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The Electric Field
A field is a mathematical concept — it is a function that decribes
the characteristics of every point in space. There are two kinds:
I
Scalar Field: A scalar field assigns a numerical value to every
point in space. Examples include: temperature distributons,
elevation maps, the magnitude of the “spring force” in an
atomic lattice.
The Electric Field
A field is a mathematical concept — it is a function that decribes
the characteristics of every point in space. There are two kinds:
I
Vector Field: A vector field assigns both a numerical
magnitude and a direction to every point in space. Examples
include: wind speed and direction, fluid flow.
The Electric Field
The electric field is obtained from Coulomb’s
law:
~ electric = 1 q1 q2 r̂
F
4π0 r 2
This is an equation describing an interaction,
requiring information about two interacting
charges q1 and q2 separated by ~r = ~r2 − ~r1 .
The Electric Field
The electric field is obtained from Coulomb’s
law:
~ electric = 1 q1 q2 r̂
F
4π0 r 2
This is an equation describing an interaction,
requiring information about two interacting
charges q1 and q2 separated by ~r = ~r2 − ~r1 .
The electric field of a single charge q1 is obtained by dividing both
sides of the equation by q2
~
~ q = Fe = 1 q1 r̂
E
1
q2
4π0 r 2
The Electric Field
The electric field is obtained from Coulomb’s
law:
~ electric = 1 q1 q2 r̂
F
4π0 r 2
This is an equation describing an interaction,
requiring information about two interacting
charges q1 and q2 separated by ~r = ~r2 − ~r1 .
The electric field of a single charge q1 is obtained by dividing both
sides of the equation by q2
~
~ q = Fe = 1 q1 r̂
E
1
q2
4π0 r 2
Question: How does this fit the definition of a vector field?
The Electric Field
~
~ q = Fe = 1 q1 r̂
E
1
q2
4π0 r 2
By letting q2 = 1 (unit charge), this equation
describes the electric field vector generated by
charge q1 at the point ~r2 . It has units of
Newtons/Coulomb (i.e., force per unit
charge), and is a physical property of the
charge q1 . Remember that r̂ points radially
outward from the source point.
r2
~r
q1
~q
E
1
The Electric Field
~
~ q = Fe = 1 q1 r̂
E
1
q2
4π0 r 2
By letting q2 = 1 (unit charge), this equation
describes the electric field vector generated by
charge q1 at the point ~r2 . It has units of
Newtons/Coulomb (i.e., force per unit
charge), and is a physical property of the
charge q1 . Remember that r̂ points radially
outward from the source point.
r2
~q
E
1
~r
q1
Using the concept of the electric field, we can express the electric
force of q1 on some arbitrary charge Q as
~ Qq = E
~q Q
F
1
1
~ — i.e., how do we know
Question: What determines the sign of E
~
whether E points toward or away from the charge?
The Electric Field
The Superposition Principle
Electric field vectors add linearly — there are no quadratic or
~ generated by a
higher-order terms. To find the electric field E
collection of point charges at some point p, we simply calculate
the linear vector sum
~ =E
~q + E
~q + · · · =
E
1
2
N
X
i=1
this is known as the superpostion principle
~i ,
E
The Electric Field
The Superposition Principle
Electric field vectors add linearly — there are no quadratic or
~ generated by a
higher-order terms. To find the electric field E
collection of point charges at some point p, we simply calculate
the linear vector sum
~ =E
~q + E
~q + · · · =
E
1
2
N
X
~i ,
E
i=1
this is known as the superpostion principle
Question: Given positive charges of charge Q
at points r1 = h−1, −1, 0i , r2 = h0, 2, 0i, and
r3 = h−2, 0, 0i, and a negative charge of equal
magnitude Q at point r4 = h0, 0, 0i, determine
the electric field at the point p = h3, 1, 0i.
~.
Draw in the resultant vector for E
q2
q3
p
q4
q1
The Electric Field
Electric Dipoles
~p
-q
+q
s
An electric dipole is created by a pair of equal
but opposite charges ±q separated by a
distance s. The dipole moment ~p is the vector
that points from the negative to the positive
charge and has magnitude |~p | = qs
The Electric Field
Electric Dipoles
~p
-q
+q
s
An electric dipole is created by a pair of equal
but opposite charges ±q separated by a
distance s. The dipole moment ~p is the vector
that points from the negative to the positive
charge and has magnitude |~p | = qs
Electric dipoles are common in nature, and
they have the characteristic form seen to the
right. Calcuating the strength of the field at
any given point in space is, in general, not
trivial, but we can gain some understanding by
studying the field parallel to and perpendicular
to the dipole axis.
The Electric Field
Electric Dipoles
Derivation of electric field parallel to dipole axis.
r−
E
s
r+
−q +q
E−
E+
The Electric Field
Electric Dipoles
Derivation of electric field along perpedicular to dipole axis.
E+
E
E−
r−
−q
r+
y
s
+q
The Electric Field
Electric Dipoles
To appreciate the radial dependence of the dipole field, we make
an approximation that is very common in physics: assume r s.
In other words, evaluate the strength of the field at a point far
enough away that the dipole acts almost like a neutral point
~ ⊥ becomes
charge. The expression for the magnitude of E
1 qs
~ .
E⊥ =
4π0 r 3
The Electric Field
Electric Dipoles
To appreciate the radial dependence of the dipole field, we make
an approximation that is very common in physics: assume r s.
In other words, evaluate the strength of the field at a point far
enough away that the dipole acts almost like a neutral point
~ ⊥ becomes
charge. The expression for the magnitude of E
1 qs
~ .
E⊥ =
4π0 r 3
This is a general feature of dipole fields — they have a radial
dependence that goes like 1/r 3 for r s in every direction.
The Electric Field
Taylor Series Approximation
Approximations are done using Taylor series expansions. The general form
of the Taylor series for a function f (x) about some point x0 is given by
f 0 (x)|x=x0 (x − x0 ) f 00 (x)|x=x0 (x − x0 )2
+
+ ···
1!
2!
∞
X
f (n) (x)|x=x0
(x − x0 )n .
=
n!
f (x) = f (x0 ) +
n=0
The Electric Field
Taylor Series Approximation
Approximations are done using Taylor series expansions. The general form
of the Taylor series for a function f (x) about some point x0 is given by
f 0 (x)|x=x0 (x − x0 ) f 00 (x)|x=x0 (x − x0 )2
+
+ ···
1!
2!
∞
X
f (n) (x)|x=x0
(x − x0 )n .
=
n!
f (x) = f (x0 ) +
n=0
~ ⊥ |, pull the r 2 out of the
Exercise: Using the general formula for |E
−3/2
(·)
term in the denominator to obtain the expression
~ E⊥ =
1 qs
1
,
4π0 r 3 1 + s 2
2r
then Taylor expand the (1 + (s/2r )2 )−3/2 term by letting x = (s/2r )2
and evaluating at x0 = 0 (i.e., r s). Think about how this
approximation fails as r approaches s.