ‬ c) At point c, both ଵ and ଶ have same magnitude, since this point is equidistant from both charges and charge magnitude are the same; ͳ ȁ‫ݍ‬ଵ ȁ ሬሬሬሬԦ ሬሬሬሬԦ ห‫ܧ‬ଵ ห ൌ ห‫ܧ‬ଶ ห ൌ ൌ Ͷߨ߳଴ ‫ ݎ‬ଶ ଽ ଶ ିଶ ሻ ȁͳʹ ൈ ͳͲିଽ ‫ ܥ‬ȁ ൌ ሺͻǤͲ ൈ ͳͲ ܰ݉ ‫ܥ‬ ሺͲǤͳ͵݉ሻଶ ൌ ͸Ǥ͵ͻ ൈ ͳͲସ ܰȀ‫ܥ‬
  • 24.
    The direction ofଵ and ଶ are shown in Figure. The x-component of the both vector as the same ே ே ହ ‫ܧ‬ଵ௫ ൌ ‫ܧ‬ଶ௫ ൌ ‫ܧ‬ଵ …‘• ߙ ൌ ͸Ǥ͵Ͳ ൈ ͳͲଷ …‘• ߙ ൌ ͸Ǥ͵Ͳ ൈ ͳͲଷ ቀ ቁ ൌ ஼ ஼ ଵଷ ே ʹǤͶ͸ ൈ ͳͲସ ஼ From symmetry the y-component are equal and opposite direction so add to zero. ே ே ‫ܧ‬௖௫ ൌ ‫ܧ‬ଵ௫ ൅ ‫ܧ‬ଶ௫ ൌ ʹሺʹǤͶ͸ሻ ൈ ͳͲଷ ൌ ͶǤͻ ൈ ͳͲଷ ሬԦ ஼ ஼ ሬԦ௖௬ ൌ ‫ܧ‬ଵ௬ ൅ ‫ܧ‬ଶ௬ ൌ Ͳ ‫ܧ‬ So at point c the total electric field has magnitude ே ͶǤͻ ൈ ͳͲଷ and its direction toward the right. ஼ ͵ܰ ௖ ‫ܥ‬ 1.6 Electric Field Lines Electric field lines can be a big help for visualizing electric fields and making them seem more real. An electric field line is an imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric field vector at that point.
  • 25.
    Figure below showssome of the electric field lines in a plane (a) a single positive charge, (b) two equal- magnitude charges, one positive and one negative (dipole), (c) two equal positive charges. Diagram called Field Map.