Electrostatics
An electric charge exherts a force on
another electric charge, according to
Coulomb’s Law . . .
1 Q1 Q2
F =
4π0 r2
2
C
where 0 = 8.85 × 10−12
N m2
is the permittivity of free space,
and r is the distance between the two
charges.
This equation only gives the
magnitude of the force.
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The corresponding vector equation
gives both the magnitude and the
direction.
The force on charge Q2 located at P~2
due to Q1 at P~1 is . . .
F~ =
1 Q1 Q2 ~
R
3
4π0 R
~ = P~2 − P~1.
where R
When several charges are acting on
another charge, Coulomb’s Law obeys
the principle of superposition, provided
we use vector addition.
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Coulomb’s Law
Suppose that charge Q1 is located at
P~1. Then, a charge Q2 at P~2 would be
subjected to a force . . .
F~ = Q2
1 Q1 ~
R
3
4π0 R
~ = P~2 − P~1.
where R
This can be viewed as a two step
process.
Firstly, Q1 creates an electric field . . .
~ 2) =
E(P
1 Q1 ~
R
3
4π0 R
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(1)
~ then
Secondly, the electric field E
exherts a force on Q2 . . .
~ 2)
F~ = Q2E(P
~ is called the Electric Field Intensity.
E
(1) is also called Coulom’s Law.
When several charges are acting on
another charge, Coulomb’s Law obeys
the principle of superposition, provided
we use vector addition.
n
X
1
Qi ~
~
E(P ) =
Ri
3
4π0
R
i=1 i
~ i = P − Pi .
where R
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Continuous Distributions of Charge
In addition to isolated point charges,
we will consider charge as being
distributed along a line, over a surface,
and throughout a volume.
Point Charges Qi
Line Charges λdl
Surface Charges σda
Volume Charges ρdτ
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To work out the electric field due to
continuous distributions of charge,
view them as being composed of many
tiny elements. Approximate each
element as a point charge, and sum.
In the limit, as the size of the elements
tends to zero, the summation becomes
an integral.
42
Example
A straight line segment of length 2L
carries a uniform electric charge of
λ C/m. Find the electric field a
distance Z above the midpoint of the
line segment.
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1 2λdx
dE =
cos θ
2
4π0 R
z
z
cos θ = = √
R
x2 + z 2
Z L
Z L
1 2λ
cos θdx
dE = E =
2
0
0 4π0 R
Z L
λ
1
=
cos θdx
2
2π0 0 R
Z L
λz
1
=
dx
2π0 0 (x2 + z 2)3/2
Since
d
x
1
√
=
2
2
2
dx z x + z
(x2 + z 2)3/2
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L
2
λz
x
√
E=
2π0 z 2 x2 + z 0
λ
L
√
=
2π0 z L2 + z 2
Hence,
λ
L
~k
~
√
E=
2π0 z L2 + z 2
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Two special cases, . . .
If z >> L, then
2λL 1
Q
E=
=
2
4π0 z
4π0z 2
as expected, since it resembles a point
charge of 2λL.
If L → ∞, then
λ
=
2π0z
So this is the field due to an infinite
line of charge.
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The Curl of E
Consider moving a charge along a
~
closed path, in an electric field E.
Since work done = force * distance
Work Done =
I
~
~ • dl
E
But the physical situation at the end is
the same as at the beginning. So the
net work done must be zero, by the
law of conservation of energy.
I
~ =0
~ • dl
E
for any path
~ =0
By Theorem 1, ∇ × E
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The Div of E
~ due to a
Consider the electric field E
charge Q located at some point P~ .
~ over the surface of the
Integrate E
sphere of radius r centered on the
charge.
~ the directions of E
~
~ • da,
~ and da
In E
~ = Eda. Also,
~ • da
are the same. So E
by spherical symmetry, E is constant
on the surface of the sphere. So
Z
Z
Z
~ = Eda = E da
~ • da
E
R
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The area of a sphere is 4πr2, so
Z
~ = E4πr2
~ • da
E
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From Coulomb’s Law,
1 Q
E=
4π0 r2
Substitute for E,
Z
1
Q
2
~ =
~ • da
E
4πr
4π0 r2
Q
=
0
This holds much more generally.
For any closed surface,
0
Z
~ = charge enclosed by the surface
~ da
E•
This is Gauss’ Law.
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When the charge is continuously
distributed over a volume,
charge enclosed by the surface
Z
=
ρ dτ
vol
One of the fundamental theorems of
vector calculus was the Divergence
Theorem,
Z Z
~
~ dτ =
~ • da
∇•A
A
vol
sur
Applying these two facts to the
integral form of Gauss’ Law, . . .
0
Z
~ = charge enclosed by the surface
~ da
E•
sur
Z
Z ~ dτ = 0
∇•E
ρ dτ
=
vol
vol
Since this holds for any volume, the
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two integrands must be equal,
ρ
~
=∇•E =
0
This is the differential form of Gauss’
Law.
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Application of Gauss’ Law
Gauss’ Law is useful for computing the
electric field, especially when the
situation has a high degree of
symmetry.
Example
Determine the electric field outside a
uniformly charges sphere of radius a,
centered at the origin, where the total
charge is Q Coulombs.
Consider a spherical surface of radius
r > a, a so-called Gaussian surface.
Z
1
~
~
E • da = Q
0
sur
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~ the directions of E
~ • da,
~ and
In sur E
~ = Eda.
~ are the same. So E
~ • da
da
Also, by spherical symmetry, E is
constant on the surface of the sphere.
So
Z
Z
Z
~ =
~ • da
E
Eda = E
da
R
sur
sur
sur
2, so
The area of
a
sphere
is
4πr
Z
~ = E4πr2
~ • da
E
1
= Q
0
so that
1 Q
E=
4π0 r2
By spherical symmetry, the direction
~ at a point P~ is the same as that
of E
of P~ , so
1 Q~
~
P
E=
3
4π0 P
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This looks exactly like Coulomb’s Law.
So a uniformly charged sphere looks
exactly like a point charge, when
outside the sphere.
55
Example
Consider two parallel infinite planes a
distance d apart.
Suppose the left plane carries a charge
of +σC/m2 and the right plane carries
a charge of −σC/m2. What is the
resulting electric field?
By symmetry, the electric fields due to
each half plane are in the (plus or
minus) x direction.
Consider a Gaussian surface, say a
cube, where each face of the cube has
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area A.
57
Using Gauss’ Law,
Z
1
~
~
E • da = Qenc
0
=
Z
=E
Eda
Z
da
= E2A since two faces contribute
σA
=
0
σ
E=
20
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Adding the contribution from each
plane,
E = 0 to the left of both planes
E = 0 to the right of both planes
σ~
~
E = i between the planes
(2)
0
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Electric Potential
~ = 0, the electric field E
~
Since ∇ × E
must be the grad of something (by
Theorem 1).
~ = −∇V
E
V is called the Electric Potential.
It is a scalar field.
The minus is a notational convention.
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Gauss’ Law in terms of V
Gauss’ Law can be rewritten in terms
of V .
ρ
~
∇ • E = = −∇ • ∇V
0
Since
∇ • ∇ ∂ ~∂ ~ ∂
∂ ~∂ ~ ∂
~
~
• i +j +k
= i +j +k
∂x
∂y
∂z
∂x
∂y
∂z
∂2
∂2
∂2
= 2+ 2+ 2
∂x
∂y
∂z
This is usually abbreviated to ∇2, and
is called the Laplacian operator.
Hence, ∇2V = − ρ
0
This is called Poisson’s Equation.
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Potential due to a point charge
The electric potential a distance R
from a point charge of Q Coulomb’s is
1 Q
V =
4π0 R
To see this, note that
1
Q
p
V =
4π0 x2 + y 2 + z 2
∂V
1
Q
1
=
− (2x)
3/2
2
2
2
∂x 4π0 (x + y + z )
2
1 Q
(−x)
=
3
4π0 R
∂V . So
and similarly for ∂V
and
∂y
∂z
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~ = −∇V
E
∂V ~ ∂V ~ ∂V ~
=− i−
j−
k
∂x
∂y
∂z
1 Q ~
~j − z~k
−x
i
−
y
=−
4π0 R3
~ =
=E
Q ~
R
3
4π0R
as expected.
63
Potential and Voltage
~ = −∇V
Integrate both sides of E
along a path from point a to point b,
Z b
a
~ =−
~ • dl
E
Z b
a
~
∇V • dl
From the gradient theorem,
= − (V (b) − V (a)) = V (a) − V (b)
This is the potential difference between
points a and b,
It is just the usual notion of voltage, as
in circuits.
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Summary
~ = 0, E
~ is the grad of
Since ∇ × E
something.
~ = −∇V
E
Gauss’ Law in terms of V is Poisson’s
Equation
ρ
2
∇ V =−
0
The potential due to a point charge is
Q
V =
4π0R
This is Coulomb’s Law in terms of V
The potential difference (voltage)
between a and b is
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Z b
a
~ = V (a) − V (b)
~ • dl
E
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So the voltage between two points is
V (a) − V (b) =
Z b
a
~
~ • dl
E
If we integrate around a closed path,
we find that
V (a) − V (a) = 0 =
I
~
~ • dl
E
and this is just Kirchhoff’s Voltage
Law.
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Potential and Power
Consider moving a charge Q in an
~ along a path from a to
electric field E
b.
Work Done = Force ∗ Distance
=
Z b
a
=Q
~
F~ • dl
Z b
~
~ • dl
E
Z b
~
∇V • dl
a
= −Q
a
Using the gradient theorem,
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= Q (V (a) − V (b))
So the voltage between two points is
the work done per coulomb when
moving charge from a to b.
Suppose charge is moving continuously
from a to b (i.e. so many Coulomb’s
per second) and the voltage between a
and b is constant. Then,
Work = QVab
d
Work = Power
dt
dQ
= Vab
= VabI
dt
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as in circuit theory.
70
The Capacitor
Consider two plates of area A a
distance d apart. Suppose the planes
carry uniform charges of plus and
minus σC/m2.
The total charge on each plate is then
Q = σA
We suppose that the plates are large
enough that the formula obtained
above for the infinite planes is a
reasonable approximation. From
eqn. (2) above, between the plates we
have
σ
E=
0
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Q
=
A0
V =
Z
~
~ • dl
E
=
Z
Edl
=E
Z
dl
= Ed
Qd
=
A0
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The quantities d, A and 0 are fixed in
the physical situation, they are
constant.
A0
Let C =
d
⇒Q=VC
dQ
dV
⇒
=C
dt
dt
dV
I=C
dt
. . . the usual equation for a capacitor.
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Conductors
For an ideal conductor,
~ =0
(a) Inside the conductor, E
(b) Inside the conductor, ρ = 0
(c) Any charge resides on the surface
(d) V is constant throughout the
conductor
~ is
(e) Just outside the conductor, E
perpendicular to the surface
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