Mathematics of the Faraday Cage
Jon Chapman, Dave Hewett and Nick Trefethen
Mathematical Institute
University of Oxford
JBK Meeting, Feb 2017
Chapman
Mathematics of the Faraday Cage
Faraday Cage
On a metal shell the potential is
constant.
Thus the potential inside is constant, so the eld inside is zero.
Faraday observed in 1836 that the
same holds (nearly) for a metal
mesh.
This eect is used in shielding (e.g.
microwave oven).
Photo from Museum of Science, Boston
Chapman
Mathematics of the Faraday Cage
Analogy
I got interested in this problem because of Trefethen.
He got interested because of an analogy.
The trapezoidal quadrature rule is exponentially accurate for
periodic analytic integrands. Perhaps the mathematics of the
Faraday cage is analogous, and the shielding eect is exponential?
In this analogy, trapezoidal quadrature points corresponds to the
cross-sections of wires of a Faraday cage.
Chapman
Mathematics of the Faraday Cage
Electrostatic case
Feynman's
gives one of the few treatments,
appearing to conrm the intuition that the eect is exponential.
Lecture Notes
The intuition is wrong.
The shielding is much weaker, just linear in the mesh spacing.
Chapman
Mathematics of the Faraday Cage
Outline
1. Electrostatic Faraday cage: numerics
2. Homogenised model via multiple scales
3. Point charges model via constrained
quadratic optimisation
Chapman
Mathematics of the Faraday Cage
1. Electrostatic Faraday cage: numerics
∇2 φ
point charge
•z
=0
s
external eld
on n disks
of radius r
φ = φ0
log |z − zs |
Boundary conditions
φ = φ0 = unknown constant on disks
φ = log |z − zs | + O(1) as z → zs
φ = log |z| + o(1)
as
z → zs
radius
r
at the
nth
roots of unity
(zero total charge on the disks)
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Mathematics of the Faraday Cage
Dependence on
r:
logarithmic (n
Chapman
= 12)
Mathematics of the Faraday Cage
Dependence on
n:
linear (r
Chapman
= 0.01)
Mathematics of the Faraday Cage
2. Multiple scales analysis
1
Inner region (x, y) = (rξ, rη).
φξξ + φηη = 0
on ξ
Complex potential
φ = φ0
r = δ
for ξ
2
2
+ η 2 > 1,
+ η 2 = 1.
w = φ+iψ = φ0 +A log ζ,
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ζ = ξ+iη.
Mathematics of the Faraday Cage
2. Multiple scales analysis
1
Middle region (x, y) = (X, X).
φXX +φY Y = 2πCδ(X, Y )
for |Y | < 1/2,
on Y = ±1/2.
C = charge on the wire.
φY = 0
w = B + C log (sinh (πZ)) .
r = δ
Inner region (x, y) = (rξ, rη).
w = φ0 + A log ζ.
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Mathematics of the Faraday Cage
Matching
1
Middle region (x, y) = (X, X).
w ∼ B + C log(πZ)
r = δ
as Z → 0.
Inner region (x, y) = (rξ, rη).
w ∼ φ0 + A log 1/δ + A log Z.
Chapman
Mathematics of the Faraday Cage
Matching
Matching gives
1
φ0 + A log 1/δ = B + C log π,
A = C.
Middle region (x, y) = (X, X).
w ∼ B + C log(πZ)
r = δ
as Z → 0.
Inner region (x, y) = (rξ, rη).
w ∼ φ0 + A log 1/δ + A log Z.
Chapman
Mathematics of the Faraday Cage
Matching
1
Middle region (x, y) = (X, X).
φ ∼ ±CπX−C log 2+B
as X → ±∞.
r = δ
Chapman
Mathematics of the Faraday Cage
Matching
Outer region
1
Matching gives
∂φ
∂n
Γ+
=
Γ−
2πC
,
φ(Γ) = B−C log 2.
Middle region (x, y) = (X, X).
φ ∼ ±CπX−C log 2+B
as X → ±∞.
r = δ
Chapman
Mathematics of the Faraday Cage
2. Multiple scales analysis
Eliminating A, B and C gives
∂φ
2π
n
= α(φ − φ ) on Γ,
α=
=
.
∂n
log(/(2πr))
log(1/(rn))
α(φ − φ ) = charge density.
α = capacitance per unit length.
Solution of homogenised problem has
+
0
−
0
|∇φ(0)| =
2
.
(2 + α)|zs |
Eective screening requires α 1, i.e.
r
e−n
.
n
Thus when there are many wires they can be very thin.
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Mathematics of the Faraday Cage
Dependence on
r (n = 12)
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Mathematics of the Faraday Cage
Dependence on
n (r = 0.01)
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Mathematics of the Faraday Cage
3. Point charges via constrained optimisation
A disk at z of radius r 1 behaves like a point charge of some
strength q . The challenge for Faraday cage analysis is, what is q ?
We can nd these charges by energy minimisation.
Key observation: a charged disk has a self-energy of − q log r.
It takes work to force charge onto a small disk.
k
k
k
1 2
2 k
Chapman
Mathematics of the Faraday Cage
Find n charges q , summing to 0, that minimise the total energy:
k
E(q) = −
n
n X
n
X
X
1X 2
qk log r−
qj qk log |zk −zj |−
qk log |zk −zs |
2
k=1
k=1 j>k
k=1
SELF-ENERGY
MUTUAL ENERGY
ENERGY OF DISKS IN
OF DISKS
BETWEEN DISKS
EXTERNAL FIELD
This is a constrained quadratic programming problem
minimise E(q) = 21 q Aq − f q, c q = 0.
T
Chapman
T
T
Mathematics of the Faraday Cage
Dependence on
n (r = 0.01)
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Mathematics of the Faraday Cage
Back to trapezoidal rule
Small charged disks can be approximated by point charges to high
accuracy.
A charge distribution on a smooth curve can be approximated by
point samples to high accuracythe trapezoidal rule.
So why doesn't the Faraday cage provide strong screening?
Because it provides an exponentially good approximation not to the
zero eld, but to the eld associated with the homogenized
boundary condition.
Chapman
Mathematics of the Faraday Cage
Chapman
Mathematics of the Faraday Cage