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Electric Charges and Forces
Electric Force
The electric force is one of the fundamental forces of nature.
Often, as you know from experience, rubbing two objects together
causes them to experience an electric force.
Examples:
• Unrolling plastic rap
• Running a comb through hair
• Rubbing rubber/plastic/glass rods with fur and silk
• Walking on carpet with slippers
•  Consider the following experiments to help determine the
nature and cause of this force:
The Charge Model
4
The charge model is a theory that accounts for the
observations:
Charge Model:
•  Frictional forces, such as rubbing, add or remove something called charge to/
from an object. More vigorous rubbing produces a larger quantity of charge.
•  There are two kinds of electric charge: positive and negative (arbitrary names)
•  Objects with like charges repel
•  Objects with opposite charge attract
•  a 3rd type of charge that is attracted to both positive and negative charges
has never been found.
•  Choice of what is a positive charge and a negative charge is arbitrary
(glass rubbed with silk is defined as positive charging).
•  Neutral objects have an equal mixture of positive and negative charge.
•  The electric force is a long-range force, but decreases with increasing
distance.
Clicker Question
Experiments with neutral object
Consider a couple experiments involving a macroscopic
neutral object:
•  How can a charged object attract a neutral, macroscopic
object?
Conductors, insulators, and dielectrics
•  Materials in which charge is free to move are conductors.
•  Materials in which charge isn’t free to move are
insulators.
Clicker Question
An electroscope consists of a neutral metallic sphere
connected to metal leaves via a metal rod.
Why do the leaves of the electroscope repel each other
when the charged rod is brought near the sphere?
(metals are good conductors)
1.  The leaves become negatively
charged
2.  The leaves become positively
charged
3.  The leaves are neutral, but are
attracted to the charged rod
Clicker question
A negatively charged glass rod is brought near a spherical
conductor, which is initially neutral. The spherical conductor is
initially touching another spherical conductor which is also
initially neutral. While the glass rod is near the conductor, the
two conductors are separated. The glass rod is then removed.
What are the charged states of the two conductors?
1.  Left is positively
charged, right is
negatively charged.
2.  Right is positively
charged, left is
negatively charged.
3.  Left is positively
charged, right is
neutral.
Coulomb’s law and the electric force
•  Like charges repel, and opposite charges attract, with a
force that depends on
•  The product of the two charges
•  The inverse square of the distance between them
•  Mathematically, the electric force is described by
Coulomb’s law:
F�12
kq1 q2
=
r̂
r2
r
Here F
F�12
12 is the force q1 exerts
on q2 and rˆ is a unit vector
pointing from q1 toward q2 .
k is the Coulomb constant ,
9
2
2
approximately 9.0 × 10 N ⋅ m / C .
Clicker question
14
What is the angle between two strings attached to
pithballs of mass m and charge Q?
The superposition principle
•  The electric force obeys the superposition principle.
•  That means the force two charges exert on a third force is just
the vector sum of the forces from the two charges, each treated
without regard to the other charge.
•  The superposition principle makes it mathematically
straightforward to calculate the electric forces exerted by
distributions of electric charge.
•  The net electric force is the sum of the individual forces.
Clicker Question
Clicker Question
Which of the arrows
best represents the
direction of the net
force on charge +Q
due to the other two
charges?
CT 25.11
Consider the charge configuration shown below.
What is the direction of the net force on
the +q charge?
B
+q
A
C
h
E
s/2 s/2
+Q
+Q
Copyright Univ. of Colorado, Boulder
D
Clicker question
A charge q is to be placed at either A or B on the
mid-line of two charges +q. Will the
force on q be greater at point A or at point B?
A) A
B) B
C) Can't tell without
knowing magnitude
of q.
2kqQy
F = 2
(a + y 2 )3/2
The Electric Field, E
•  The electric field at a point in space (due to a distribution
of charge) is the force per unit charge that a test-charge q’
placed at that point would experience:
�
F
� =
E
q�
The electric field exists
in space (even if no charge
exists there).
22
An electron is
placed at the
position marked
by the dot. The
force on the
electron is
A. to the left.
B. to the right.
C. zero.
D. There s not enough information to tell.
Fields of point charges and charge
distributions
•  Based on Coulomb’s law, the
field of a point charge is radial,
outward for a positive charge
and inward for a negative
charge.
� pt
E
charge
kq
= 2 r̂
r
•  The superposition principle states that, for a collection of
charges, the total electric field at a point is simply the
summation of the individual fields due to each charge
� =
E
�
kq
i
�
Ei = 2 r̂i
ri
Clicker Question
The dipole: an important charge distribution
•  An electric dipole consists of two point charges of equal
magnitude but opposite signs, held a short distance apart.
•  The dipole is electrically
neutral, but the separation of
its charges results in an
electric field.
•  Many charge distributions,
especially molecules, behave
like electric dipoles.
•  The product of the charge and
separation is the dipole
moment: p = qd.
•  Far from the dipole, its
electric field falls off as the
inverse cube of the distance.
3
(E ∝ 1/r )
Field lines
•  Field lines show the direction of the electric field. The
density of field lines is proportional the the strength of
the field.
•  Examples:
phet: charges
Continuous Charge Distribution
•  Macroscopic objects typically contain so many excess protons or
electrons (say, over 1012), it is a very good approximation to
describe the distribution to be continuous.
•  Instead of summing over every single point charge, we integrate over the
infinitesimal spatial chunks of charge.
E field at P due to Δq
∆q
�
∆E = k 2 r̂
r
In limit that volume becomes
infinitesimally small:
∆q → dq
� =
E
�
� =
dE
�
dq
�
dE = k 2 r̂
r
k r̂
dq
2
r
�
�
k r̂
dq
2
r
•  What is dq in terms of charge density?
� =
E
� =
dE
1D
2D
3D
dq=!dA
dq=!dx
dx
dq = λdl
� =
E
�
dq = σda
kλ dl r̂ E
� =
r2
�
kσ da r̂
r2
dq = ρdV
� =
E
�
kρ dV r̂
r2
Note that, in general, r̂ changes direction during the integral!
Do integral one component at a time.
Example: Line Charge
Let’s calculate the electric field a distance a to the right of
the right end of a uniform line charge of length L and charge
Q.
Q
�
�
++++++++++
L
a
� =
E
k r̂ dq
=
2
r
Q
λ=
L
Since charge distribution is uniform:
Pick a coordinate system: say the x-axis along the rod
with origin at left end
Ex =
�
L
0
kλ dx
kQ
= ... =
2
(a + L − x)
a(L + a)
Note: as a becomes much bigger than L,
kQ
Ex → 2
a
kλ dl r̂
r2
•  The electric field of an infinite line of charge:
•  The line carries charge density λ (units are C/m):
What is the vertical component, dEy, due to
the little chunk of lengthT dx?
26.11e
dE
y=H
x
+ + + + + + + + + + + + + + + + + +dx+ + + ++ +
dEy=
A) kλdx/(x2 + H 2 )
C) kλH dx/(x2 + H 2 )
E) kλx
|
dx/(x2 + H 2 )
2
2 3/2
B) kλx dx/(x + H )
D) kλH dx/(x2 + H 2 )3/2
from CU Boulder
Ey =
�
∞
−∞
2kλ
Ey =
y
kλy dx
(x2 + y 2 )3/2
2kλyx
∞
= �
|0
y 2 (x2 + y 2 )
Figure 21.49
The electric field at point P is
1.  kQ/a2
2.  less than kQ/a2 but greater than zero
3.  zero
4.  Depends on the value of the charge at point P
•  The electric field on the axis of a charged ring:
Ex =
�
k dq
cos θ =
2
r
cos θ = x/r = x/
�
2π
0
�
k λ a dφ
cos θ
2
2
x +a
x2 + a2
kQx
Ex = 2
(x + a2 )3/2
•  The electric field on the axis of a charged ring:
•  Notice that since all chunks of charge contribute equally,
there really isn’t a need to integrate spatially:
�
k dq
k Qx
Ex =
cos θ = 2
2
r
r r
kQx
Ex = 2
(x + a2 )3/2
•  Electric field due to disk of radius R and charge Q (a
distance x from center):
E=
E=
�
�
dERing =
R
0
�
k dQRing x
(x2 + r2 )3/2
k σ 2πr dr x
(x2 + r2 )3/2
1
R
E = 2πσkx 2
|
(x + r2 )1/2 0
�
x
E = 2πσk 1 − √
x2 + R 2
�