Prof. Sang-Yong Jung - contents.kocw.netcontents.kocw.net/KOCW/document/2014/sungkyunkwan/jungsangyon… · - This method employs these properties of conductors and fields: 1)

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Engineering Electromagnetics (Week 10)

Prof. Sang-Yong Jung

( 23432B 031-299-4952 www.emlab.kr [email protected] )

L o g o

Ch6. Capacitance

6.1_ Capacitance Defined

6.2_ Parallel-Plate Capacitor

6.3_ Several Capacitance Examples

6.4_ Capacitance of a Two-Wire Line

6.5_ Using Field Sketches to Estimate Capacitance in Two-Dimensional Problems

2

전기기기연구실 (www.emlab.kr) Electric Machines LAB (www.emlab.kr)

□ Capacitance Defined

- 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 in 𝐄 (and in 𝐃) by the

same factor.

𝑄 = 𝐃 ∙ 𝑑𝐒

𝑆

Consequently, the potential difference between conductors:

- Capacitance : the ratio of the magnitude of

the total charge on either conductor to

the magnitude of the potential difference

between conductors, or

Units are Coul/V or farads

6.1 CAPACITANCE DEFINED 3

𝑉0 = − 𝐄 ∙ 𝑑𝐋𝐴

𝐵

𝐶 =𝑄

𝑉0


6.2 PARALLEL-PLATE CAPACITOR 4

□ Parallel-Plate Capacitor

- Assumptions

1) The horizontal dimensions : much grater than the plate separation, 𝑑.

2) Electric field : in the 𝑧 direction, and potential varies only with 𝑧.

- Apply the boundary for D at the surface of a perfect conductor:

Lower plate : 𝐃 ∙ 𝐧𝑙 𝑧=0 = 𝐃 ∙ 𝐚𝑧 = 𝜌𝑠 𝐃 = 𝜌𝑠𝐚𝑧

Upper plate : 𝐃 ∙ 𝐧𝑙 𝑧=𝑑 = 𝐃 ∙ (−𝐚𝑧) = −𝜌𝑠 𝐃 = 𝜌𝑠𝐚𝑧

The electric field between plates:

Plate area = S

Same result

𝐸 =𝜌𝑆𝜖𝑎𝑧



□ Capacitance of a Parallel Plate Capacitor

With , the voltage between plate can be found through:

With 𝑄 = 𝜌𝑆𝑆

finally:

Plate area = S

𝐶 =𝑄

𝑉0=𝜖𝑆

𝑑

𝐄 =𝜌𝑆𝜖𝐚𝑧

𝑉0 = − 𝐄 ∙ 𝑑𝐋lower

upper

= − 𝜌𝑆𝜖𝑑𝑧 =

𝜌𝑆𝜖𝑑

0

𝑑



□ Stored Energy in a Capacitor

- Stored energy is found by integrating the energy density in the electric field over

the capacitor volume.

or

- Three ways of writing the energy:

𝑊𝐸 =1

2𝐶𝑉0

2 =1

2𝑄𝑉0 =

1

2

𝑄2

𝐶

𝑊𝐸 =1

2 𝜖𝐸2𝑑𝑣

vol

=1

2

𝜖𝜌𝑠2

𝜖2𝑑𝑧𝑑𝑆

𝑑

0

𝑆

=1

2

𝜌𝑠2

𝜖𝑆𝑑 =

1

2

𝜖𝑆

𝑑

𝜌𝑠2𝑑2

𝜖2

𝑊𝐸 = 1

2𝐃 ∙ 𝐄𝑑𝑣

vol

C V02


6.3 SEVERAL CAPACITANCE EXAMPLES 7

□ Coaxial Transmission Line

- Using Gauss’s Law:

The potential difference between conductors:

The charge on the inner conductor per unit length:

Finally:

L

𝐄 𝜌 =𝑎𝜌𝑠𝜖𝜌𝐚𝜌 V m (𝑎 < 𝜌 < 𝑏)

𝑉0 = − 𝐄 ∙ 𝑑𝐋𝑎

𝑏

= − 𝑎𝜌𝑠𝜖𝜌𝐚𝜌 ∙ 𝐚𝜌𝑑𝜌

𝑎

𝑏

=𝑎𝜌𝑠𝜖ln𝑏

𝑎

𝑄 = 2𝜋𝑎(1)𝜌𝑠 𝐶 =𝑄

𝑉0=

2𝜋𝜖

ln 𝑏 𝑎 F/m


□ Isolated Sphere with a Dielectric Coating

- Assumptions

1) Two concentric spherical conductors, having radii a and b.

2) Equal and opposite charges, Q, are on the inner

and outer conductors.

Using Gauss’s Law:

The potential difference between inner and outer spheres:

The capacitance:

Note that as 𝑏 → ∞ (isolated sphere)

𝐶 → 4𝜋𝜖𝑎


a

b

Q E

-Q

𝐄 = 𝐸𝑟𝐚𝑟 =𝑄

4𝜋𝜖𝑟2𝐚𝑟

𝑉0 = − 𝐄 ∙ 𝑑𝐋𝑎

𝑏

= − 𝑄

4𝜋𝜖𝑟2𝐚𝑟 ∙ 𝐚𝑟𝑑𝑟

𝑎

𝑏

=𝑄

4𝜋𝜖

1

𝑎−1

𝑏

𝐶 =𝑄

𝑉0=

4𝜋𝜖

1 𝑎 − 1 𝑏



□ Capacitor with a Two-Layer Dielectric

- Assumption : No surface charge is present

- D is normal to an interface between two dielectrics.

𝐷𝑁1 = 𝐷𝑁2 and 𝜖1𝐸1 = 𝜖2𝐸2

The potential difference

between bottom and top plates:

which leads to:

𝐸1 =𝑉0

𝑑1 + 𝑑2(𝜖1 𝜖2 )

𝑉0 = 𝐸1𝑑1 + 𝐸2𝑑2

E2

E1 DN1

DN2



□ Capacitor with a Two-Layer Dielectric (cont.)

the bottom plate:

The capacitance:

E2

E1 DN1

DN2

𝜌𝑆1 = 𝐷1 = 𝜖1𝐸1 =𝑉0

𝑑1𝜖1+𝑑2𝜖2

𝐶 =𝑄

𝑉0=𝜌𝑆𝑆

𝑉0=

1

𝑑1𝜖1𝑆

+𝑑2𝜖1𝑆

1

1𝐶1+1𝐶2

=


6.4 CAPACITANCE OF A TWO-WIRE LINE 11

□ Two-Wire Transmission Line

- The potential field of a single line charge on the z axis, with a zero reference

at 𝜌 = 𝑅0

Use this result to write the potential at point 𝑃 from two line charges of

opposite sign, positioned as shown:

𝑉 =𝜌𝐿2𝜋𝜖

ln𝑅10𝑅1− ln

𝑅20𝑅2


ln𝑅0𝑅

=𝜌𝐿2𝜋𝜖

ln𝑅10𝑅2𝑅20𝑅1


□ Two-Wire Transmission Line (cont.)

Using

Choose 𝑅10 = 𝑅20, thus placing the zero reference at equal distances from

each line.

This surface is the 𝑥 = 0 plane.

Expressing 𝑅1 and 𝑅2 in terms of 𝑥 and 𝑦.


ln𝑅10𝑅1− ln

𝑅20𝑅2

=𝜌𝐿2𝜋𝜖

ln𝑅10𝑅2𝑅20𝑅1



ln(𝑥 + 𝑎)2+𝑦2

(𝑥 − 𝑎)2+𝑦2

=𝜌𝐿4𝜋𝜖

ln(𝑥 + 𝑎)2+𝑦2

(𝑥 − 𝑎)2+𝑦2



Using

Choose an equipotential surface on which 𝑉 = 𝑉1, and define the dimensionless

parameter:

therefore,


ln(𝑥 + 𝑎)2+𝑦2

(𝑥 − 𝑎)2+𝑦2=𝜌𝐿4𝜋𝜖

ln(𝑥 + 𝑎)2+𝑦2

(𝑥 − 𝑎)2+𝑦2


𝐾1 =(𝑥 + 𝑎)2+𝑦2

(𝑥 − 𝑎)2+𝑦2

𝐾1 = 𝑒4𝜋𝜖𝑉1 𝜌𝐿



This is the equation of the equipotential surface on which the voltage is 𝑉1.

To better identify the surface, expand the squares, and collect terms:


𝑥2 − 2𝑎𝑥𝐾1 + 1

𝐾1 − 1+ 𝑦2 + 𝑎2 = 0

𝑥 − 𝑎𝐾1 + 1

𝐾1 − 1

2

+ 𝑦2 =2𝑎 𝐾1𝐾1 − 1

2

where 𝑏 =2𝑎 𝐾1𝐾1 − 1



The equation of an equipotential surface:

This is the equation of a circle (actually a cylinder), displaced along the 𝑥 axis

by distance ℎ, and having radius 𝑏, where

and


𝑥 − 𝑎𝐾1 + 1

𝐾1 − 1

2

+ 𝑦2 =2𝑎 𝐾1𝐾1 − 1

2

𝑏 =2𝑎 𝐾1𝐾1 − 1

ℎ = 𝑎𝐾1 + 1

𝐾1 − 1



□ Capacitance of a Cylinder Above a Plane

- Consider a grounded conductor in the 𝑦 − 𝑧 plane, and a conducting cylinder,

parallel to the 𝑧 axis, positioned as shown.

The cylinder has radius 𝑏 and center

location ℎ on the 𝑥 axis

Eliminate 𝑎 between the two equations

to get the quadratic:

y

x h

b

V = V0

V = 0

𝑏𝐾1 − 2ℎ 𝐾1 + 𝑏 = 0

ℎ = 𝑎𝐾1 + 1

𝐾1 − 1, 𝑏 =

2𝑎 𝐾1𝐾1 − 1



□ Capacitance of a Cylinder Above a Plane (cont.)

solution of this equation is:

substitute 𝐾1 back into either

starting equation to find:

y

x h

b

V = V0

V = 0

𝐾1 =ℎ ± ℎ2 − 𝑏2

𝑏

Choose positive sign because it leads to

a positive value for 𝑎

𝑎 = ℎ2 − 𝑏2



- The expression 𝑎 = ℎ2 − 𝑏2 gives the location of an equivalent line charge, 𝜌𝑙, as shown:

From the original definition,

we now have:

from which:

Now, given ℎ, 𝑏, and 𝑉0, it is possible

to find 𝑎, 𝜌𝑙 , and 𝐾1.

y

x h

b

V = V0

V = 0

l

a

𝐾1 = 𝑒2𝜋𝑉0 𝜌𝐿

𝜌𝐿 =4𝜋𝜖𝑉0ln𝐾1



For a length 𝐿 in the 𝑧 direction,

finally:

𝐶 =𝜌𝐿𝐿

𝑉0=4𝜋𝜖𝐿

ln𝐾1=2𝜋𝜖𝐿

ln 𝐾1

𝐶 =2𝜋𝜖𝐿

ln ℎ + (ℎ2−𝑏2)/𝑏

2𝜋𝜖𝐿

cosh−1(ℎ 𝑏 ) =

y

x h

b

V = V0

V = 0

l

a


6.5 USING FIELD SKETCHES TO ESTIMATE CAPACITANCE

IN TWO-DIMENSIONAL PROBLEMS 20

□ Using Field Sketches to Estimate Capacitance

- This method employs these properties of conductors and fields:

1) A conductor boundary is an equipotential surface.

2) The electric field intensity and electric flux density are both perpendicular

to the equipotential surfaces.

3) 𝐄 and 𝐃 are therefore perpendicular to the conductor boundaries and

possess zero tangential values.

4) The lines of electric flux, or streamlines, begin and terminate on charge

and hence, in a charge-free, homogeneous dielectric, begin and terminate

only on the conductor boundaries.




□ 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 line at right-angles.




□ Sketching Lines of Electric Flux

- Two red line : D

The length of the line joining A to B : ∆𝐿𝑡

All field lines must intersect equipotentials at 90 ̊ .

- The volume between the two red field lines forms a “tube” of flux, of amount ∆ψ.

This is the same as the charge on the conductor that is bounded by the tube.

- The electric field:

assuming potential difference ∆𝑉

between adjacent equipotentials,

the electric field:

𝐸 =1

𝜖

∆Ψ

∆𝐿𝑡

𝐸 =∆𝑉

∆𝐿𝑁




□ Graphical Relationship Between Incremental Potential Difference and Electric Flux

- having two expressions for electric field, based on the line spacings in the sketch:

Set the two expressions equal:

The ratio ∆𝐿𝑡/∆𝐿𝑁 is fixed.

Each grid segment is approximately square.

𝐸 =1

𝜖

∆Ψ

∆𝐿𝑡, 𝐸 =

1

𝜖

∆𝑉

∆𝐿𝑁

∆𝐿𝑡∆𝐿𝑁

= constant =1

𝜖

∆Ψ

∆𝑉

∆𝐿𝑡 = ∆𝐿𝑁




□ Determining Capacitance from the Field Sketch

- The region is divided into curvilinear squares.

Each segment along the conductor walls contains flux ∆ψ, (or charge ∆𝑄).

The total charge on either conductor:

The potential difference between conductor:

The capacitance:

∆𝑉 ∆𝑉

∆𝑄

∆𝑄

∆𝑄

∆𝑄

𝑁𝑉∆𝑉

𝑁𝑄∆𝑄

𝑄 = 𝑁𝑄∆𝑄 = 𝑁𝑄∆Ψ

𝑉0 = 𝑁𝑉∆𝑉

𝐶 =𝑁𝑄∆𝑄

𝑁𝑉∆𝑉

𝐶 =𝑁𝑄𝑁𝑉𝜖∆𝐿𝑡∆𝐿𝑁

= 𝜖𝑁𝑄𝑁𝑉

𝑁𝑄: the number of segment along the conductor wall

𝑁𝑉: the number of segment between conductors

Documents

Prof. Sang-Yong Jung - contents.kocw.netcontents.kocw.net/KOCW/document/2014/sungkyunkwan/jungsangyon… · - This method employs these properties of conductors and fields: 1)