Capacitance and Laplace’s Equation

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

• Line (coaxial)

• Sheet

• 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: 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

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.

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

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

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

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

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)

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:

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:

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.

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.

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 :

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 )

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

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

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

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

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¿¿

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

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

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 crossesequipotential lines at right-angles.

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

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)

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)

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 fieldare known at its boundaries.

Start with Maxwell’s first equation:

where

and

so that

or finally:

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

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:

Laplacian Operator in Three Coordinate Systems

(Laplace’s equation)

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

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

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

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

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

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

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

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

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

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

Cone and Plane Geometry II

r1

r2

Potential:

Electric field:

Get electric field

Check symboliccalculators

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

Example 6 – Product Solution in 2 Dimensions

𝛻2𝑉=0 [ 𝜕2

𝜕 𝑥2 +𝜕2

𝜕 𝑦2 + 𝜕2

𝜕𝑧 2 ] [𝑋 (𝑥 )𝑌 ( 𝑦 )𝑍 (𝑧 ) ]=0

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 fieldApplied 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