Transcript Document

Chapter 21 Electric Field and
Coulomb’s Law (again)
•
•
•
Electric fields and forces
(sec. 21.4)
Electric field calculations
(sec. 21.5)
Vector addition (quick review)
C 2012 J. Becker
Learning Goals - we will learn:
• How to use Coulomb’s Law (and vector
addition) to calculate the force between
electric charges.
• How to calculate the electric field caused
by discrete electric charges.
• How to calculate the electric field caused
by a continuous distribution of electric
charge.
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.
Figure 21.14
SYMMETRY!
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 is
vector E is the electric field caused by q2.
The direction of the E field is determined by
the direction of the F, or the E field lines are
directed
away from positive q2 and toward -q2.
The F on a charge q in an E field is
F = q E and |E| = (k q2/r2)
Fig. 21.15 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 ]
E1
C
See
Lab #2
ET
E2
See Fig. 21.23: Electric field at
“C” set up by charges q1 and q1
Calculate E1, E2, and ETOTAL
at point “C”:
q = 12 nC
A
(an “electric dipole”)
At “C” E1= 6.4 (10)3 N/C
E2 = 6.4 (10)3 N/C
ET = 4.9 (10)3 N/C
in the +x-direction
Need TABLE of ALL
vector component
VALUES.
dq
Fig. 21.24 Consider symmetry! Ey = 0
 dEx = Ex
o
|dE| = k dq / r2
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 and a stay
constant as we add all the dq’s ( dq = Q)
in the integration: Ex = k xo Q/[xo2+a2]3/2
y
Consider symmetry! Ey = 0
dq
|dE| = k dq / r2
Xo
Fig. 21.25 Electric field at P caused by a line
of charge uniformly distributed along y-axis.
|dE| = k dq / r2
cos a = xo/ r
and
and
r = (xo2+ y2)1/2
cos a = dEx / dE
dEx = dE cos a
Ex =  dEx =  dE cos a
Ex = [k dq /r2] [xo / r]
Ex = [k dq /(xo2+y2)] [xo /(xo2+ y2)1/2]
Linear charge density = l
l = charge / length = Q / 2a = dq / dy
dq = l dy
Ex =  [k dq /(xo2+y2)] [xo /(xo2+ y2)1/2]
Ex =  [k l dy /(xo2+y2)] [xo /(xo2+ y2)1/2]
Ex = k lxo  [dy /(xo2+y2)] [1 /(xo2+ y2)1/2]
Ex = k lxo  [dy /(xo2+y2) 3/2]
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
 dy / (Xo2+y2) 3/2 = y / Xo2 (Xo2+y2) 1/2
Ex = k lxo
a
2+y2) 3/2]

[dy
/(x
o
-a
Ex = k(Q/2a) Xo [y
/Xo2 (Xo2+y2) 1/2
Ex = k (Q /2a) Xo [(a –(-a)) /
]
a
-a
Xo2 (Xo2+a2) 1/2
Ex = k (Q /2a) Xo [2a / Xo2 (Xo2+a2) 1/2 ]
Ex = k (Q /
Xo) [1
/ (Xo2+a2) 1/2 ]
]
Tabulated integral:
 dz / (c-z) 2 = 1 / (c-z)
l is uniform (= constant)
+Q
b
Fig. 21.47 Calculate the electric field at the
proton caused by the distributed charge +Q.
Tabulated integrals:
 dz / (z2 + a2)3/2 = z / a2 (z2 + a2) ½
for calculation of Ex
 z dz / (z2 + a2)3/2 = -1 / (z2 + a2) ½
for calculation of Ey
l is uniform (= constant)
Fig. 21.48 Calculate the electric field at -q
caused by +Q, and then the force on –q: F=qE
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
ELECTRIC DIPOLE
MOMENT is
p = qd
t=rxF
t=pxE
Fig. 21.32 Net force on an ELECTRIC DIPOLE
is zero, but torque (t) is into the page.
Review
see
www.physics.sjsu.edu/Becker/physics51
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 decomposed
into 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.
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
Right-hand
rule for
DIRECTION
of vector
cross product.
PROFESSIONAL FORMAT