Lecture 6
Capacitance
In this lecture you will learn:
• Capacitance
• Capacitance of Some Simple Structures
• Some Simple Solutions of Laplace’s Equation
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance
Consider two metallic objects connected by a voltage source
+
+
+
+ + + + + +
+
total charge +Q
+
+
++
+
V
-
-
-
-
- - - -
- -
- -
-
total charge -Q
• The total charge on both metal objects have the same magnitude but opposite sign
• The charge is proportional to the applied voltage: Q ∝ V
• The constant of proportionality is called the capacitance C :
Q = CV
• Capacitance has units Coulombs/Volts or Farads
• More generally, capacitance is also defined as: C =
dQ
dV
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
1
Capacitance of Parallel Plates
From previous lectures:
area = A
σ L = εo
Ex =
V
d
φ=0
φ=V
+
+
+
+
+
+
+
+
V
d
-
d
+
σ R = −ε o
V
d
V
Q=
C=
εo A
d
V
dQ Q ε o A
= =
dV V
d
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Concentric Cylinders
From previous lectures:
+
E r (r ) = −
+
V
⎛b⎞
r ln⎜ ⎟
⎝a⎠
b
+
+
-- - a---
+
σ out = ε o
V
b ln(b a )
σ in = −ε o
V
a ln(b a )
+
+
+
+
V
-
Since the structure is infinite in the z-direction (i.e. in the direction coming out of
the slide) one can only talk about capacitance per unit length (units: Farads/m)
Q = charge per unit length = ( 2π b ) ε o
C = capacitanc e per unit length =
V
b ln(b a )
dQ Q
2π ε o
= =
dV V ln(b a )
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
2
Capacitance of Parallel Metal Wires - I
Consider two infinitely long metal wires connected by a voltage source
(the wires are infinitely long in the z-direction)
y
d
x
metal wires of radius a
d >> a
+
V
-
Need to find the capacitance per unit length between them under the
assumption that d >> a
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Line Charge Revisited
Start from the line charge ………….
r
φ (r ) =
λ
2π ε o
⎛r ⎞
ln⎜ o ⎟
⎝r ⎠
r
r
y
x
+λ Coulombs/m
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
3
Line Charge Dipole Revisited
Then consider the line dipole……………
r
r
y
r+
rd
x
+λ Coulombs/m
-λ Coulombs/m
r
φ (r ) =
λ
2π ε o
⎛r ⎞
ln⎜⎜ − ⎟⎟
⎝ r+ ⎠
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - II
Look at the E-fields of the line dipole:
y
x
+λ Coulombs/m
-λ Coulombs/m
d
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
4
Capacitance of Parallel Metal Wires - III
Draw imaginary cylinders of radius a around each wire:
y
Note that the E-field
lines are approximately
(but not exactly) normal
to the surface of the
cylinders
(as long as d >> a)
x
+λ Coulombs/m
-λ Coulombs/m
d
The total electric flux per unit length coming out of the left cylinder is +λ
The total electric flux per unit length coming out of the right cylinder is -λ
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - IV
Now if one replaces the cylinders and the line charges with metal wires of radius a
that carry +λ and –λ coulombs of charge per unit length then the E-field outside
remains approximately unchanged y
r
r
r+
rd >> a
x
- --
+
+ +
+
+λ Coulombs/m
-λ Coulombs/m
d
The total charge carried by the left wire is +λ
The total charge carried by the right wire is -λ
r
φ (r ) =
λ
2π ε o
⎛r ⎞
ln⎜⎜ − ⎟⎟
⎝ r+ ⎠
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
5
Capacitance of Parallel Metal Wires - V
r
r
Now come back to the original problem with a voltage source:
y
r+
rd
- --
+
+ +
+
x
metal wires of radius a
d >> a
+
r
φ (r ) =
⇒
⇒
⎛r ⎞
λ
ln⎜ − ⎟
2π ε o ⎜⎝ r+ ⎟⎠
-
V
λ
⎛d ⎞ V
ln⎜ ⎟ =
⎝a⎠ 2
λ
V
⎛a⎞
Potential at the right cylinder =
ln⎜ ⎟ = −
2π ε o ⎝ d ⎠
2
Potential at the left cylinder =
2π ε o
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - VI
Finally calculate the capacitance per unit length:
y
r
r
r+
rd
- --
+
+ +
+
x
metal wires of radius a
d >> a
+
Potential difference = V =
V
-
λ
⎛d ⎞
ln⎜ ⎟
π εo ⎝ a ⎠
Capacitance per unit length = C =
dQ Q
= =
dV V
λ
π εo
=
λ
d
ln(d a ) ln⎛⎜ ⎞⎟
π εo
⎝a⎠
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
6
Solutions of Laplace’s Equation in Spherical Coordinates - I
r
Following are the solutions of ∇ 2φ (r ) = 0
Spherically Symmetric Point Charge Solution:
r
φ (r ) =
+q
A
r
z
A Constant Uniform Electric Field Solution
r
θ
φ (r ) = −Eo r cos(θ ) = −Eo z
r r
r
⇒ E (r ) = −∇ φ (r ) = Eo zˆ
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Solutions of Laplace’s Equation in Spherical Coordinates - II
A Point Charge Dipole Oriented Along z-Axis Solution:
r
φ (r ) = A
cos(θ )
r2
z
+q
θ
-q
A Constant Potential Solution (Trivial Solution):
r
φ (r ) = A
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
7
Solutions of Laplace’s Equation in Cylindrical Coordinates - I
y
r
Following are the solutions of ∇ 2φ (r ) = 0
+λ
Cylindrically Symmetric Line Charge Solution:
r
φ (r ) = A ln(r )
x
y
A Constant Uniform Electric Field Solution
r
φ (r ) = −Eo r cos(φ ) = −Eo x
φ
r r
r
⇒ E (r ) = −∇ φ (r ) = Eo xˆ
x
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Solutions of Laplace’s Equation in Cylindrical Coordinates - II
y
A Line Dipole Oriented Along x-Axis Solution:
r
φ (r ) = A
cos(φ )
r
φ
-λ
+λ
x
d
r
φ (r ) =
⎛
d
⎞
⎝
2
⎠
⎜ r + cos(φ ) ⎟
⎛r ⎞
λ
λ
λ d cos(φ )
2
⎟≈
ln⎜⎜ − ⎟⎟ ≈
ln⎜
d
2π ε o ⎝ r+ ⎠ 2π ε o ⎜ r − cos(φ ) ⎟ 2π ε o
r
⎜
⎟
A Constant Potential Solution (Trivial Solution):
r
φ (r ) = A
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
8