# Chapter 21 Electric Field and Coulomb�s Law (again)

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```					         Chapter 21 Electric Field and
Coulomb’s Law (again)

• Coulomb’s Law              (sec. 21.3)
• Electric fields & forces   (sec. 21.4 & -6)

C 2009 J. F. Becker
INTRODUCTION: see Ch. 1
(Volume 1)
Vectors (Review)
Used extensively throughout the course

C 2009 J. Becker
Vectors are quantities that
have both magnitude and
direction.
An example of a vector
quantity is velocity. A
velocity has both magnitude
(speed) and direction, say
60 miles per hour in a
DIRECTION due west.
(A scalar quantity is
different; it has only
magnitude – mass, time,
temperature, etc.)
A vector may
be composed of
its x- and y-
components as
shown.
Ax  A cos 
Ay  A sin 
A  Ax  Ay
2     2        2
The scalar (or dot) product of two
vectors is defined as

A  B  AB cos  Ax Bx  Ay By  Az Bz

Note: The dot product of two
vectors is a scalar quantity.

C 2009 J. F. Becker
The vector (or cross) product of
two vectors is a vector where the
direction of the vector product is
given by the right-hand rule.
The MAGNITUDE of the vector
product is given by:

A  B  AB sin 
Note: The dot product of two
vectors is a scalar quantity.
C 2009 J. F. Becker
Right-hand
rule for
DIRECTION
of vector
cross product.
Coulomb’s Law
Coulomb’s Law lets us calculate the FORCE
between two ELECTRIC CHARGES.
Coulomb’s Law
Coulomb’s Law lets us calculate the force
between MANY charges. We calculate
the forces one at a time and ADD them
AS VECTORS.
(This is called “superposition.”)

THE FORCE ON q3 CAUSED BY q1 AND q2.
Coulomb’s Law -forces
21-9 Coulomb’s Law – between two charges
vector problem

Net force on charge Q is the vector sum of
the forces by the other two charges.
Recall GRAVITATIONAL FIELD near Earth:
F = G m1 m2/r2 = m1 (G m2/r2) = m1 g
where the vector g = 9.8 m/s2 in the downward
direction, and F = m g.
ELECTRIC FIELD is obtained in a similar way:
F = k q1 q2/r2 = q1 (k q2/r2) = q1 (E) where the
vector E is the electric field caused by q2.
The direction of the E field is determined by
the direction of the F, or as you noticed in lab
#1, the E field lines are directed away from a
positive q2 and toward a -q2.
The F on a charge q in an E field is
C 2009 J. F. Becker   F = q E and |E| = (k q2/r2)
A charged body
creates an electric field.
Coulomb force of repulsion
between two charged bodies at A
and B, (having charges Q and qo
respectively) has magnitude:
F = k |Q qo |/r2 = qo [ k Q/r2 ]
where we have factored out
the small charge qo.
We can write the force in
terms of an electric field E:
F = qo E
Therefore we can write for
the electric field
E = [ k Q / r2 ]
Electric field at“A” and
C        “C” set up by charges q1 and q1
Lab #1                 Calculate E1, E2, and ETOTAL
at points “A” & “C”:

a) E1= 3.0 (10)4 N/C
E2 = 6.8 (10)4 N/C
EA = 9.8 (10)4 N/C
q = 12 nC
c) E1= 6.4 (10)3 N/C
A
E2 = 6.4 (10)3 N/C
EC = 4.9 (10)3 N/C
in the +x-direction
(an electric dipole)
y                    Consider symmetry!
Ey = 0

Xo

Electric field at P caused by a line of charge
along the y-axis.
dq               Consider symmetry! Ey = 0

|dE| = k dq / r2
o

Xo

cos a =xo / r
dEx = dE cos a =[k dq /xo2+a2][xo /(xo2+ a2)1/2]
Ex = k xo  dq /[xo2 + a2]3/2 where xo is
constant as we add all the dq’s (=Q) in the
integration: Ex = k xo Q/[xo2+a2]3/2
Tabulated integral:  dz / (c-z)   2   = 1 / (c-z)

d

b

Calculate the electric field at +q caused by
the distributed charge +Q.
y             Consider symmetry! Ey = 0

Xo

Electric field at P caused by a line of charge
along the y-axis.
y              Consider symmetry! Ey = 0

Xo

Tabulated integral:
 dz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2

Electric field at P caused by a line of charge
along the y-axis.
Tabulated integral: (Integration variable “z”)

 dz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2
 dy / (c2+y2) 3/2 = y / c2 (c2+y2) 1/2                Notation
change
 dy / (Xo2+y2) 3/2 = y / Xo2 (Xo2+y2) 1/2
Our integral=k (Q/2a) Xo 2[y                 /Xo2 (Xo2+y2) 1/2   ]0   a

Ex = k (Q /2a) Xo 2 [(a –0) / Xo2 (Xo2+a2) 1/2 ]

Ex = k (Q /2a) Xo 2 [a / Xo2 (Xo2+a2) 1/2 ]

Ex = k (Q /           Xo) [ 1   / (Xo2+a2) 1/2 ]
C 2009 J. F. Becker
Tabulated integral:
 dz / (z2 + a2)3/2 = z / a2 (z2 + a2) 1/2

 z dz / (z2 + a2)3/2 = -1 / (z2 + a2) 1/2

Calculate the electric field at -q
caused by +Q, and then the force on -q.
An ELECTRIC DIPOLE consists of a
+q and –q separated by a distance d.
ELECTRIC DIPOLE MOMENT is p = q d

ELECTRIC DIPOLE in E experiences a torque:
t=pxE

ELECTRIC DIPOLE in E has potential energy:
U=-p                  E

C 2009 J. F. Becker
ELECTRIC DIPOLE
MOMENT is
p = qd

t=rxF
t=pxE
Net force on an ELECTRIC DIPOLE is zero,
but torque (t) is into the page.
Review
See www.physics.edu/becker/physics51

C 2009 J. F. Becker

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