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The Electric Field
Read: Chapter 14
A field is a mathematical concept — it is a function that decribes the characteristics of every point in space. There are two
kinds:
• 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.
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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:
• 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.
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The Electric Field
The electric field is obtained from Coulomb’s law:
F~electric =
1 q1 q2
r̂
4π0 r2
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 r2
Question: How does this fit the definition of a vector field?
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The Electric Field
~
~ q = Fe = 1 q1 r̂
E
1
q2
4π0 r2
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
Using the concept of the electric field, we can express the electric force
of q1 on some arbitrary charge Q as
~q Q
F~Qq1 = E
1
~r
q1
~ — i.e., how do we know
Question: What determines the sign of E
~
whether E points toward or away from the charge?
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~q
E
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The Electric Field
The Superposition Principle
~ generated by
Electric field vectors add linearly — there are no quadratic or higher-order terms. To find the electric field E
a 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
p
q3
q4
q1
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The Electric Field
Electric Dipoles
p~
-q
+q
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
s
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.
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The Electric Field
Electric Dipoles
Derivation of electric field parallel to dipole axis.
r−
E
s
−q
r+
+q
E−
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E+
The Electric Field
Electric Dipoles
Derivation of electric field along perpedicular to dipole axis.
E+
E
E−
r−
r+
y
s
−q
+q
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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
~ ⊥ becomes
point charge. The expression for the magnitude of E
1 qs
~ .
E⊥ =
4π0 r3
This is a general feature of dipole fields — they have a radial dependence that goes like 1/r3 for r s in every direction.
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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!
n=0
f (x) = f (x0 ) +
~ ⊥ |, pull the r2 out of the (·)−3/2 term in the denominator to obtain the expression
Exercise: Using the general formula for |E
~ E⊥ =
1 qs
1
,
4π0 r3 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.
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