Electric Forces & Electric Fields
Chapter 20
Structure of atom
Proton
Mass (kg)
Charge
(Coulombs)
1.673X10-27
+1.60X10-19
=+e
Neutron
1.675X10-27
0
Electron
9.11 X 10-31
 0 mass
-1.60X10-19
=-e
Example:
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A rod has a charge of -4.0 C. How
many electrons must be removed so
that the charge becomes +2.0 C?
1 C = 1 X 10-6 C
Solution…
The total amount of electrons that
need to be removed is 6.0 C or
6.0X10-6 C.
6.0X10-6 C (1 e-/1.60X10-19C) =
=3.75 X 1013 electrons
Law of Conservation
of Electrical Charge
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Electrons may be
transferred from one
object to another, but
they will never
disappear or appear
from nowhere.
Electron abundant =
negative charge
Electron deficient =
positive charge
Electrically charged objects...
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Exert a force on each other
Unlike charges attract
Like charges repel
Electrical Conductors
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In most metal atoms, the outer electrons are not
held tightly by the nucleus—these electrons can
freely roam/move.
Generally, good thermal conductors are good
electrical conductors.
Examples: silver, copper, aluminum
Electrical Insulators
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In materials where the outer electrons are held
tightly by the nucleus—these electrons cannot
freely roam/move.
Generally, good thermal insulators are good
electrical insulators.
In a power line, charge flows much more easily
through hundred of kilometers of metal wire than
through the few centimeters of insulating
material that separates the wire from the
supporting tower.
In a common appliance cord, charges will flow
through several meters of wire to the appliance,
and then through its electrical network, and then
back through the return wire rather than flow
directly across from one wire to the other
through the tiny thickness of rubber insulation.
In general, electrons follow path of least
resistance.
Charging by Friction
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It is possible to transfer electric charge from one
object to another.
Walk across carpeted floor with wool socks and
you can build up an excess of charge (either on
you or the carpet) and therefore, become
charged.
Can use friction to remove electrons from fur and
deposit onto rod.
Fur becomes positively charged. Why?
Rod becomes negatively charged. Why?
Charging by Contact
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The process of
giving one object
a net electric
charge by placing
it in contact
w/another object
that is already
charged is known
as charging by
contact.
Charging by Induction
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The process of giving one object a net
electric charge without touching the object
to a second charged object is called
charging by induction.
Temporarily Charging by Induction
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An object may
become polarized
(opposite ends) while
a charged object is
brought near. But,
the overall charge on
this piece of plastic
(for example) is
neutral.
When the charged rod
goes away, the
electrons rearrange
and disperse evenly.
Object is still neutral.
Coulomb’s Law
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When oppositely charged
objects are near each other,
there is a force experienced
by each attracting them to
one another. True for
oppositely charged objects
too… except the force is
pushing the objects apart
and a repelling force exists.
According to Newton’s 3rd
law, these forces are
equal and opposite.
Coulomb’s Law
The electrostatic force that stationary charged
objects exert on each other depends on the amount
of charge (q) on the objects and the distance (r)
between them.
 Since these objects are so small compared to the
distance between them, they may be regarded as
“point charges.”
 F = k (q1q2)/r2
 F = the amount of force that particle #1 exerts on
#2 and vice versa (same force!) (Unit: Newtons)
 k = proportionality constant = 8.99 X 109 Nm2/C2
 q1 = amount of charge of particle #1 (Unit:
Coulombs)
 q2 = amount of charge of particle #2 (Unit:
Coulombs)
 r = the distance between q1 and q2 (Unit: meters)
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You can have more than two charged
particles interacting.
To solve for the net force… recall that
net force is the sum of all the
forces.
Solve for the x and y components if the
forces are not in one plane.
Then add the x-components, add the ycomponents, and solve for the resultant,
using the Pythagorean theorem.
Don’t forget direction!
Next slides show two examples…
What’s the net electrostatic force on q1?
q1 is being
attracted
(pulled
toward) both
q2 and q3.
Solution…
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F12 = (kq1q2)/(r122) = plug in values
F12 = 2.7 N
F13 = (kq1q3)/(r132) = plug in values
F13 = 8.4 N
Since F12 points in the –x direction,
and F13 points in the +x direction:
F = -F12 + F13 = -2.7N + 8.4N =
F = +5.7 N
The net force is 5.7 N pointing to the
right.
What’s the net electrostatic force on q1?
Solution:
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F12 = (kq1q2)/(r122) = plug in values
F12 = 9.6 N directed 73o above +x axis, having
both an x-component and y-component
F12x : cos 73o = F12x / F12
F12x=(F12 )(cos 73o)=9.6N(cos 73o) = +2.8N
F12y : sin 73o = F12y / F12
F12y=(F12 )(sin 73o)=9.6N(sin 73o) = +9.2N
F13 = (kq1q3)/(r132) = plug in values
F13 = 18 N directed along +x axis
F =[(F12x + F13 )2 + (F12y )2] =
F =[(2.8N+ 18N)2 + (9.2N)2] = 23 N
 = tan-1 (Fy/Fx) = tan-1 (9.2N/20.8N) = 24o