HS 2010
Physics III
Simon Lilly
Chapter 2: CONDUCTORS
1. Conductors and Insulators
The key difference between conductors and insulators is the mobility of the charges. In a
conductor, the individual charges (i.e. the electrons) are highly mobile: in an insulator, each
charge is essentially fixed in one location. The difference in the mobility of charges is about
a factor of 1020 between an insulator and a conductor.
Because charges can move freely in a conductor, two important things result:
a) E = 0 everywhere within a conductor. If not, charges would move along field lines, and
thereby reduce the field, eventually to zero.
b) From Gauss' Law, this implies that the charge density ρ within a conductor is everywhere
zero. The only place that there can be an excess of negative or positive charge is on the
surface.
In insulators, charges are fixed and it is easy, in principle, to calculate the φ(x,y,z) and E(x,y,z)
etc. In contrast, a conductor will adapt to any externally applied charges, fields, or potentials,
and we must be able to calculate the response of the conductor. Fortunately, although the
actual response is very difficult to follow, it is usually easy to see what the final state will be,
once the charges have stopped moving.
2. Conditions of a conductor
We can consider some system of one or more conductors, surrounded by vacuum. Each
conductor must satisfy three conditions (once it stablizes):
1.
The electrical potential φ will be the same at all points within and on the surface of
a given conductor, i.e. the kth conductor will exhibit φ = φk everywhere.
2.
The E-field near the surface of the conductor will be zero just inside the conductor,
and just outside will be perpedicular to surface with a value = σ/ε0, where is the
local charge density at that point of the surface.
3.
The total charge on the kth conductor will therefore be given by integrating over
the surface of that conductor:
Qk =
∫ σ da = ε ∫ E. da
0
Sk
€
Sk
It should be noted that general solutions for Qk, φk, E etc may or may not be analytically tractable.
Many problems must be solved numerically although there are a whole host of matehmatical tools to
solve this sort of problem.
3. The General Electrostatic Problem
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HS 2010
Physics III
Simon Lilly
Assuming that the conductors are in a vacuum, we need solutions to Laplace's equation
with a set of boundary conditions. The boundary conditions are imposed by the
situation (or by the experimenter) and can be one of three forms:
a) all the φk are defined: this is known as the Dirichlet problem;
b) all the Qk defined: this is known as Neumann boundary conditions;
c) mixed φ and Q defined: Note that it is easy (in setting questions but not experimentally!)
to over-define the problem, in which case no solutions will exist.
It is worth thinking how an experimenter could set either φ or Q of a conductor to a particular
value. To set Qk, we could simply transfer the required charge to the conductor using some
mechanical device. The conductor would then end up with some (unknown, but calculable)
potential φk, which would depend on the strength of the E fields established in the system by
the charges.
To set φk, we would connect the conductor to some potential source that has an inexhaustible
supply of charge, using another conductor (i.e. a wire). Some (unknown but calculable)
amount of charge will then flow onto or off the conductor until the E-fields in the system
result in the conductor reaching the desired potential, likely producing a non-zero net charge
on the conductor. A common example of this procedure is to connect the conductor to the
Earth (also known as "grounding" the conductor). The Earth may be safely assumed to have
an inexhaustible supply of charge at zero potential.
4. Uniqueness Theorem
The Uniqueness Theorem is very powerful. It states that there is at most one solution to the
general problem for a given set of applied boundary conditions. As noted above, there will be
no solution if the problem is over-constrained by the boundary conditions. The Uniqueness
Theorem is easy to prove:
Proof: Suppose φ(x,y,z) is a solution satisfying a particular set of boundary conditions. If
ψ(x,y,z) is also a solution, then because
is a linear equation, a combination w = φ−ψ
2
must be a solution to ∇ w = 0 , but with all boundary conditions equal to zero (because w
will by construction be equal to zero on the boundaries, because on the boundaries φ = ψ). If
w is zero on the boundaries and ∇ 2 w =0 , then it is easy to see that w = 0 everywhere, since
any local minimum or maximum in w would have non-zero second derivatives. Thus φ = ψ
€
everywhere. QED
€
There is an extensive mathematics
of solving Laplace's equation subject to the above
boundary conditions, which we will not explore in this introductory course. Rather we will
look at a few simple cases and learn a few tricks to solve simple cases. If we find any
solution (no matter how we do it) we can use the Uniqueness Theorem to establish that this
must be the only solution.
5. Faraday Cages
(a) Empty cavity enclosed by conductor
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HS 2010
Physics III
Simon Lilly
Cosider an empty cavity inside a conductor. The surface of the cavity is a conductor, it must
have constant potential. By the same reasoning as in Section 4, if
(because there is
no charge within the cavity) this implies that the potential must be constant throughout the
volume of the cavity. We can then conclude that there can be no E-field within such a cavity,
regardless of what is happening elsewhere, or outseide, the conductor.
(b) Conductor completely surrounding a charge
Now consider the case of a conductor surrounding a charge Q. We know that E is zero in the
conductor. Therefore from Gauss' Law, thetotal charge inside a surface within the conductor
must be zero. Charge will have rearranged itself within the conductor so that -Q lies on the
inner surface ("cancelling out" the charge Q), and the outer surface of the conductor will
therefore be charged with +Q (assuming the conductor had net zero charge - if the conductor
carried some net charge Q', the charge on the surface would be Q'+Q).
The distribution of +Q on the outer surface of the conductor will adjust itself to conform to
any external boundary conditions while satisfying the three conditions in Section 2 (especially
φ = constant throughout the conductor). The distribution of charge will be exactly the same as
if there was no inner charge and the conductor itself carried a net charge of Q. The
distribution of charge on the outer surface cannot be affected by the distribution of charge on
the inner surface, or by the location or motion of the inner charge Q in the cavity. To the
outside world, the conductor looks like it simply carries a net charge +Q.
This is the principle of the so-called Faraday cage. Enclosing a region by a conductor
(a) isolates it from all external influences, E = 0 everywhere in the region, regardless of what
is happening externally.
(b) shields the outside world from all knowledge of what is happening inside that region in
terms of the motion of charges.
Faraday cages are extensively used to shield unwanted electromagnetic radiation caused by
rapidly oscillating electromagnetic fields. In this case, the "cage" can have quite large holes
in it!
6. Some tricks
It is often possible to use some short-cuts to obtain a solution to Laplace's equation, and then
use the Uniqueness Theorem to establish that this is the solution. These tricks generally
involve recognizing that some other charge distribution (possibly extending beyond the
system in question) will have a field geometry that will satisfy the boundary conditions that
we have imposed.
(a) Mirror charges
Can we place extra charges outside of the system that will, by symmetry arguments,
reproduce the required boundary conditions on the boundaries?
Example 1:
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HS 2010
Physics III
Simon Lilly
Imagine a charge q located some distance h from an infinite conducting surface. Charge in
the conductor will rearrange itself to satisfy the conditions of a conductor. In particular a total
charge -q will be found on the surface of the conductor. But how is it distributed in the final
state?
The E-field must be perpendicular to the surface of the conductor, i.e. the surface of the
conductor must have constant potential φ. This is the boundary condition to be satisfied.
Now imagine a second, different, problem, in which there is no conductor, but two equal and
opposite charges q and -q, separated by a distance 2h. On symmetry grounds, we can argue
that the E-field on the plane that bisects the line between the charges (i.e. the "mid-plane"
between the charges) will be parallel to the line joining the charges and therefore
perpendicular to the mid-plane. The solution to φ, E etc in this second situation therefore
automatically satisfies the boundary conditions of the first problem. From the uniqueness
theorem, we know that φ, E as computed (quite easily) in the second case will be exactly the
same as in the first case in the vacuum region outside of the conductor, and that this will be
the correct solution.
E z,q + =
−qcosθ
4 π ε0 ( r 2 + h 2 )
E z,q − =
−qcosθ
4 π ε0 ( r 2 + h 2 )
E z = E z,q + + E z,q −
=
=
−2qcosθ
4 π ε0 ( r 2 + h 2 )
−q h
2π ε0 ( r 2 + h 2 )
32
Therefore the charge distribution on the conducting plane will be
σ = ε0 E z =
€
−q h
2π ( r 2 + h 2 )
32
As a check, the total charge on the sheet can be obtained by integrating:
Q
=€
∞
∫
σ 2πr dr
0
∞
=
∫
0
−q h
( r2 + h2)
32
dr
∞


qh


=
 ( r 2 + h 2 )1 2 

0
= −q
€
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HS 2010
Physics III
Simon Lilly
The additional imagined charge -q that we placed outside of the system so as to reproduce the
boundary conditions is often called a "mirror charge".
Example 2:
Another example will be explored further in a problem sheet. We want to know what
happens if we place an uncharged conducting sphere into a uniform external E-field.
In brief, it is easy to show that a "dipole" (formed by equal and opposite charges q
separated by a small distance l) p = ql, with the direction of p being from the negative
to positive charge, has a E-field and potential in polar coordinates given by
1
4 πε0
1
Eθ =
4 πε0
Eφ = 0
Er =
φ=
€
2p
cos θ
r3
p
sin θ
r3
(2.1)
1 p
cosθ
4 πε0 r 2
where θ is the angle to the direction of p. It is also easy to see that in a uniform field
E0, the potential will be given by
(2.2)
φ = E 0 r cosθ
Both of the above potentials of course satisfy Laplace's equation (away from the origin).
€
If we therefore place an uncharged conducting sphere of radius R in an external uniform
field, then the resulting potential on the surface of the sphere after charge redistribution within the sphere, must of course be constant. We cannot alter the external
field, but adding the above potentials will make φ = 0 at r = R if, and only if,
(2.3)
p = 4 πε0 E 0 R 3
€
The sum of these two fields, the original uniform one plus the one from a dipole of the
correct size given by (2.3) therefore (a) satisfies the boundary condition at the surface of
the conducting sphere and (b) satisfies Laplace's equation at all points outside of the
sphere. It is therefore the correct and only solution to the problem of the field around a
conducting sphere placed in a uniform field.
It is then easy to calculate σ around the surface of the sphere and so on.
These two particular examples are example of a general principle:
If we can find any system of charges such that an equipotential surface describes the surface of a
conductor, then that E(x,y,z) and φ(x,y,z) will also be the solution for the case of a conductor of
that shape.
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HS 2010
Physics III
Simon Lilly
7. Capacitance and Capacitors
Consider a conductor k of some size and shape and size, carrying a charge Q at some potential
, relative to = 0. From superposition, it is clear that φ and Q must be proportional, i.e.
(2.4)
Q = Cφ
€
C is called the Capacitance of the conductor. The SI units of capacitance are Coulombs/Volt,
which is given the name of Farad, i.e. F = CV-1.
It is more usual to have two (or more) conductors in close proximity to each other, which
carry equal and opposite charge, +Q and -Q and which have a potential difference V
(measured in Volts).
A simple case is two parallel disks of conducting material, or area A and separated by a small
distance, s. If we neglect edge effects (because s << A1/2), the field between these will be
given by
E =
(φ1 − φ 2 ) V
=
s
s
So the charge density on each plate will be
€
σ = ε0 E = ε0
V
s
and the total charge will be
€
Q = Aσ =
Aε0
V
s
so the capacitance of this system is
€
€
C=
Aε0
s
(2.5)
8. Energy stored in Capacitor
Consider the process of "charging" a Capacitor by adding charge to it. The work dW to add a
small increment of charge dq is
dW = φ dq
q
= dq
C
€
6
HS 2010
Physics III
Simon Lilly
so the total work is just
W =
1
C
Q
∫ q dq
0
2
=
€
Q
CV 2 QV
=
=
2C
2
2
(2.6)
The factor of 1/2 is of course the same factor of 1/2 that entered in our computation of the
total energy of a system of charges.
9. General capacitor systems
Consider a general set of k conductors. We have seen that, from uniqueness, specifying φ1,
φ2, φ3... φk defines E throughout the system, and hence also Q1, Q2, Q3... Qk, and vice versa.
We can therefore imagine various different states of this general system defined by φi and
associated Qi.
First, imagine the state obtained by setting all potentials except φ1 to zero. By superposition,
we can then write, in this instance:
Q1 = C11 φ1
Q2 = C21 φ1
Q3 = C31 φ1
.........
Qi = Ci1 φ1
€
Similarly, if we set all potentials to zero except φ2 we will get
Q1 = C12 φ 2
Qi = Ci2 φ 2
And so on.
€
Now, suppose that we have a general state characterized by a particular set of φi. By
superposition, the sum of the simple states considered above is a solution, i.e.
Q1 = C11φ1 + C12 φ 2 + C13 φ 3 + ..... + C1k φ k
Q2 = C21φ1 + C22 φ 2 + C23 φ 3 + .....+ C2k φ k
i.e. the Capacitance of this general system is a matrix quantity,
€
Qi = Cij φ j
(2.7)
€
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HS 2010
Physics III
Simon Lilly
whose elements can be determined through the above thought experiment of sequentially
setting all the potentials except one to zero.
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