PHY2061 Enriched Physics 2 Lecture Notes
Electric Field
Electric Fields
Disclaimer: These lecture notes are not meant to replace the course textbook. The
content may be incomplete. Some topics may be unclear. These notes are only meant to
be a study aid and a supplement to your own notes. Please report any inaccuracies to the
professor.
2
Fields
1
?
How does one charged particle attract or repel another particle
at a distance? ⇒ Action at a distance
This is more prevalent than you think. Consider how you touch something and cause it to
move by applying a force. In fact, you never actually “touch” the object. The electrons in
the atoms of your hand start repelling the electrons in the atoms of the object. The
electrons never really touch! (they are point particles anyway…) So touching something
is Coulomb repulsion at a distance.
In physics, we say that the force exerted by one object onto another a distance away is
conveyed through a field.
Electric Field
The electric field is defined as the force acting on a positive test charge, per unit charge.
E≡
F
q0
q0 > 0
E points in direction of F
Units are thus N/C for the electric field.
It is similar to the gravitational field on the surface of the Earth for a test mass m0 :
F
m0
The electric field is a vector field. It has a magnitude and a direction. Another example
vector field is the wind distribution. Temperature distribution is an example of a scalar
field.
g=
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Electric Field
Field Lines
Consider a positive test charge immersed in the electric field of a positive charge.
q0
+
•
•
•
•
•
Lines indicate the direction of F if a test charge is placed there
Field lines point away from positive charge, and toward negative charge
Density of lines indicates the strength of the field
Field lines cannot cross. If they did, there would be more than one possible
direction for a particle to travel at a given point.
The lines themselves have no meaning. The field is present everywhere.
The magnitude of the electric field a distance r away from a point charge q:
E=
F
q
=K 2
q0
r
i.e. dropped the q0 from Coulomb’s Law
Electric Field of Several Point Charges
Apply the superposition principle. This principle states that the resulting electric field is
the sum of all fields, without any interference of one field upon another. It is generally
true for electromagnetism at least for fields that are not enormously strong.
For example, the total electric field at some point A from N charges is:
E A = ∑ i =1 EiA = ∑ i =1 K
N
N
qi
rˆiA
r12A
In other words, the fields from each point particle i a distance riA away from point A all
add together, without any field affecting another.
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Electric Field
Electric Dipole
Let’s consider the simplest multi-charge system: just two oppositely charged particles—a
system referred to as an electric dipole. Let’s calculate the electric field anywhere along
the x axis (result will actually apply anywhere in the x-z plane for y=0).
y
Let the top particle have charge q,
And the bottom particle charge −q.
The separation of charges is d.
x
Fig. from HRW 7/e
By symmetry, E must point in the −yˆ direction.
Apply the superposition principle to determine it, by adding the electric field from each
charge separately to get the net electric field.
Consider a point along the x axis. The distance from either charge is
r = x2 + ( d / 2)
2
The magnitude of the electric field from each charge separately is
q
q
E+ = E− = K 2 = K 2
2
r
x + ( d / 2)
But the vector force points in a direction away from the
positive charge and toward the negative one. i.e.
y
+q
θ
r
d
E + = E + sin θ xˆ − E + cos θ yˆ
x
−q
E − = − E − sin θ xˆ − E − cos θ yˆ
E−
E+
So we see explicitly how the x-component of the total field cancels, and we are left with:
ETOT = E+ + E− = −2 E+ cos θ yˆ
since E+ = E−
Now,
cos θ =
d /2
=
r
d /2
x2 + ( d / 2)
2
So,
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⎡
⎤⎡
q
d /2
⎢
ETOT = −2 ⎢ K 2
2⎥
2
⎢⎣ x + ( d / 2 ) ⎥⎦ ⎢ x + ( d / 2 )2
⎣
⇒ ETOT =
Kqd
⎡ x 2 + ( d / 2 )2 ⎤
⎣
⎦
By the binomial expansion,
2
⇒ ⎡1 + ( d / 2 x ) ⎤
⎣
⎦
⇒ ETOT K
−3/ 2
(1 + x )
1−
=
3/ 2
n
Electric Field
⎤
⎥ yˆ
⎥
⎦
1
Kqd
3
x ⎡1 + ( d / 2 x )2 ⎤ 3/ 2
⎣
⎦
1 + nx + " for small x.
3 d
1 for x d
2 2x
qd
x3
Let’s define the electric dipole moment p ≡ qd
⇒ ETOT K
p
r3
where we have generalized from anywhere along the x axis to anywhere in the x-z plane,
a distance r away from the dipole.
Note that this field falls off faster than a single point charge, r −2 , because the oppositesign charges contribute to the canceling of the electric field. This is a general behavior
for the dipole field strength in any direction. For example, along the y axis, one gets a
dipole field 2 times larger than the expression above (but all other terms the same).
Continuous Charge Distributions
Consider the electric field arising from an infinitesimal charge dq at a point a distance r
away:
dq
y
dE = K 2 rˆ
r
r
dE
dq
x
Projecting along each axis yields the following components of the infinitesimal electric
field:
dEx = dE ⋅ xˆ
dE y = dE ⋅ yˆ
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Electric Field
Field from a Continuous Line Charge
Now consider electric charge distributed uniformly along a 1-dimensional line from −L to
L along the z-axis.
+L
z
dq
r
θ
dE
y
dE’
x
dq'
-L
The charge per unit length is λ (units: C/m)
q
In terms of the total charge q, λ =
2L
So an infinitesimal length of the line, dz, has a charge dq = λdz.
Let’s calculate the electric field along the y-axis (any axis perpendicular to the line will
do).
By symmetry, only the ŷ component of the field (that is, a field pointing perpendicular to
the line) can survive when we integrate the field contributions from each infinitesimal
charge dq along the length.
E = E y = ∫ dE y = ∫ K
dq
cos θ
r2
where
r 2 = y2 + z2
cos θ =
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y
Important to remember to project
the field along the appropriate axis!
y2 + z2
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L
λ dz
−L
y +z
⇒ E =∫ K
= Kλ y∫
(y
2
⋅
y
y2 + z2
dz
L
−L
2
Electric Field
2
+ z2 )
3/ 2
z
1
= Kλ y 2
y ( y 2 + z 2 )1/ 2
L
−L
by integral tables
Appendix I of HRK
5/e, Appendix E of
HRW 7/e
Consider y L , alternatively that L → ∞
Then:
E=
2K λ
λ
=
y
2πε 0 y
where y can be replaced by the distance from the line charge to where the field is
measured.
We can consider another limit, the one when L y (i.e. very far away from a finite
length line charge). Then:
2λ L
E =K 2
y
But since q = λ ⋅ 2 L represents the total charge of the line, then we can re-write this as:
q
E =K 2
y
That is, when viewed far away, the field is just that due to a point charge.
Field of a Continuous Ring of Charge
dE′ z
θ
dE
dq'
r
R
dq
Let’s find the field along the z-axis only. Ring has radius R, charge per unit length λ.
By symmetry, only Ez is non-zero (the x-y components cancel)
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Electric Field
An infinitesimal length along the ring, ds, has a charge dq = λ ds.
It generates an electric field of:
dq
λ ds
dE = K 2 rˆ = K 2 rˆ
r
r
Only the z-component matters, so:
dEz = K
λ ds
r2
cos θ
and cosθ =
where r = R 2 + z 2
E = ∫ dEz = ∫ K
E=
(R
λ ds
2
+z
2
)
⋅
z
R2 + z 2
The integration is a line integral of the
function around the ring. ds represents an
infinitesimal length around this path.
z
R + z2
2
Kλz
ds
( R2 + z 2 ) ∫
since R and z are held constant when integrating around the ring.
But ∫ ds = 2π R , the circumference of the circle. We can see this explicitly by switching
∫ ds = ∫
to polar coordinates:
So:
E=
2π
0
Rdθ = 2π R
2π RK λ z
(R
2
+ z2 )
3/ 2
Now, 2π Rλ = q , the total charge of the ring. So we can re-write the field arising from a
ring of charge as:
E =K
(R
qz
2
+ z2 )
3/ 2
q
, the field of a point charge as expected when the ring
z2
becomes very small when observed far away.
If z R
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⇒ E →K
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Electric Field
Field of a Charged Disk
Fig. from HRW 7/e
Let’s find the field along the z-axis again. Disk has radius R,
and a charge per unit area of σ, such that an infinitesimal area has a charge of
dq = σ dA = σ dxdy
We can solve for the electric field by first determining the electric field arising from a
ring of infinitesimal thickness dr and radius r. (Note that this is a different definition of
“r” from the previous example).
In fact, we solved for the field of a charged ring already in the previous example! Just
make the following substitutions:
R→r
q → dq = σ dA = σ ( 2π rdr )
So the electric field of this thin ring points in the z-direction and has a magnitude of:
dE =
(r
z dq
2
+z
)
2 3/ 2
=K
zσ 2π rdr
(r
2
+ z2 )
3/ 2
The total field arising from the entire disk is then:
E = ∫ dE = Kσ zπ ∫
2rdr
R
0
(r
2
+ z2 )
3/ 2
Let’s change variables:
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Electric Field
u = (r2 + z2 )
du = 2rdr
Then:
R2 + z 2
E = Kσ zπ ∫ 2
z
u −3/ 2 du
R2 + z 2
E = Kσ zπ
u −1/ 2
1
−
2
z2
⎛
⎞
z
E = 2π Kσ ⎜1 −
⎟
2
2
R +z ⎠
⎝
If the disk becomes very large, R z , then we can simplify by the approximation:
E = 2π Kσ =
σ
2ε 0
This is the electric field from an infinite sheet of charge, and you can see that it is
independent of the distance, z, away from the sheet.
Now you should also be able to solve problems with non-uniform charge densities (i.e.
λ ( z ) , σ ( x, y ) , or ρ ( x, y, z ) . Only the integrals become slightly more complicated, but
the techniques are the same.
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