Capacitance and Laplace’s Equation
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Transcript of Capacitance and Laplace’s Equation
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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
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Potential of various charge arrangements• Point
• Line (coaxial)
• Sheet
• V proportional to Q, with some factor involving geometry
• Define
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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: Q
-QE, D
S
B
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
Units are Coul/V or Farads
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Example 1 - Parallel-Plate Capacitor - I
Plate area = S
Applying boundary conditions for D at surface of a perfect conductor:
Lower plate:
Upper Plate:
Same result either way!
Electric field between plates is therefore:
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.
Boundary conditions needed at only one surface to obtain total field between plates.
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Example 1 - Parallel-Plate Capacitor - II
Combining with capacitance is
Plate area = S
With Electric Field
The voltage between plates is:
NoteIn region between plates𝜵 ∙𝑫=0
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Energy Stored in parallel-plate CapacitorStored energy is found by integrating the energy density in the electric field over the capacitor volume.
C V02
S
Gives 3 ways of stored energy:
Rearranging gives
𝑊 𝐸=12 ∫𝑣𝑜𝑙𝜌𝑣𝑉 𝑑𝑣=¿ 1
2 ∫𝑣𝑜𝑙
(𝛻 ∙𝐷 )𝑉 𝑑𝑣¿
𝑊 𝐸=12 ∫𝑣𝑜𝑙𝐷 ∙𝛻𝑉 𝑑𝑣=1
2 ∫𝑣𝑜𝑙𝐷∙𝐸 𝑑𝑣
From Chapter 4 page 102
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Example 2 - Coaxial Transmission Line - I
E = 0 elsewhere, assuming hollow inner conductor, equal and opposite charges on inner and outer conductors.
Coaxial Electric Field using Gauss’ Law:
S
E
1
assume a unit length in z
𝐸=𝜌𝑙
2𝜋𝜌𝜀 𝒂𝝆
Writing with surface-charge density
𝐸=2𝜋 𝑎𝜌 𝑠2𝜋𝜌𝜀 𝒂𝝆
Simplifying
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Example 2 - Coaxial Transmission Line - IIElectric Field between conductors
S
E
1
Potential difference between conductors:
Charge per unit length on inner conductor
Gives capacitance:
assume unit length in z
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Example 3 – Concentric Spherical CapacitorTwo concentric spherical conductors of radii a and b, with equal and opposite charges Q on inner and outer conductors.
a
b
QE -Q
From Gauss’ Law, electric field exists only between spheres and is given by:
Potential difference between inner and outer spheres is
Capacitance is thus:Note as (isolated sphere)
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Example 4 - Sphere with Dielectric Coating
a
r1
E1
E2
Q
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
The potential at the sphere surface (relative to infinity) is:
= V0
The capacitance is:
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Example 5 – Parallel Capacitor with 2-Layer Dielectric
Surface charge on either plate is normal displacement DN through both dielectrics:
Potential between top and bottom surfaces
<< Rule for 2 capacitors in series
𝜌𝑆1=𝑫𝑵 𝟏=𝑫𝑵𝟐=𝜌𝑆 2
𝑉 𝑜=𝑬𝟏𝑑1+𝑬𝟐𝑑2
𝑉 𝑜=𝑫𝟏
𝜀1𝑑1+
𝑫𝟐
𝜀2𝑑2
𝜀1 𝑬𝟏=𝑫𝑵 𝟏𝑫𝑵𝟐=𝜀2 𝑬𝟐
𝑉 𝑜=𝜌𝑆(𝑑1
𝜀1+𝑑2
𝜀2)=𝑄𝑆 (𝑑1
𝜀1+𝑑2
𝜀2)
𝐶=𝑄𝑉 𝑜
=1
1𝑆 (𝑑1
𝜀1+𝑑2
𝜀2 )=
1
( 𝑑1
𝜀1𝑆+𝑑2
𝜀2𝑆 )=
1
( 1𝐶1
+ 1𝐶2 )
The capacitance is thus:
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y
xh
b
V = V0
V = 0
.l
a
Example 6Two Parallel Wires vs. Conducting-Cylinder/Plane
Two parallel wires 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.
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Example 6 - Two Parallel Wires and Conducting-Cylinder/Plane
• Parallel Wires1. 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/Plane1. 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.
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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
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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.
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2 Parallel Wires – Using Parameter K
Two opposite line charges in rectangular coordinates :
Corresponding to equipotential surface V = V1 for dimensionless parameter K = K1
Write ln( ) term as parameter K1:
Corresponding to potential V = V1 according to:
𝑉=𝜌𝐿
4𝜋𝜀 𝑙𝑛(𝑥+𝑎 )2+𝑦 2
(𝑥−𝑎)2+ 𝑦2 =𝜌𝐿
4𝜋𝜀 𝑙𝑛 (𝐾 )
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2 Parallel Wires – Getting Streamlines for KFind streamlines for constant parameter K1 where voltage is constant V1
To better identify surface, expand the squares, and collect terms:
Equation of circle (cylinder) with radius b and displaced along x axis h
y
xh
b
V = V0
V = 0
l
a
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2 Parallel Wires - Substituting Conducting-Cylinder/Plane
Find physical parameters of wires (a, ρL, K1) from streamline parameters (h, b, Vo)
Eliminate a in h and b equations to get quadratic
Substitution above gives image wire position as function of cylinder diameter/offset
and
Solution gives K parameter as function of cylinder diameter/offset
Choose positive sign for positive value for a
y
xh
b
V = V0
V = 0
l
a
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Getting Capacitance of Conducting-Cylinder/Plane
Equivalent line charge l for conducting cylinder is located at
From original definition or
Capacitance for length L is thus
y
xh
b
V = V0
V = 0
.l
a
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Example 1 - Conducting Cylinder/Plane
y
xh
b
V = V0
V = 0
.l
a
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.
mm
Charge density and capacitance
Results unchanged so long as relative proportions maintained
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Example 2 - Conducting Cylinder/Plane
𝜌𝐿=4𝜋𝜀𝑉 𝑜
ln (𝐾 1)=3.46𝑛𝐶 /𝑚
For V0 = 50-volt equipotential surface we recalculate cylinder radius and offset
mm
mm
The resulting surface is the dashed red circle
𝐶=2𝜋𝜀
h𝑐𝑜𝑠 − 1¿¿
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Getting Fields for Conducting Cylinder/Plane• Gradient of Potential
• Electric Field
• Displacement
• For original 5 mm cylinder diameter, 13 mm offset, and 12 mm image-wire offset
• Where max and min are between cylinder and ground plane, and opposite ground plane
y
xhb
V = V0
V = 0
l
a
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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. x
b
h
L
Finally, if the cylinder (wire) dimensions are muchless than their spacing (b << h), then
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Using Field Sketches to Estimate Capacitance
This method employs these properties of conductors and fields:
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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 crossesequipotential lines at right-angles.
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Total Capacitance as # of Flux/Voltage Increments
For conductor boundaries on left and right, capacitance is
𝐶=𝑄𝑉 =
Ψ𝑉
Writing with # flux increments and # voltage increments
𝐶=𝑁𝑄∆𝑄𝑁 𝑉 ∆𝑉
Electrode
Electrode
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Capacitance of Individual Flux/Voltage Increments
Writing flux increment as flux density times area (1 m depth into page)
∆𝑄=∆ Ψ=𝐷1 ∆ 𝐿𝑄=𝜀𝐸 ∆ 𝐿𝑄
Writing voltage increment as Electric field times distance
∆𝑉=𝐸∆ 𝐿𝑉
Forming ratio
∆𝑄∆𝑉 =
𝜀 𝐸∆ 𝐿𝑄𝐸∆ 𝐿𝑉
=𝜀∆ 𝐿𝑄∆ 𝐿𝑉
=𝜀(𝑠𝑖𝑑𝑒𝑟𝑎𝑡𝑖𝑜)
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Total Capacitance for Square Flux/Voltage Increments
Capacitance between conductor boundaries
𝐶=𝑁𝑄∆𝑄𝑁 𝑉 ∆𝑉
Combining with flux/voltage ratio
𝐶=𝑁𝑄
𝑁 𝑉𝜀
∆𝐿𝑄∆ 𝐿𝑉
=𝜀𝑁𝑄
𝑁 𝑉
Provided ΔLQ = ΔLV (increments square)
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Field sketch example I
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Field Sketch Example II
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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.
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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 fieldare known at its boundaries.
Start with Maxwell’s first equation:
where
and
so that
or finally:
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Poisson’s and Laplace’s Equations (continued)
Recall the divergence as expressed in rectangular coordinates:
…and the gradient:
then:
It is known as the Laplacian operator.
𝜵=𝜕𝜕 𝑥 𝒂𝒙+
𝜕𝜕 𝑦 𝒂𝒚+
𝜕𝜕 𝑧 𝒂𝒛 →𝛻2=𝜵 ∙𝜵=
𝜕2
𝜕𝑥2 +𝜕2
𝜕 𝑦 2 +𝜕2
𝜕 𝑧2
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Summary of Poisson’s and Laplace’s Equations
we already have:
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:
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Laplacian Operator in Three Coordinate Systems
(Laplace’s equation)
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Example 1 - Parallel Plate Capacitor
d
0
x
V = V0
V = 0
Plate separation d smaller than plate dimensions. Thus V varies only with x. Laplace’s equation is:
Integrate once:
Integrate againBoundary 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.
Get general expression for potential function
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Parallel Plate Capacitor II
General expression:
Boundary condition 1:
0 = A(0) + B
Boundary condition 2:
V0 = Ad
Finally:
d
0
x
V = V0
V = 0
Boundary conditions:
1. V = 0 at x = 0
2. V = V0 at x = d
EquipotentialSurfaces
Apply boundary conditions
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Parallel Plate Capacitor III
Potential
Electric Field
Displacement
d
0
x
V = V0
V = 0
EquipotentialSurfaces
E+ + + + + + + + + + + + + +
- - - - -- - - - - - - - -
Surface Area = S
n
At the lower plate n = ax
Conductor boundary condition
Total charge onlower plate capacitance
Getting 1) Electric field, 2) Displacement, 3) Charge density, 4) Capacitance
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Example 2 - Coaxial Transmission Line
V0
E
L
V = 0
Boundary conditions:
1. V = 0 at b2. V = V0 at a
V varies with radius only, Laplace’s equation is:
(>0)
Integrate once:
Integrate again:
Get general expression for potential
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Coaxial Transmission Line II
V0
E
L
V = 0
Boundary conditions:
1. V = 0 at b2. V = V0 at a
General Expression
Boundary condition 1:
Boundary condition 2:
Combining:
Apply boundary conditions
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Coaxial Transmission Line III
V0
E
L
V = 0
Potential:
Electric Field:
Charge density on inner conductor:
Total charge on inner conductor:Capacitance:
Getting 1) Electric field, 2) Displacement, 3) Charge density, 4) Capacitance
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Example 3 - Angled Plate Geometry
Cylindrical coordinates, potential varies only with
x
Boundary Conditions:
1. V = 0 at 02. V =V0 at
Integrate once:
Integrate again:
Boundary condition 1:
Boundary condition 2:
Potential: Field:
Get general expression, apply boundary conditions, get electric field
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Example 4 - Concentric Sphere Geometry
a
b
V0
E
V = 0
Boundary Conditions:
1. V = 0 at r = b2. V = V0 at r = a
V varies only with radius. Laplace’s equation:
or:
Integrate once: Integrate again:
Boundary condition 1:
Boundary condition 2:
Potential:
Get general expression, apply boundary conditions
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Concentric Sphere Geometry II
a
b
V0
E
V = 0Potential: (a < r < b)
Electric field:
Charge density on inner conductor:
Total charge on inner conductor:
Capacitance:
Get 1) electric field, 2) displacement, 3) charge density, 4) capacitance
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Example 5 – Cone and Plane Geometry
V varies only with only, Laplace’s equation is:
Integrate once:
R, >
Integrate again
Boundary Conditions:
1. V = 0 at 2. V = V0 at
Boundary condition 1:
Boundary condition 2:
Potential:
Get general expression, apply boundary conditions
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Cone and Plane Geometry II
r1
r2
Potential:
Electric field:
Get electric field
Check symboliccalculators
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Cone and Plane Geometry III
r1
r2
Charge density on cone surface:
Total charge on cone surface:
Capacitance: Neglects fringing fields, important for smaller .
Note capacitance positive (as should be).
Get 1) charge density, 2) capacitance
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Example 6 – Product Solution in 2 Dimensions
𝛻2𝑉=0 [ 𝜕2
𝜕 𝑥2 +𝜕2
𝜕 𝑦2 + 𝜕2
𝜕𝑧 2 ] [𝑋 (𝑥 )𝑌 ( 𝑦 )𝑍 (𝑧 ) ]=0
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Product Solution in 2 Dimensions II
Paul Lorrain and Dale Corson, “Electromagnetic Fields and Waves” 2nd Ed, W.H. Freeman, 1970
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Product Solution in 2 Dimensions III
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Product Solution in 2 Dimensions IV
This problem just keeps on going!
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Product Solution in 2 Dimensions V
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Product Solution in 2 Dimensions VI
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Example 7 – Another Product Solution in 2 Dimensions
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Another Product Solution in 2 Dimensions II
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Another Product Solution in 2 Dimensions III
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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
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p-n junction – zero bias
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p-n junction – forward/reverse bias
Applied fieldApplied field
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p-n junction – charge, field, potential
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p-n junction – Obtaining Electric Field
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p-n junction – Obtaining Potential
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p-n junction - Obtaining Charge
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p-n junction Obtaining Junction Capacitance