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Topic 5: Electricity and Magnetism.
IB PHYSICS
Electric Charges
• Evidence for electric charges is everywhere.
(static electricity and lightning)
• Objects may become charged by contact and
frictional forces. (clothes in dryer)
• Benjamin Franklin (1700’s) discovered that there
are two types of charges:
(positive and negative)
• Franklin also discovered the Rules of Charge.
(like charges repel and unlike charges attract)
• Electric charge was found to be conserved and
quantized. (Franklin, late 1700’s and Millikan, early 1900’s)
Classes of Materials
• CONDUCTORS are materials in which charges
may move freely (Cu, Hg, Ag, Au, Al, Fe).
• INSULATORS are materials in which charges
cannot move freely (rubber, plastic, wood, glass).
• SEMICONDUCTORS are materials in which
charges may move under some conditions (e.g. Si,
Gd, Ge, Y).
Charles Coulomb measured the force between charged
objects. Later, the unit of charge, the Coulomb, C, was
named after him. 1 C = 6.25 x 1018 electrons, which is very
large. We usually have mC, μC, nC, and pC.
F=k·
(q1∙q2)
r2
Coulomb’s Constant
Charles Coulomb
(1736-1806)
France
k = 8.99 x 109 N∙m2/C2
SAMPLE PROBLEMS: (Draw the diagram for each.)
1. Charge is conserved.
A small conducting sphere contains a net surface charge of 8 pC. A
larger sphere has a surface charge of -18 pC. The spheres are
touched together and then separated. What is the charge on each?
2. Coulomb’s Law.
Calculate the electric force between two charges of q1 = 2.5 nC, and
q2 = 10.0 nC, if they are separated by a distance of 2.5 cm.
Benjamin Franklin (1706-1790), USA, determined that
there are only 2 types of charge.
Rubber Rod - Negative
Glass Rod - Positive
Unlike Charges Attract.
A charge in space determines an electric field.
Electric field lines are
lines of force.
Electric Field Lines: Conventions
Positive Point Charge Negative Point Charge
(Both can set-up an Electric Field.)
Electric Field lines always point away from the
positive and toward the negative.
How to Measure Electric Field Strength:
Use a test charge q0 to probe the field which exerts Force.
(Based on Coulomb’s Law.) And we see, Field does Work.
E = F q0
and F =
F
E = k · Q d2
later, k = 1/4πε0
ε0 = 8.85x10-12 C2/Nm2
Permittivity of Free Space
d
qE
• Electric charge is quantized. The elementary unit of
charge is the charge on one electron or one proton:
•
1e = 1.602 x 10-19 C
Robert Millikan (1868-1953)
In 1900, while a professor at
the University of Chicago.
Famous for the “Oil-drop”
Experiment.
Millikan’s Oil-Drop Experiment
(Diagram the forces on a drop.)
Consider a charge, Q, and its Electric Field,
E, and test charge, q, placed in the field.
E
Q
+
a|
q
d
F
b
|
Calculate work done by the field as charge, q, moves from
a to b: F = qE W = Fd or W = qEd . Better, would
be Work per unit charge, so W/q = Ed . We call W/q,
Voltage, V . Now W = U = qV and V = Ed . Voltage
also: Potential Difference or Electric Potential. What is U?
3/31/2017
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Ex: What is the speed of a proton accelerated from rest by a voltage of
4.0 kV? ( proton mass = 1.67x10-27 kg, charge = 1.6x10-19 C )
a. 1.1 x 106 m/s
b. 9.8 x 105 m/s
c. 8.8 x 105 m/s
Ex: To recharge a 12-V battery, a battery charger must move a charge
of 3.6 x 105 C from the negative to the positive terminal. What amount
of work is done by the battery charger? How many kilowatt-hours is
this? 1 Kwh costs $.13, how much to charge the battery?
Ans: 4.32x 106 J
1.2 kWh
$.0156
Consider a conducting wire in an Electric Field:
Electric Current, I:
I = n∙A∙vd∙q
Defined as: I = q/t
Unit: C/s or Ampere,
Amp, A
Named after:
Andre Ampere
France(1775-1836).
Ex: A current of 5.0 A is flowing in a Cu wire with a cross-section of
.5 mm2. For Cu, n =8.5x1028 per m3. and the charge on an electron is
1.6x10-19 C. Find the drift velocity of the electrons.
a. 7.35x10-4 m/s
b. 2.46x104 m/s
c. 8.68x10-2 m/s
Ex. A charged particle (q = –8.0 mC), which moves in a region where
the only force acting on the particle is an electric force, is released
from rest at point A. At point B kinetic energy of the particle is equal
to 4.8 J. What is the electric potential difference VB – VA?
a. –0.60 kV b. +0.60 kV c. +0.80 kV d. –0.80 kV e. +0.48 kV
SUMMARY: 5.1. Electric fields
F = kq1q2/r2 F = qE
Electric force
k = 1/(40) = 8.99x109 Nm2/C2 (Coulomb’s constant)
0 = 8.85x10-12 C2/Nm2
V = W/q= Ed
U = W = qV
(Permittivity constant)
Voltage
Potential Energy, Work
E = F/q
Electric field strength
I = nAvdq
Movement of charge
I = q/t
Electric current