Download 2.11 Capacitance

2.11 Capacitance
One application of Gauss’s law is the analysis of capacitors. The definition of capacitance is
C=
Q
V
(2.121)
with units of Farads = Coulombs/Volt, where V is the potential difference between the conductor with charge
+Q and the conductor with charge −Q.
To compute the capacitance of a given structure, the basic procedure is to assume a charge Q on the positive
conductor, then use Gauss’s law to find the electric field and the voltage between the conductors:
1. Assume a charge Q on the conductors
2. Find D from Gauss’s Law
3. Find E = D/².
4. Find V from E
5. Find the capacitance from C = Q/V
2.11.1
Parallel Plate Capacitor
z
Area = LxLy
+Q
d
E
-Q
y
Figure 2.11: Parallel plate capacitor.
1. Assume the charge on the upper plate is Q. We also assume that the charge will evenly distribute itself
over the conducting plates, and neglect fringing fields near the edges of the plates.
2. Use Gauss’s law to find D. The Gaussian surface is a cube that contains the top plate. Since we are
neglecting fringing fields outside the plates, the only contribution to the surface integral is from the part of
the Gaussian surface that is between the plates:
I
D · ds = Q
Z Ly Z Lx
Do dxdy = Do Lx Ly
0
0
Q
Q
=
Lx Ly
A
Q
D = − ẑ
A
Do =
77
3. The electric field intensity is
E=−
Q
ẑ
²A
4. The potential between the two plates is
Z
d
V =−
−
0
5. Finally, the capacitance is
C=
Q
Qd
ẑ · ẑdz =
²A
²A
Q
²A
=
V
d
(2.122)
Question: Why does larger ² increase the capacitance?
Differential Forms
The electric flux density two-form is D = −Do dx ∧ dy, so
I
D = Q
Z Ly Z Lx
Do dx ∧ dy = Do Lx Ly
0
0
Q
Q
=
Lx Ly
A
Q
D = − dx ∧ dy
A
Do =
The electric field intensity one-form is
E=
Q
1
? D = − dz
²
²A
The potential between the two plates is
Z
V =−
d
−
0
Q
Qd
dz =
²A
²A
and we obtain the same result, C = ²A/d, for the capacitance.
2.11.2
Energy
Electromagnetic fields store energy. For electric fields, the energy density is
1
we = E · D
2
(2.123)
which has units of Joules/m3 . With differential forms, this becomes a three-form, (1/2)E ∧ D. The total
energy in a volume V is
Z
1
We =
E∧D
(2.124)
2 V
78
Figure 2.12: Tubes of the two-form D intersect with surfaces of the one-form E to produce boxes, representing energy stored in the capacitor.
For the parallel plate capacitor,
We =
=
=
=
2.11.3
Z Z
Z
1 d Ly Lx Q2
dx ∧ dy ∧ dz
2 0 0
²A2
0
1 Q2
Ad
2 ²A2
1 C 2V 2d
2 ²A
1
CV 2
2
(2.125)
Spherical Shell Capacitor
y
-Q
a +Q
x
ε
b
Figure 2.13: Spherical shell capacitor geometry. Solid lines represent conducting spheres of radius a and b.
79
The charge will evenly distribute itself over the conducting spherical shells.
I
Gauss’s law:
D · ds = Q
Z 2π Z π
Do R2 sin θ dθdφ = Q
0
(2.127)
0
4πR2 Do = Q
Do =
Flux density: D =
Field intensity: E =
Potential: Vab = Va − Vb =
=
Capacitance: C =
2.11.4
(2.126)
Q
4πR2
Q
R̂
4πR2
Q
R̂
2
4π²R
Z a
Q
−
R̂ · R̂dR
2
b 4π²R
¯
µ
¶
Q ¯¯a
Q 1 1
=
−
4π²R ¯b
4π² a b
Q
4π²
= 1 1
Vab
a − b
(2.128)
(2.129)
(2.130)
(2.131)
(2.132)
(2.133)
(2.134)
Electric Field Boundary Conditions
At a boundary between two different dielectrics, the tangential components of E have to be equal on either
side of the interface:
E1t = E2t
(2.135)
If this were not the case, then the closed path integral of E would not be zero around a small closed path
at the boundary. Similarly, the normal components of D have to be equal on either side of the boundary, or
else Gauss’s law would not give zero charge inside a small closed surface at the boundary:
D1n = D2n
(2.136)
If there were charge stored on the boundary, then this changes to D1n − D2n = ρs , where ρs is the surface
charge density. (A dielectrics will in general have a nonzero surface charge at the boundary due to polarization by an external electric field, but we have already taken this into account when we changed the value of
the dielectric constant ² in Eq. (2.119), so we do not include it on the right-hand side of Gauss’s law.)
A2
A2
ε1
ε2
Figure 2.14: Parallel plate capacitor with two different dielectrics.
We can use this to analyze a parallel plate capacitor filled with two different dielectrics, with constants ²1
and ²2 , each occupying half of the area of the plates. Because the electric field intensity E is tangential to
80
the interface between the dielectrics, E must be the same in both dielectrics. So, the flux density is of the
form
½
−²1 Eo ẑ region 1
(2.137)
D=
−²2 Eo ẑ region 2
Applying Gauss’s law,
Z
Z
²1 Eo ẑ · dxdyẑ +
A/2
A/2
²2 Eo ẑ · dxdyẑ = ²1 Eo A/2 + ²2 Eo A/2 = Q
Solving for Eo ,
Eo =
2Q
A(²1 + ²2 )
The potential is
Z
V
= −
E · d`
Z
d
−2Q
ẑ · dz ẑ
A(²
1 + ²2 )
0
2Qd
A(²1 + ²2 )
= −
=
and the capacitance is
A(²1 + ²2 )
(2.138)
2d
If the dielectric layers were layered one on top of the other instead of side by side, then the flux density
would have to be the same in both materials, and the solution would change correspondingly.
C=
81