Electricity _ Magnetism by wanghonghx

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									Electricity
If you rub a piece of amber




With the fur of a rabbit




It will attract bits of stuff (paper, leaves, etc)
• Equal charges repel.
• Opposite charges attract.

The force between two charges:
• Proportional to the magnitude
  of the charges
• Decreases as the square of the
  distance
(a) electrons are
transferred to the metal
sphere by rubbing the
negatively charged rod
(b) when the rod is
removed, electrons
distribute themselves
uniformly over the
surface
Five Styrofoam balls are suspended from insulating threads.
Several experiments are performed on the balls and the
following observations are made:
    I. Ball A attracts B and repels C.
    II. Ball D attracts B and has no effect on E.
    III. A negatively charged rod attracts both A and E.
What are the charges, if any, on each ball?


     A B C       D E
a.   +   -   +   0   +
b.   +   -   +   +   0
c.   +   -   +   0   0
d.   -   +   -   0   0
e.   +   0   -   +   0
Answer:
     The first statement indicates that A attracts B and
     repels C. This indicates that both A and C are
     charged, but B could be charged by induction.
     The second statement indicates that D attracts B, but
     has no effect on E. This means that D and E must
     not be charged because if either were charged and
     the other not, then they would by attracted to each
     other by induction.
     A negative rod attracts A and E. We know then that A
     is positively charged. This then tells us from
     statement one that B is negatively charged. C is also
     positively charged.
     Therefore, choice (C) is the correct answer.
COULOMB’S LAW

The magnitude of the electrostatic force exerted by one point charge
on another point charge is directly proportional to the magnitude of the
charges and inversely proportional to the square of the distance between
them.
              q1 q2
         F k                                1
                r2                   k
                                          4 o
                                                   8.99  109 N  m 2 C 2


                                   8.85 1012 C2  N  m2 
            mp  1.6731027 kg
            mn  1.675 1027 kg
             me  9.1110 31 kg
                           19
           e  1.60 10          C

An atom contains a small, positively charged nucleus,
about which negatively charged electrons move. The
orbits shown are only symbolic. In reality electrons,
occupy a region of space that encircles the nucleus.
In nature, atoms are normally
found with equal numbers of protons
and electrons, so they are electrically
neutral.

By adding or removing electrons
from matter it will acquire a net
electric charge with magnitude equal
to e times the number of electrons
added or removed, N.



            q  Ne
An electrical conductor is a material in which
electrons move easily and are readily conducted.

metals:
silver, copper, gold, aluminum, tungsten, iron, etc.
An electrical insulator is a
material in which electrons are
poorly conducted.

insulators:
mica, rubber, teflon, wood, paper

semi-conductors:
silicon, germanium, carbon
(semi-metal)
DEFINITION OF ELECRIC FIELD

The electric field that exists at a point is the electrostatic force experienced
by a small test charge placed at that point divided by the charge itself:


         F              SI Units of Electric Field: newton per coulomb (N/C)
      E
         qo
       Electric fields from different sources
       add as vectors.




The Electric Fields from Separate
Charges May Cancel
THE PARALLEL PLATE CAPACITOR




        Parallel plate
        capacitor
WAB  EPE A  EPE B
WAB EPE A EPE B
        
 qo   qo   qo
The potential energy per unit charge
is called the electric potential.


           EPE
        V
            qo
 SI Unit of Electric Potential:
           joule/coulomb = volt (V)
Work, Potential Energy, and Electric Potential

The work done by the electric force as a test +2.0x10-6 C charge moves
   from point A to point B is + 5.0x10-5J.

(a) Find the difference in EPE between these
points.

(b) Determine the potential difference between
these points.


WAB  EPE A  EPE B
                                                                    5
     EPE B  EPE A  WAB  5.0  10 J
                                                 5
          WAB 5.0 10 J
VB  VA                  25 V
           qo   2.0 10 C
                       -6
 Batteries


Within a battery, a chemical reaction occurs that transfers
electrons from one terminal to another terminal.
The electric current is the amount of charge per unit time that passes
through a surface that is perpendicular to the motion of the charges.




                             q
                          I
                             t
          One coulomb per second equals one ampere (A).
If the charges move around the circuit in the same
direction at all times, the current is said to be direct
current (dc).

If the charges move first one way and then the
opposite way, the current is
said to be alternating current (ac).
A Pocket Calculator

The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour
of operation, (a) how much charge flows in the circuit and (b) how much energy
does the battery deliver to the calculator circuit?



 (a)        q  I  t    0.17 103 A   3600 s   0.61 C




                               Energy
 (b)     Energy  Charge               0.61 C  3.0 V   1.8 J
                               Charge
The resistance (R) is defined as the
ratio of the voltage V applied across
a piece of material to the current I through
the material.



   OHM’S LAW

   The ratio V/I is a constant, where V is the
   voltage applied across a piece of material
   and I is the current through the material:

 V
    R  constant             or        V  IR
 I
SI Unit of Resistance:
          volt/ampere (V/A) = ohm (Ω)
A Flashlight

The filament in a light bulb is a resistor in the form
of a thin piece of wire. The wire becomes hot enough
to emit light because of the current in it. The flashlight
uses two 1.5-V batteries to provide a current of
0.40 A in the filament. Determine the resistance of
the glowing filament.


            V 3.0 V
          R         7.5 
            I 0.40 A
Solar
Array
ELECTRIC POWER

When there is current in a circuit as a result of a voltage, the electric
power delivered to the circuit is:

                              P  IV
 SI Unit of Power: watt (W)



  Many electrical devices are essentially resistors:



                         P  I  IR   I 2 R


                              V    V2
                          P   V 
                              R    R
The Power and Energy Used in a
Flashlight

In the flashlight, the current is 0.40A and the voltage
is 3.0 V. Find (a) the power delivered to the bulb and
(b) the energy dissipated in the bulb in 5.5 minutes
of operation.


    P  IV   0.40 A3.0 V  1.2 W



   E  Pt  1.2 W330 s   4.0 102 J
             100 watt bulb
What is the resistance
and how much
current flows through
a 100 W bulb?
Note: The wattage on
a bulb is its power
output and assumes
that you will use it in
the US where the
voltage in 110 V.
              60 W bulb
Redo the calculations
for a 60 W bulb.

								
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