### Engr302 - Lecture 6

```Capacitance and Laplace’s Equation
• Capacitance Definition
• Simple Capacitance Examples
• Capacitance Example using Streamlines & Images
– Two-wire Transmission Line
– Conducting Cylinder/Plane
• Field Sketching
• Laplace and Poison’s Equation
• Laplace’s Equation Examples
• Laplace’s Equation - Separation of variables
• Poisson’s Equation Example
Potential of various charge arrangements
• Point
=

4
• Line (coaxial)
=
• Sheet

2

=

• V proportional to Q, with some factor involving geometry
• Define
=

Basic Capacitance Definition
A simple capacitor consists of two oppositely charged conductors surrounded by a uniform dielectric.
An increase in Q by some factor results in an increase D (and E) by same factor.
With the potential difference between conductors:
S
Q
.A
increasing by the same factor -- so the ratio Q to V0 is constant.
We define the capacitance of the structure as the ratio of stored
charge to applied voltage, or
-Q
.
B
E, D
Example 1 - Parallel-Plate Capacitor - I
The horizontal dimensions are assumed to be much greater than the plate separation, d. The
electric field thus lies only in the z direction, with the potential varying only with z.
Plate area = S
Applying boundary conditions for D at surface of a perfect conductor:
Lower plate:
Same result either way!
Upper Plate:
Electric field between plates is therefore:
Boundary conditions needed
at only one surface to obtain
total field between plates.
Example 1 - Parallel-Plate Capacitor - II
With Electric Field
Plate area = S
The voltage between plates is:
Combining with
capacitance is
Note  ∙  = 0
In region between plates
Energy Stored in parallel-plate Capacitor
Stored energy is found by integrating the energy density in the electric field over the capacitor volume.
=
1
2

=
1
2
∙

From Chapter 4 page 102
1
=
2
1
∙   =
2

∙

Rearranging gives
S
C
Gives 3 ways of stored energy:
V02
Example 2 - Coaxial Transmission Line - I
Coaxial Electric Field using Gauss’ Law:

=

2
Writing with surface-charge density
=
2

2
Simplifying
E = 0 elsewhere, assuming hollow inner conductor, equal
and opposite charges on inner and outer conductors.
S
E
1
assume a unit length in z
Example 2 - Coaxial Transmission Line - II
Electric Field between conductors
S
Potential difference between conductors:
E
1
assume unit length
in z
Charge per unit length on inner conductor
Gives capacitance:
Example 3 – Concentric Spherical Capacitor
Two concentric spherical conductors of radii a and b, with
equal and opposite charges Q on inner and outer conductors.
From Gauss’ Law, electric field exists only between spheres
and is given by:
E
Q
a
-Q
b
Potential difference between inner and outer spheres is
Capacitance is thus:
Note as
(isolated sphere)
Example 4 - Sphere with Dielectric Coating
A conducting sphere of radius a carries charge Q. A
dielectric layer of thickness (r1 – a) and of permittivity 1
surrounds the conductor. Electric field in the 2 regions is
found from Gauss’ Law


a
E2
Q
E1
r1
The potential at the sphere surface (relative to infinity) is:
= V0
The capacitance is:
Example 5 – Parallel Capacitor with 2-Layer Dielectric
Surface charge on either plate is normal displacement DN through both dielectrics:
1 =  =  = 2
1  =
= 2
Potential between top and bottom surfaces
=  1 +  2
=

1 +

1
2 2
=
1 2
1 2
+
=
+
1 2
1 2
The capacitance is thus:
=

1
1
1
=
=
=
1
1
1
2
1 1 + 2
+
+
1 2
1 2
1  2
<< Rule for 2
capacitors in series
Example 6
Two Parallel Wires vs. Conducting-Cylinder/Plane
Parallel wires on left substitute conducting cylinder/plane on right
• Equipotential streamline for wires on left match equipotential surface for cylinder on right.
• Image wire (-a) on left emulates vertical conducting plane on right.
y
V=0
V = V0
.
b
l
a
Two parallel wires
h
x
Conducting cylinder/plane
Example 6 - Two Parallel Wires and
Conducting-Cylinder/Plane
• Parallel Wires
1.
Superimpose 2 long-wire potentials at x = +a and x = -a.
2.
Translate to common rectangular coordinate system.
3.
Define parameter K (=constant) for V (=constant) equipotential.
4.
Find streamlines (x,y) for constant K and constant V.
• Conducting-Cylinder/Plane
1.
Insert metal cylinder along equipotential (constant K) streamline.
2.
Work backward to find long-wire position, charge density, and K parameter from cylinder
diameter, offset, and V potential.
3.
Calculate capacitance of cylinder/plane from long-wire position and charge density
4.
Write expression for potential, D, and E fields between cylinder and plane.
5.
Write expression for surface charge density on plane.
Two Parallel Wires – Basic Potential
Begin with potential of single line charge on z axis,
with zero reference at = R0
Then write potential for 2 line charges of opposite sign positioned at x = +a and x = -a
2 Parallel Wires – Rectangular Coordinates
2 line charges of opposite sign:
Choose a common reference radius R10 = R20 . Write R1 and R2 in terms of common rectangular
coordinates x, y.
2 Parallel Wires – Using Parameter K
Two opposite line charges in rectangular coordinates :

+
=

4
−
2
+ 2

=

2 + 2
4
Write ln( ) term as parameter K1:
Corresponding to potential V = V1 according to:
Corresponding to equipotential surface V = V1 for dimensionless parameter K = K1
2 Parallel Wires – Getting Streamlines for K
Find streamlines for constant parameter K1 where voltage is constant V1
y
V=0
To better identify surface, expand the squares, and collect terms:
V = V0
l
a h
Equation of circle (cylinder) with radius b and displaced along x axis h
b
x
2 Parallel Wires - Substituting Conducting-Cylinder/Plane
Find physical parameters of wires (a, ρL, K1) from streamline parameters (h, b, Vo)
y
and
V=0
Eliminate a in h and b equations to get quadratic
V = V0
l
a h
Solution gives K parameter as function of cylinder diameter/offset
Choose positive sign for
positive value for a
Substitution above gives image wire position as function of cylinder diameter/offset
b
x
Getting Capacitance of Conducting-Cylinder/Plane
Equivalent line charge l for conducting cylinder is located at
y
From original definition
or
V=0
Capacitance for length L is thus
V = V0
.
b
l
a
h
x
Example 1 - Conducting Cylinder/Plane
Conducting cylinder radius b = 5 mm, offset h = 13 mm, potential V0 = 100 V. Find
offset of equivalent line charge a, parameter K, charge density l , and capacitance C.
y
V=0
mm
V = V0
.
b
l
a
h
x
Charge density and capacitance
Results unchanged so long as
relative proportions maintained
Example 2 - Conducting Cylinder/Plane
For V0 = 50-volt equipotential surface we recalculate cylinder radius and offset
mm
mm
The resulting surface is the dashed red circle
=
=
4
= 3.46
ln(1 )
2
≈ 34.6
ℎ−1 (ℎ )
Getting Fields for Conducting Cylinder/Plane
•
•
•

+
= −

4
−
•
+ 2
2 + 2
y
V=0
Electric Field
2  +   + 2 2  −   + 2
=−
−
4
+  2 + 2
−  2 + 2
V = V0
l b
a h
Displacement
=−
•
2
−   +
+   +
−
2
+  2 + 2
−  2 + 2
For original 5 mm cylinder diameter, 13 mm offset, and 12 mm image-wire offset
, = −,=−,=0
3.46 ∙ 10−9
=
2
13 − 5 + 12
13 − 5 − 12
−
= 0.165  3
2
2
13 − 5 + 12
13 − 5 − 12
, = −,=+,=0
3.46 ∙ 10−9
=
2
13 + 5 + 12
13 + 5 − 12
−
= 0.073  3
2
2
13 + 5 + 12
13 + 5 − 12
Where max and min are between cylinder and ground plane, and opposite ground plane
x
Getting Capacitance of 2-Wire or 2-Cylinder Line
With two wires or cylinders (and zero potential
plane between them) the structure represents two
wire/plane or two cylinder/plane capacitors in
series, so the overall capacitance is half that
derived previously.
b
x
h
Finally, if the cylinder (wire) dimensions are much
less than their spacing (b << h), then
L
Using Field Sketches to Estimate Capacitance
This method employs these properties of conductors and fields:
Sketching Equipotentials
Given the conductor boundaries, equipotentials may be sketched in. An attempt is made to
establish approximately equal potential differences between them.
A line of electric flux density, D, is then started (at point A), and then drawn such that it crosses
equipotential lines at right-angles.
Total Capacitance as # of Flux/Voltage Increments
For conductor boundaries on
left and right, capacitance is
Ψ
= =

Writing with # flux increments
and # voltage increments
∆
=
∆
Electrode
Electrode
Capacitance of Individual Flux/Voltage Increments
Writing flux increment as flux density
times area (1 m depth into page)
∆ = ∆Ψ =  1 ∆ = ∆
Writing voltage increment as
Electric field times distance
∆ = ∆
Forming ratio
∆
∆ ∆
=
=
=  ( )
∆
∆
∆
Total Capacitance for Square Flux/Voltage Increments
Capacitance between
conductor boundaries
∆
=
∆
Combining with flux/voltage ratio
∆

=

=
∆

Provided ΔLQ = ΔLV (increments square)
Field sketch example I
Field Sketch Example II
Laplace and Poisson’s Equation
1. Assert the obvious
– Laplace - Flux must have zero divergence in empty
space, consistent with geometry (rectangular,
cylindrical, spherical)
– Poisson - Flux divergence must be related to free
charge density
2. This provides general form of potential and
field with unknown integration constants.
3. Fit boundary conditions to find integration
constants.
Derivation of Poisson’s and Laplace’s Equations
These equations allow one to find the potential field in a region, in which values of potential or electric field
are known at its boundaries.
where
and
so that
or finally:
Poisson’s and Laplace’s Equations (continued)
Recall the divergence as expressed in
rectangular coordinates:
then:
. It is known as the Laplacian operator

=
+
+

→
2
2
2

2 =  ∙  = 2 + 2 + 2

Summary of Poisson’s and Laplace’s Equations
which becomes:
This is Poisson’s equation, as stated in rectangular coordinates.
In the event that there is zero volume charge density, the right-hand-side becomes zero, and we obtain
Laplace’s equation:
Laplacian Operator in Three Coordinate Systems
(Laplace’s equation)
Example 1 - Parallel Plate Capacitor
Get general expression for potential function
Plate separation d smaller than plate dimensions.
Thus V varies only with x. Laplace’s equation is:
x
V = V0
d
Integrate once:
0
V=0
Integrate again
Boundary conditions:
1. V = 0 at x = 0
2. V = V0 at x = d
where A and B are integration constants evaluated according to boundary conditions.
Parallel Plate Capacitor II
Apply boundary conditions
General expression:
x
V = V0
Boundary condition 1:
d
Equipotential
Surfaces
0 = A(0) + B
0
Boundary condition 2:
V=0
Boundary conditions:
1. V = 0 at x = 0
2. V = V0 at x = d
Finally:
Parallel Plate Capacitor III
Getting 1) Electric field, 2) Displacement, 3) Charge density, 4) Capacitance
Potential
x
Electric Field
Surface Area = S
V = V0
d
+ + + + + + + + + + + + + +
E
n
Displacement
0
- - - - - - - - - - - - - -
V=0
At the lower plate
n = ax
Conductor boundary condition
Total charge on
lower plate
capacitance
Equipotential
Surfaces
Example 2 - Coaxial Transmission Line
Get general expression for potential
V varies with radius only, Laplace’s equation is:
(> 0)
V0
V=0
E
L
Integrate once:
Boundary conditions:
Integrate again:
1.
2.
V = 0 at b
V = V0 at a
Coaxial Transmission Line II
Apply boundary conditions
General Expression
Boundary condition 1:
V0
V=0
E
L
Boundary condition 2:
Combining:
Boundary conditions:
1.
2.
V = 0 at b
V = V0 at a
Coaxial Transmission Line III
Getting 1) Electric field, 2) Displacement, 3) Charge density, 4) Capacitance
Potential:
Electric Field:
V0
V=0
E
Charge density on inner conductor:
Total charge on inner conductor:
L
Capacitance:
Example 3 - Angled Plate Geometry
Get general expression, apply boundary conditions, get electric field
Cylindrical coordinates, potential varies only with 
Integrate once:
Integrate again:
x
Boundary Conditions:
Boundary condition 1:
1.
2.
Boundary condition 2:
Potential:
Field:
V = 0 at 0
V =V0 at 
Example 4 - Concentric Sphere Geometry
Get general expression, apply boundary conditions
V varies only with radius. Laplace’s equation:
V=0
E
or:
V0
a
b
Integrate once:
Integrate again:
Boundary Conditions:
Boundary condition 1:
1.
2.
Boundary condition 2:
Potential:
V = 0 at r = b
V = V0 at r = a
Concentric Sphere Geometry II
Get 1) electric field, 2) displacement, 3) charge density, 4) capacitance
Potential:
V=0
(a < r < b)
Electric field:
E
V0
a
b
Charge density on inner conductor:
Total charge on inner conductor:
Capacitance:
Example 5 – Cone and Plane Geometry
Get general expression, apply boundary conditions
V varies only with  only, Laplace’s equation is:
R,  >
Integrate once:
Integrate again
Boundary Conditions:
Boundary condition 1:
1.
2.
Boundary condition 2:
Potential:
V = 0 at 
V = V0 at 
Cone and Plane Geometry II
Get electric field
Potential:
r2
Electric field:
r1
Check symbolic
calculators
Cone and Plane Geometry III
Get 1) charge density, 2) capacitance
Charge density on cone surface:
r2
Total charge on cone surface:
r1
Capacitance:
Note capacitance positive (as should be).
Neglects fringing fields, important
for smaller .
Example 6 – Product Solution in 2 Dimensions
2
=0
2
2
2
+
+
() = 0
2  2  2
Product Solution in 2 Dimensions II
Paul Lorrain and Dale Corson, “Electromagnetic Fields and Waves” 2nd Ed, W.H. Freeman, 1970
Product Solution in 2 Dimensions III
Product Solution in 2 Dimensions IV
This problem just keeps on going!
Product Solution in 2 Dimensions V
Product Solution in 2 Dimensions VI
Example 7 – Another Product Solution in 2 Dimensions
Another Product Solution in 2 Dimensions II
Another Product Solution in 2 Dimensions III
Poisson’s equation example
p-n junction – zero bias
•
p-type for x < 0, n-type for x > 0.
•
holes diffuse to right, electrons diffuse to left.
•
creates electric field to left (depletion layer).
•
electric field to left inhibits further hole movement right, electron movement left.
en.wikipedia.org/wiki/P-n_junction
p-n junction – zero bias
p-n junction – forward/reverse bias
Applied field
Applied field
p-n junction – charge, field, potential
p-n junction – Obtaining Electric Field
p-n junction – Obtaining Potential
p-n junction - Obtaining Charge
p-n junction Obtaining Junction Capacitance
```