Electric Force

					    Electric Charge
   and Electric Force
     Based on a Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
             R. Morgan
                2009
 Objectives: After finishing this
 lesson, you should be able to:
 • Explain and discuss the fundamental properties
   of charge, the charging processes: contact &
   induction, and the conservation of electric charge.
• Write Coulomb’s Law and
  apply it to problems involving
  electric forces.
• Define the electron, the
  coulomb, and the microcoulomb
  as units of electric charge.
Operational Definition of the
Charging Process

• When amber is rubbed, it obtains the
  ability to attract bits of dust and straw.
• Similarly, when other objects are rubbed
  they can exhibit an ability to attract or
  repel other objects. This process is
  called charging of the object.
Operational Definition of the
Charging Process (cont’d)

• When two like objects are rubbed with
  like materials, they will have a like
  charge (as with two glass rods after
  each is rubbed with silk cloth).
Operational Definition of the
Charging Process (cont’d)

• When a charged object touches another
  object, the second object can also
  become charged. This process is called
  conduction.
• When a charged object touches another
  object or other objects, these objects will
  have like charge.
The Most Fundamental
Operational Property of Charges

• Like charges repel
• Opposite charges attract
     The Electroscope
  Laboratory devices used to study the
existence of two kinds of electric charge.




     Pith-ball            Gold-leaf
   Electroscope          Electroscope
The two parts of an electroscope
 will receive a like charge when
 touched by a charged object.
  They then repel each other.




       Pith-ball     Gold-leaf
     Electroscope   Electroscope
   When two objects are charged by
rubbing together and then one is used
 to charge a pith ball electroscope the
  other rubbed object will attract the
 pith balls (and then discharge them).




         Pith-ball     Gold-leaf
       Electroscope   Electroscope
   When two objects are charged by
rubbing together and then one is used
to charge a metal leaf electroscope the
   other rubbed object will cause the
   leaves to return back to their rest
                position.




          Pith-ball     Gold-leaf
        Electroscope   Electroscope
Classification of Charges
• Experimentation can show that there are
  only two charge conditions.
• By using electroscopes we can now
  classify all charges into these two types.
• Benjamin Franklin defined “positive” as
  the charge that glass obtains when
  rubbed with silk (and all like charges).
• Opposite charges are called “negative”.
Charge at the atomic level.

• Based on J. J. Thomson’s experiments, we
  know that the flow of charge is generally the
  flow of electrons.
• Atomic protons are generally fixed in place and
  do not flow.
• Positive charges can flow too as with positive
  ions in a salt water solutions, the acid in lead-
  acid batteries, and in living cells.
Charge at the atomic level.
• Charged is “quantized” meaning that it is
  “grainy” made up of “building blocks” all
  having magnitudes equal to the charge
  of a proton or an electron.
• Conservation of charge: charges can not
  be created or destroyed. When we
  charge an object, we are just moving
  charges around.
Charge at the atomic level.
• Conductors are materials that allow
  electrons to flow relatively easily
• Insulators hold all electrons tightly
  preventing flow.
• Metals are the best conductors because
  the metallic bonds allow free movement
  of electrons.
  Electric Charge: Atomic Level
When a glass rod is rubbed against silk, electrons are
removed from the glass and deposited on the silk.

  glass            Electrons               positive
                   move from                   + +
                   glass to the   negative    + +
          silk     silk cloth.     - - - -


The glass is said to be positively charged and has a
deficiency of electrons. The silk is said to be
negatively charged and has an excess of electrons.
  Electric Charge: Atomic Level
 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.
Charging by Rubbing

• Electrons are rubbed off from one
  material to the other.
• The material gaining the electrons does
  so because it has a stronger attraction to
  electrons than does the material losing
  the electrons.
Two Negative Charges Repel
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.
   Two Positive Charges Repel
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.
  The Two Types of Charge


      Rubber                         glass
                    Attraction

         fur                       silk

Note that the negatively charged (green) ball is
attracted to the positively charged (red) ball.

         Opposite Charges Attract!
The Fundamental (Operational)
   Characteristic of Charge
 Like charges repel; unlike charges attract.




  Neg         Neg                Neg Pos
  Pos         Pos
         Charging by Contact
1. Take an uncharged electroscope as shown below.
2. Bring a negatively charged rod into contact with knob.

                     --- - -              -
                                              -
                                                  -

                       -- --              -- -
                                          - -
                       - -                   -


3. Electrons move down on leaf and shaft, causing them
   to separate. When the rod is removed, the scope
   remains negatively charged.
Charging Electroscope Positively
 by Contact with a Glass Rod:
 Repeat procedures by using a positively charged
 glass rod. Electrons move from the ball to fill
 deficiency on glass, leaving the scope with a net
 positive charge when glass is removed.

                       ++++                   +
                      +                   +       +
                         +                +
                        ++               + +
                        + ++             + ++
Charging Conducting Spheres by
Induction: Not Touching
                        --- - -
                                            - - Electrons
                               ++
                               ++           -- Repelled

   Uncharged Spheres     Separation of Charge
--- - -
      ++
      ++
              --              +
                            + +     -
                                        -
                                            -
              --             +          -

 Isolation of Spheres     Charged by Induction
Induction for a Single Sphere
                                        Induction
                       --- - -
                               + ----
                             + --
                             ++ --

Uncharged Sphere         Separation of Charge

--- - -
         -- - - - -
      ++ -
      ++ -                   + +
                                +
                              +

 Electrons move       Charged by Induction
   to ground.
         Charge Distribution
           on Conductors
• When an excess of electrons is located
  on a conductor each electron will repel
  every other electron. The result is that
  all net negative charge will become
  evenly distributed on the very outer
  surface of the object.
• In a similar way, if an object is positively
  charged, all net charge will exist on the
  outer surface of the object.
Charge on Nonconductors
• On non-conductors, the charge (excess/
  shortage of electrons) remains in place:
  does not distribute across the object.
• When brought near a charged object,
  surface charges in a non-conductor can
  become polarized (charge polarization).
• The surface can exhibit charged
  properties (attraction/repulsion) even
  though the net charge of the non-
  conductor is zero.
Charge Polarization
in a non-conductor
Charge Polarization
 Causing Lightning
       The Quantity of Charge
  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. A
  temporary definition might be as given below:

     The Coulomb: 1 C = 6.25 x 1018 electrons

Which means that the charge on a single electron is:

            1 electron: e- = -1.6 x 10-19 C
           Units of Charge
The coulomb (selected for use with electric
currents) is actually a very large unit for static
electricity. Thus, we often encounter a need 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
Example 1. If 16 million electrons are
removed from a neutral sphere, what is
the charge on the sphere in coulombs?
  1 electron: e- = -1.6 x 10-19 C
                                           + +
                                          + + +
                     -1.6 x 10-19 C    + + + +
  q  (16 x 106e- )          -          + + +
                          1e             + +
       q = -2.56 x 10-12 C
  Since electrons are removed, the charge
  remaining on the sphere will be positive.
  Final charge on sphere:      q = +2.56 pC
           COULOMB’S LAW
The force of attraction or repulsion between two
point charges is directly proportional to the product
of the two charges and inversely proportional to the
square of the distance between them.

              F
    - q                q’   +
                                       qq '
              r                      F 2
F                            F          r
      q                q’
      -                -
         Coulomb’s Law:
     Calculating Electric Force
The proportionality constant k for Coulomb’s
law depends on the choice of units for charge.

           kqq '        Fr 2
        F  2 where k 
            r           qq '
When the charge q is in coulombs, the distance r is
in meters and the force F is in newtons, we have:

                          9 Nm
                 2                   2
            Fr
         k       9 x 10
            qq '             C2
   Example 2. A –5 mC charge is placed
   2 mm from a +3 mC charge. Find the
   force between the two charges.

                               -5 mC      F        +3 mC
   Draw and label
   givens on figure:
                           q     -        r         +   q’
                                         2 mm

   kqq ' (9 x 10
                       9 Nm2
                               )(5 x 10 C)(3 x 10 C
                                              -6             -6

 F 2                    C2
    r                          (2 x 10-3m)2
         F = 3.38 x 104 N;             Attraction

Note: Signs are used ONLY to determine force direction.
   Problem-Solving Strategies
1. Read, draw, and label a sketch showing all
   given information in appropriate SI units.
2. It is generally easiest to determine the
   direction of the force based on attraction/
   repulsion and then ignore signs of charges.
3. Superposition Principle: The Resultant force
   caused by multiple charges is found by consid-
   ering force due to each charge independently
   and then adding the force vectors.
4. For forces in equilibrium: SFx = 0 = SFy = 0.
Example 3. A –6 mC charge is placed 4 cm
from a +9 mC charge. What is the resultant
force on a –5 mC charge located midway
between the first charges? 1 nC = 1 x 10-9 C
1. Draw and label.                                    F1
                             -6 mC                      F2 +9 mC
2. Draw forces.                           q3
3. Find resultant;
                           q1   -    r1        -     r2    + q2
                                     2 cm          2 cm
   right is positive.

    kq1q3 (9 x 109 )(6 x 10-6 )(5 x 10-6 )
F1  2                       2
                                           ;         F1 = 675 N
     r1             (0.02 m)
    kq2 q3 (9 x 109 )(9 x 10-6 )(5 x 10-6 )
F2  2                        2
                                            ;        F2 = 1013 N
     r1              (0.02 m)
Example 3. (Cont.) Note that direction
(sign) of forces are found from attraction-
repulsion, not from + or – of charge.

          +             -6 mC
                                                 F1
                                     q3
                                                   F2 +9 mC
  F1 = 675 N          q1   -    r1        -     r2    + q2
  F2 = 1013 N                   2 cm          2 cm

The resultant force is sum of each independent force:

 FR = F1 + F2 = 675 N + 1013 N;               FR = +1690 N
Example 4. Three charges, q1 = +8 mC,
q2 = +6 mC and q3 = -4 mC are arranged
as shown below. Find the resultant force
on the –4 mC charge due to the others.
   +6 mC 3 cm q        Draw free-body diagram.
  q2 +       - -4 mC
                3


                              F2           q3
 4 cm        5 cm                        - -4 mC
                            53.1o
          53.1o
      +                             F1
 q1     +8 mC


 Note the directions of forces F1and F2 on q3
 based on attraction/repulsion from q1 and q2.
Example 4 (Cont.) Next we find the forces
F1 and F2 from Coulomb’s law. Take data
from the figure and use SI units.
       kq1q3             kq2 q3            +6 mC 3 cm q
   F1  2 ;          F2  2
                                          q2 + F2    - -4 mC
                                                        3
        r1                r2
     (9 x 109 )(8 x 10-6 )(4 x 10 -6 )   4 cm F1      5 cm
F1 
               (0.05 m) 2
                                                  53.1o
     (9 x 109 )(6 x 10-6 )(4 x 10-6 )         +
F2                                      q1   +8 mC
               (0.03 m) 2
 Thus, we need to find resultant of two forces:

 F1 = 115 N, 53.1o S of W                F2 = 240 N, West
Example 4 (Cont.) We find components of
each force F1 and F2 (review vectors).
 F1x = -(115 N) Cos 53.1o            F2
     = - 69.2 N                    240 N F1x           q3
                                                     - -4 mC
                                             53.1o
 F1y = -(115 N) Sin53.1o    =        F1y
     - 92.1 N
Now look at force F2:                        F1= 115 N

 F2x = -240 N; F2y = 0          Rx = SFx ;      Ry = SFy

Rx = – 69.2 N – 240 N = -309 N         Rx= -92.1 N

Ry = -69.2 N – 0 = -69.2 N             Ry= -240 N
Example 4 (Cont.) Next find resultant R from
components Fx and Fy. (review vectors).

  Rx= -309 N       Ry= -69.2 N
                                 Rx = -309 N q
  We now find resultant R,q:          f      - -4 mC
                                               3



                          Ry           R
   R  R  R ; tan f =
          2
          x
               2
               y
                          Rx          Ry = -69.2 N

      R  (309 N)2  (69.2 N)2  317 N

   Thus, the magnitude of        R = 317 N
   the electric force is:
Example 4 (Cont.) The resultant force is
317 N. We now need to determine the
angle or direction of this force.

     R  Rx  Ry  317 N
          2    2
                               -309 N     q
                                    f -
            Ry   309 N              R
     tanf                             -69.2 N
                                         -62.9 N
            R x -69.2 N
    The reference angle is: f = 77.40S of W
Or, the polar angle q is: q = 1800 + 77.40 = 257.40

     Resultant Force: R = 317 N, q = 257.40
        Summary of Formulas:
     Like Charges Repel; Unlike Charges Attract.

            kqq '                       Nm      2
          F 2               k  9 x 10 9
                                           2
             r                           C

     1 mC = 1 x 10-6 C       1 nC = 1 x 10-9 C


1 pC = 1 x 10-12 C    1 electron: e- = -1.6 x 10-19 C
CONCLUSION: Electric Force

				
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