Chapter 21 Electric Field and Coulomb’s Law (again)
• Coulomb’s Law (sec. 21.3)• Electric fields & forces (sec. 21.4 & -6)• Vector addition
C 2009 J. F. Becker
Vectors (Review) Used extensively throughout the course
INTRODUCTION: see Ch. 1 (Volume 1)
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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.
2 2 2
cos
sinx
y
x y
A A
A A
A A A
Note: The dot product of two vectors is a scalar quantity.
cos x x y y z zA B AB A B A B A B
The scalar (or dot) product of two vectors is defined as
C 2009 J. F. Becker
sinA B AB
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:
C 2009 J. F. Becker
Note: The dot product of two vectors is a scalar quantity.
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.
21-9 Coulomb’s Law – vector problem
Net force on charge Q is the vector sum of the forces by the other two charges.
Coulomb’s Law -forces between 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
F = q E and |E| = (k q2/r2)C 2009 J. F. Becker
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:
Therefore we can write for
F = qo E
the electric field
E = [ k Q / r2 ]
Calculate E1, E2, and ETOTAL
at points “A” & “C”:
q = 12 nC
Electric field at“A” and “C” set up by charges q1 and
q1
(an electric dipole)
a) E1= 3.0 (10)4 N/C E2 = 6.8 (10)4 N/C
EA = 9.8 (10)4 N/C
c) E1= 6.4 (10)3 N/C E2 = 6.4 (10)3 N/C
EC = 4.9 (10)3 N/C in the +x-direction
A
C
Lab #1
Electric field at P caused by a line of charge along the y-axis.
Consider symmetry! Ey = 0
Xo
y
Consider symmetry! Ey = 0
Xo
dq
o
dEx = dE cos =[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
|dE| = k dq / r2
cos =xo / r
Calculate the electric field at +q caused by the distributed charge +Q.
Tabulated integral: dz / (c-z) 2 = 1 / (c-z)
d
b
Electric field at P caused by a line of charge along the y-axis.
Consider symmetry! Ey = 0
Xo
y
Electric field at P caused by a line of charge along the y-axis.
Consider symmetry! Ey = 0
Xo
y
Tabulated integral: dz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2
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
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 ]
Notation change
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Calculate the electric field at -q
caused by +Q, and then the force on -q.
Tabulated integral: dz / (z2 + a2)3/2 = z / a2 (z2 + a2) 1/2 z dz / (z2 + a2)3/2 = -1 / (z2 + a2) 1/2
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:
= p x E
ELECTRIC DIPOLE in E has potential energy:
U = - p EC 2009 J. F. Becker
Net force on an ELECTRIC DIPOLE is zero,
but torque () is into the page.
= r x F = p x E
ELECTRIC DIPOLE MOMENT is p = qd
See www.physics.edu/becker/physics51
Review
C 2009 J. F. Becker
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