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ECE 303 – Fall 2005 – Farhan Rana – Cornell University
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
CapacitanceConsider two metallic objects connected by a voltage source
V+
-
++ + + + + + + + + ++
++
--
- - - - - - - - - --
-
• The total charge on both metal objects have the same magnitude but opposite sign
• The charge is proportional to the applied voltage: VQ ∝
total charge +Q
total charge -Q
• The constant of proportionality is called the capacitance C : CVQ =
• Capacitance has units Coulombs/Volts or Farads
• More generally, capacitance is also defined as:dVdQC =
2
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Plates
V+ -
φ = V φ = 0area = AdVEx =
dV
oL εσ =dV
oR εσ −=
++++++++
--------
dA
VQ
dVdQC
Vd
AQ
o
o
ε
ε
===
=
d
From previous lectures:
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Concentric Cylinders
+
+
++
+
+
+
+
---
- ----
V+ -
( )⎟⎠⎞
⎜⎝⎛
−=
abr
VrErln
( )abbV
oout lnεσ =
( )abaV
oin lnεσ −=
( ) ( )
( )abVQ
dVdQC
abbVbQ
o
o
ln2 lengthunit per ecapacitanc
ln2 lengthunit per charge
επ
επ
====
==
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)
ba
From previous lectures:
3
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - IConsider two infinitely long metal wires connected by a voltage source(the wires are infinitely long in the z-direction)
d
x
y
metal wires of radius a
V+ -
Need to find the capacitance per unit length between them under the assumption that d >> a
d >> a
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Line Charge Revisited
Start from the line charge ………….
( ) ⎟⎠⎞
⎜⎝⎛=
rrr o
oln
2 επλφ
r
x
y
+λ Coulombs/m
rr
4
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Line Charge Dipole Revisited
Then consider the line dipole……………
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
−
rrr
oln
2 επλφ
r
d
x
y
+λ Coulombs/m -λ Coulombs/m
r+
r-
rr
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - II
Look at the E-fields of the line dipole:
d
x
y
+λ Coulombs/m-λ Coulombs/m
5
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - III
Draw imaginary cylinders of radius a around each wire:
x
y
+λ Coulombs/m -λ Coulombs/m
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
Note that the E-field lines are approximately (but not exactly) normal to the surface of the cylinders (as long as d >> a)
d
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - IVNow if one replaces the cylinders and the line charges with metal wires of radius athat carry +λ and –λ coulombs of charge per unit length then the E-field outside remains approximately unchanged
x
y
+λ Coulombs/m -λ Coulombs/m
The total charge carried by the left wire is +λ
The total charge carried by the right wire is -λ
d
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
−
rrr
oln
2 επλφ
r
r+
r-
rr
+++ +-- --
d >> a
6
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - VNow come back to the original problem with a voltage source:
d
x
y
metal wires of radius a
V+ -
r+
r-
rr
( )
2ln
2 cylinderright theat Potential
2ln
2 cylinderleft theat Potential
ln2
Vda
Vad
rrr
o
o
o
−=⎟⎠⎞
⎜⎝⎛=⇒
=⎟⎠⎞
⎜⎝⎛=⇒
⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
−
επλεπ
λ
επλφ
r
++ ++ - ---
d >> a
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Capacitance of Parallel Metal Wires - VI
d
x
y
metal wires of radius a
V+ -
r+
r-
rr
⎟⎠⎞
⎜⎝⎛==
adV
oln difference Potential
επλ
( ) ⎟⎠⎞
⎜⎝⎛
=====
adadV
QdVdQC o
olnln
lengthunit per eCapacitanc επ
επλ
λ
++ ++ - ---
Finally calculate the capacitance per unit length:
d >> a
7
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Solutions of Laplace’s Equation in Spherical Coordinates - I
( )rAr =
rφ
Spherically Symmetric Point Charge Solution:
Following are the solutions of ( ) 02 =∇ rr
φ
+q
A Constant Uniform Electric Field Solution
( ) ( ) zErEr oo −=−= θφ cosr
z
θ
( ) ( ) zErrE o ˆ=−∇=⇒rrr
φ
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:
( ) ( )2
cosr
Ar θφ =r
z
θ+q
-q
A Constant Potential Solution (Trivial Solution):
( ) Ar =r
φ
8
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Solutions of Laplace’s Equation in Cylindrical Coordinates - I
( ) ( )rAr ln=r
φ
Cylindrically Symmetric Line Charge Solution:
Following are the solutions of ( ) 02 =∇ rr
φ
+λ
A Constant Uniform Electric Field Solution
( ) ( ) xErEr oo −=−= φφ cosr
y
φ
( ) ( ) xErrE o ˆ=−∇=⇒rrr
φ
x
y
x
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Solutions of Laplace’s Equation in Cylindrical Coordinates - II
A Line Dipole Oriented Along x-Axis Solution:
( ) ( )r
Ar φφ cos=
r
y
φ
-λ +λ
A Constant Potential Solution (Trivial Solution):
( ) Ar =r
φ
x
( )( )
( )( )
rd
dr
dr
rrr
ooo
φεπ
λ
φ
φ
επλ
επλφ cos
2cos2
cos2ln
2ln
2≈
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
+≈⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+
−r
d
