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Coulomb’s Law
Physics 102
Professor Lee Carkner
Lecture 9
PAL #8
3m
+5e
F12
-2e
F23
2.1 m
+7e
 Force on 2 from 1:
 F12 = kq1q2/r2
 F12 = (8.99X109)(5)(1.6X10-19)(2)(1.6X10-19)/(32) = 2.56X10-28 N
 Force is to left, make negative
 Force on 2 from 3:
 F23 = (8.99X109)(7)(1.6X10-19)(2)(1.6X10-19)/(32) = 7.31X10-28 N
 Force is to right, make positive
 Fnet = F12 + F23 = - 2.56X10-28 N + 7.31X10-28 N = 4.75X10-28 N
 Direction is to the right
Inverse Square Law
Electric force:
Gravitational force:
Both are inverse square laws

Distance is more important than charge or mass
Decreasing the distance will greatly strengthen a
force, increasing the distance will greatly weaken the
force
Computing Electrical Force
F12
+q1
F21
-q2
Consider a pair of charges

Draw an arrow from the experiencing charge
either towards or away from the proving
charge
 Note that the forces are equal in magnitude
and opposite in direction (F21 =-F12)
Resultant Force
To find the net force from several forces add the
force from each vectorially:
Put an x-y axis on the system

draw as arrow
Find the magnitude each force
F1, F2, F3 ….

F1x, F1y …
Sum the x and y components
Fx = F1x + F2x …

F2 = Fx2 + Fy2
X and Y
Remember vector
addition rules:
Components along
axis computed
from:
F
Fy = F sin q


Total F:

Angle to x axis:
tan q = (Fy/Fx)
q
q
Fx = F cos q
Spherical Charge Distribution

An electron is a good example

but,
A uniform spherical distribution can be
treated as a point source located at its
center
The Electric Field
Electrical and gravitational forces act at a
distance

The area near the charge is said to be
occupied by an electric field

The test charge is small enough so that its field
does not affect the main one
Defining the Electric Field
The electric field is defined as the force on the test
charge per unit charge or:
E = F/q0

E = k q/r2
i.e., the electric field at any point a radial distance r
away from a charge q

We define the direction of the electric field based
on the direction a positive test charge would move
Positive charges move with the field, negative against
Calculating the Electric Field

Need to find the magnitude and
direction of the field from each charge
and add vectorially
Note that once you find the value of the
electric field, the force on a charge at
that point is just F=Eq0

Next Time
Read 16.8-16.10
Homework Ch. 16: P 14, 17, 26, 30
Plates A, B and D are charged plastic plates. C is an
originally uncharged metal plate. The electrostatic
forces between some of the plates are shown. Will the
remaining two pairs attract or repel each other?
A)
B)
C)
D)
E)
CD repel, BD attract
CD attract, BD repell
Both repel
Both attract
You cannot tell from the information given
If you rub a balloon on your head and the
balloon becomes negatively charged, your
head becomes,
A)
B)
C)
D)
E)
Negatively charged
Positively charged
Neutral
Its charge does not change
It depends on the type of balloon
How could you make a ordinary object
positively charged?
A)
B)
C)
D)
E)
By adding protons
By subtracting electrons
By transforming neutrons
By adding nuclei
Objects cannot become positively charged
A negatively charged rod is brought near one
end of an uncharged metal bar. What will
happen to the end of the metal bar furthest
from the rod?
A)
B)
C)
D)
E)
It will become positive
It will become negative
It will become neutral
It depends on the type of metal
Nothing