CONDUCTORS, DIELECTRICS, AND CAPACITANCE.pdf

CONDUCTORS, DIELECTRICS,

AND CAPACITANCE

CURRENT AND CURRENT

DENSITY

Current - is defined as electric charges in motion or defined as a rate of movement

of a charge passing in a given reference

point (or crossing in a given reference

plane) of one coulomb per second

CURRENT AND CURRENT

DENSITY

Current - is defined as the motion of positive charges (even though conduction

in metals takes place through the motion

of electrons). The unit of current is ampere

(A) and current is symbolized by I,

dt

dQI ( convention current)

CURRENT AND CURRENT

DENSITY

Consider the field theory, the events occurring at a point was the main concerned rather than within the large region, and in this case the concept of current density were important matter.

Current density - is defined as a vector represented by J and measured in amperes per square meter (A/m2).

The increment of current I crossing an incremental surface, S, is given as;

CURRENT AND CURRENT

DENSITY

( normal to the current

density)

or,

( current density is not

perpendicular to the

surface)

SJI N

SJI

CURRENT AND CURRENT

DENSITY

And the total current is obtained by integrating,

( convention current) S dSJI

CURRENT AND CURRENT

DENSITY

Current density can be related to the velocity of volume charge density at a given point. Consider the element of charge Q = vv = vSL, as shown in the figure .

CURRENT AND CURRENT

DENSITY

Assume that the charge element is

oriented within the

edges and parallel to

the coordinate axes,

and it shall only

posses an x

component of

velocity.

CURRENT AND CURRENT

DENSITY

With respect to the time interval t, the element of charge has moved a distance x, as shown in the figure.

CURRENT AND CURRENT

DENSITY

Therefore a charge Q = vSx has moved through a reference plane perpendicular to the direction of motion in a time increment t, and the resultant current is

t

xS

t

QI

CURRENT AND CURRENT

DENSITY

And taking the limit with respect to time,

where: vx represents the x component of the velocity, v.

And in terms of current density,

xSI

xx vJ

CURRENT AND CURRENT

DENSITY

and in general

where: J or vv is the convention current density

vJ

Sample Problem 1

The vector current density is given as J =

(4/r2)cos ar + 20 e-2r sin a r sin cos

a A/m2. (a) Find J at r = 3, = 0, = .

(b) Find the total current passing through

the spherical cap r = 3, 0 < < 20O, 0 < < 2, in the ar direction.

CONTINUITY OF CURRENT

The principle of conservation of charge states simply that charges can be neither created nor destroyed, although equal amounts of positive and negative charge may be simultaneously created, obtained by separation, destroyed, or lost by recombination.

The continuity equation, follow this principle when considering of any region bounded by a closed surface. And the current through the closed surface is


the outward flow of positive charge must

be balanced by a decrease of positive

charge (or perhaps an increase of

negative charge) within the closed surface.

S dSJI


The current at closed surface, however, is an outward-flowing current and it is the integral form of the continuity equation, and the differential, or point, form is obtained by using the divergence theorem to change the surface integral into a volume integral:

volS dJdSJ )(


Next represent the enclosed charge Qi by the volume integral of the charge density,

volvol ddt

ddJ )(


Keeping the surface constant, the derivative becomes a partial derivative and

may appear within the integral,

For an incremental volume,

volvol

dt

dJ

)(

dt

J

)(


And the point from the continuity equation,

tJ

)( (the current, or charge

per second, diverging

from a small volume per

unit volume is equal to

the time rate of

decrease of charge per

unit volume at every

point)

Sample Problem 2

Assume that an electron beam carries a total current of 500 A in the az direction, and has a current density Jz that

is not a function of or in the region 0 < < 10-4 m and is zero for < 10-4 m. If the electron velocities are given by vz = 8 x

107 z m/s, calculate v at = 0 and z =:(a) 1 mm; (b) 2 cm: (c) 1 m.

Seatwork

1. Given the vector current density J =

102za - 4cos2a A/m

2: (a) find the

current density at P( = 3, = 30O, z = 2); (b) determine the total current flowing

outward through the circular band = 3, 0 < < 2, 2 < z < 2.8.

Answer 180a 9a A/m2; 518 A

Seatwork

2. Current density is given in cylindrical coordinates as J = - 106z1.5az A/m

2 in the region 0 20m; for 20 m, J = 0. (a) Find the total current crossing the surface z = 0.1m in the az direction. (b) If the charge velocity is 2 x 10

6 m/s at z = 0.1 m, find v there. (c) If the volume charge density at z = 0.15 m is -2000 C/m3, find the charge velocity there.

Answer -39.7 mA; - 15.81 kC/m3; -2900 m/s

METALLIC CONDUCTOR

Metallic Conductor permit a higher-energy level in the valence band to merges smoothly, to a conduction band by the help of kinetic energy produce by an external field that will result in an electron flow.

METALLIC CONDUCTOR

The valence electrons, or conduction, or free, electrons, having a charge Q = -e will move under the influence of an electric field, E, and will experience a force:

F = - eE

METALLIC CONDUCTOR

And the valence electron velocity (drift velocity) is linearly related to the electric field intensity by the mobility of the electron in a given materials (i.e. free space or crystalline)

vd = - eE (electron velocity is opposite in

direction to E)

where is the mobility of an electron (positive)

METALLIC CONDUCTOR

In terms of current density, J:

where:

Therefore;

EJ ee

ee

EJ

METALLIC CONDUCTOR

Consider a uniform current density J and electric field intensity E in a cylindrical region of length L and cross-sectional area S.

METALLIC CONDUCTOR

Therefore;

or

where:

or

but

JSdSJIS

ba

a

b

a

bab LEdLEdLEV

baab LEV IRV

ELV

L

VE

S

IJ S

LR

IS

LV

METALLIC CONDUCTOR

Insulator did not permit any electron flow due to an existing gap between the valence band and the conduction band. In this case the electron cannot accept any additional amounts of energy.

METALLIC CONDUCTOR

Semiconductors had a small forbidden region that separates the valence band and the conduction band. A small amounts of energy in the form of heat, light, or an electric field may raise the energy of the electrons and provide a conduction.

Sample Problem 3

Find the magnitude of the electric field intensity in a sample of silver having = 6.17 x 107 mho/m and e = 0.0056 m

2/V.s if: (a) the drift velocity is 1 mm/s; (b) the current density is 107 A/m2; (c) the sample is a cube, 3 mm on a side, carrying a total current of 80 A; (d) the sample is a cube, 3 mm on a side, having a potential difference of 0.5 mV between opposite faces.

Sample Problem 4

An aluminum conductor is 1000 ft long and

has a circular cross section with a

diameter of 0.8 in. If there is a dc voltage

of 1.2 V between the ends, find: (a) the

current density; (b) the current; (c) the

power dissipated, using your vast

knowledge of circuit theory.

Conductivity of Metallic Conductor

Seatwork

1. Find the magnitude of the current density in a

sample of silver for which = 6.17 x 107 S/m and e = 0.0056 m2/Vs if: (a) the drift velocity is 1.5 m/s; (b) the electric field intensity is 1 mV/m; (c) the sample is a cube 2.5 mm on a

side having a voltage of 0.4 mV between

opposite faces; (d) the sample is a cube 2.5 mm

on a side carrying a total current of 0.5A.

Answer: 16.53 kA/m2; 61.7 kA/m2; 9.87MA/m2;

80.0 kA/m2

Seatwork

2. A copper conductor has a diameter of 0.6 in and

it is 1200 ft long. Assume that it carries a total dc

current of 50 A. (a) Find the total resistance of

the conductor. (b) What current density exists in

it? (c) What is the dc voltage between the

conductor ends? (d) How much power is

dissipated in the wire?

Answer: 0.0346; 2.74 x 105 A/m2; 1.729 V; 86.4 W

CONDUCTOR PROPERTIES AND

BOUNDARY CONDITIONS

Conductor Characteristics:

1. It has a surface charge density that

resides on the exterior surface and within

has zero charge density.

2. In static conditions, in which no current

will flow, follows directly Ohms law: the electric field intensity within the conductor

is zero.


BOUNDARY CONDITIONS

The principles applied to conductors in electrostatic fields:

1. The static electric field intensity inside a conductor is zero.

2. The static electric field intensity at the surface of a conductor is everywhere directed normal to the surface.

3. The conductor surface is an equipotential surface.


BOUNDARY CONDITIONS

Prove:

Consider a closed surface conductor in a

free space boundary

( In static conditions, tangential

electric field intensity and electric

flux density are zero.)

0 dLE


BOUNDARY CONDITIONS

For the tangential field around the small closed path at the surface:

and E = 0,

where: a to b = c to d = w and b to c = d to a = h

0 a

d

d

c

c

b

b

a

02

1

2

1,, hEhEwE ataNatbNt


BOUNDARY CONDITIONS

( h = 0 and w 0(finite)) 0wEt0tE

0 tt ED


BOUNDARY CONDITIONS

Consider the normal field in a small cylinder as the Gaussian surface;

(Using Gauss Law)

(Cylinder Sides)

where: bottom and

sides integral were equal to zero

QdSDS

Qsidesbottomtop

SQSD SN


BOUNDARY CONDITIONS

or SND

SNN ED 0

SAMPLE PROBLEM 5

A potential field is given as V = 100e-5x sin

3y cos 4z V. Let point P(0.1,/12,/24) be located at a conductor-free space

boundary. At point P, find the magnitude

of: (a) V; (b) E; (c) EN; (d) Et; (e) S

Methods of Images

Dipole Characteristics infinite plane at zero potential that exists midway between the two charges.

Consider a vanishingly thin conducting plane that is infinite in extent and have an equipotential surface at a potential V = 0, and electric field intensity normal to the surface.

Methods of Images

This can be represented by a single charge (image) above the plane and maintain the same fields, removing the plane and locating a negative charge at a symmetrical location below the plane.

SAMPLE PROBLEM 6

A point charge of 25 nC is located in free

space at P(2,-3,5), and a perfectly

conducting plane is at z = 0. Find: (a) V at

(3,2,4); (b) E at (3,2,4); (c) S at (3,2,0).

Solution to Problem 6

Given:

z = 0

V = 0 at z = 0 at (3,2,4)

R = (3,2,4) (2,-3,5) = (1,5,-1) R = (3,2,4) (2,-3,-5) = (1,5,9)

(a)

(b)

Solution to Problem 6 (b)

(c)

SEMICONDUCTORS

Intrinsic Semiconductor pure material (i.e. germanium or silicon)

Two types of current carriers:

1. Electrons are those from the top of the filled valence band which have received sufficient energy (usually thermal) to cross the relatively small forbidden (1 electron volt energy gap) band into the conduction band.

SEMICONDUCTORS

2. Holes are vacancies left by the electrons represent unfilled energy states in the valence band which may also move from atom to atom in the crystal.

Both carriers move in an electric field, and they move in opposite directions; hence each contributes a component of the total current which is in the same direction as that provided by the other.

SEMICONDUCTORS

The conductivity is a function of both hole and electron concentrations and mobilities,

where:

= conductivity

e and h = electron and hole charge densities

e and h = mobility of electrons and holes

hhee

SEMICONDUCTORS

or

= e + h where: e = Neee h = Nheh Ne and Nh = electron and holes

concentrations

Electron charge, e = 1.602 x 10-19 C

e = mobility of electron (0.12, silicon; 0.36, germanium)

h = mobility of holes (0.025, silicon; 0.17, germanium )

SEMICONDUCTORS

Doping the process of adding impurities to a pure semiconductor.

Acceptors furnish extra holes and form p-type materials.

Donor provide additional electrons and form n-type materials.

SAMPLE PROBLEM 7

Using the values given in this section for

the electron and hole mobilities in

germanium at 300 K, and assuming hole

and electron concentrations of 2.7 x 1019

m-3, find: (a) the component of the

conductivity due to holes; (b) the

component of the conductivity due to

electrons; (c) the conductivity.


1 = Neeh = (2.7 x 1019)( 1.602 x 10-19)(0.17)

= 0.735 mho/m

2 = Neee = (2.7 x 1019)( 1.602 x 10-19)(0.36)

= 1.557 mho/m

= 1 + 2 = 0.735 + 1.557 = 2.29 mho/m

NATURE OF DIELECTRIC

MATERIALS

The characteristic which all dielectric materials have in common, whether they are solid, liquid, or gas, and whether or not they are crystalline in nature, is their ability to store electric energy.

This storage takes place by means of a shift in the relative positions of the internal, bound positive and negative charges against the normal molecular and atomic forces.


MATERIALS

A non-polar molecule does not have this dipole arrangement until after a field is applied. The negative and positive charges shift in opposite directions against their mutual attraction and produce a dipole which is aligned with the electric field.


MATERIALS

Qdp

n

i

itotal pp1

n

i

ipP1

0

1lim


MATERIALS

Consider a dielectric containing non-polar molecules, no dipole moment and P =0. In the interior, an incremental surface element , S and an electric field, E, produces a moment p = Qd in each molecule, such that p and d make an angle with S.


MATERIALS

Net total charge that crosses the elemental surface in an upward direction,

where: Qb = bound charge

Bound charge polarization,

Bound charge within the close surface,

SnQdQb

SPQb

Sb dSPQ


MATERIALS

Total enclosed charge,

where:

Q = total free charge enclosed by the surface S.

Enclosed free charge,

ST dSEQ 0

QQQ bT

SbT dSPEQQQ )( 0

PED N 0


MATERIALS

For Polarizable Material

where Q = free charge enclosed

In terms of volume charge densities

Equivalent Divergence relationships,

S dSDQ

vTT

vv

vbb

dvQ

dvQ

dvQ

T

b

E

P

0

vD


MATERIALS

In ferroelectric materials the relationship between P and E is not only nonlinear, but

also shows hysteresis effects.

The linear relationship between P and E is

where: e (chi) = a dimensionless quantity (electric susceptibility or dielectric

constant)

EP e 0


MATERIALS

Substituting to the value of D:

where:

Therefore

where: = permittivity

EEED ee 000 )1(

1 eR

R 0

SAMPLE PROBLEM 8

A certain homogenous slab of lossless

dielectric material is characterized by an

electric susceptibility of 0.12 and carries a

uniform electric flux density within it of 1.6

nC/m2. Find: (a) E; (b) P;(c) the average

dipole moment if there are 2 x 1019 dipoles

per cubic meter; (d) the voltage between

two equipotentials 1 in. apart.


Given:

Xe = electric susceptibility = 0.12

D = electric flux density = 1.6 nC/m2

dipoles/cu.m = 2 x 1019

a) D = 0E + Xe 0E = 0E(Xe + 1)

E = D/ [0(Xe + 1)] = (1.6 x 10-9)/[(8.854 x 10-12)(0.12 +1)]

= 161.35 V/m

b) P = Xe 0E = (0.12)(8.854 x 10-12)(161.35)

= 171.4 x 10-12 C/m2 or 171.4 pC/m2

c) Average dipole moment = (171.4 x 10-12)/(2 x 1019)

= 8.57 x 10-30 C-m

d) V = ES = (161.35)(0.0254) = 4.09829 V or 4.10 V

BOUNDARY CONDITIONS FOR

PERFECT DIELECTRIC MATERIALS

Boundary conditions existing at the interface between a conductor and a

dielectric:

1. D and E are both zero inside the

conductor.

2. The tangential E and D field

components must both be zero.

0 tt ED



3. The normal electric flux density is equal

to the surface charge density on the

conductor.

SNN ED



Consider the interface between two dielectrics having permitivities 1 and 2 and occupying regions 1 and 2 and the tangential components are given by,

0S dLE

02tan1tan wEwE (around the close path)

2tan1tan EE (h = negligible)



If the tangential electric field intensity is continuous across the boundary, then the

tangential D is discontinuous, for

or

2

2tan2tan1tan

1

1tan

DEE

D

2

1

2tan

1tan

D

D



Applying the Gausss law,

QdSDS

SQSDSD SNN 21

SNN DD 21



(normal E is discontinuous)

(S = 0, the normal component of D is

continuous) or

SNN DD 21

21 NN DD

2211 NN EE

SAMPLE PROBLEM 9

The region y < 0 contains a dielectric

material for which R1 = 2.5, while the region y > 0 is characterized by R2 = 4. Let E1 = -30ax + 50ay + 70az V/m, and

find: (a) EN1; (b) Et1; (c)Et1; (d) E1; (e) 1


Given:

R1 = 2.5

R2 = 4

E1 = -30ax + 50ay + 70az V/m; y = normal

a) EN1 = E1ay = (-30ax + 50ay + 70az) ay = 0 +

50 + 0 = 50 V/m

b) Et1 = E1 EN1 = -30ax + 50ay + 70az - 50ay = -30ax + 70az V/m

c) Et1 = |Et1| =


d) E1 = |E1| =

e)

1 = cos-1 (E1ay)/|E1| = 56.71

O

SAMPLE PROBLEM 10

Continue problem 9 by finding: (a) DN2; (b)

Dt2; (c) D2; (d) P2; (e) 2


E1 = -30ax + 50ay + 70az E2 = -30ax + Ey2ay + 70az D1 = 0R1E1 = (8.854 x 10-12)(2.5)(-30ax + 50ay + 70az) = - 0.664 x 10

-

9ax + 1.107 x 10-9ay + 1.55 x 10

-9az D2 = Dx2ax + 1.107 x 10

-9ay + Dz2az

D2 = 0R2E2 Dx2ax + 1.107 x 10

-9ay + Dz2az = (8.854 x 10-12)(4)( -30ax +

Ey2ay + 70az)

Dx2 = -1.062 x 10-9

Ey2 = 31.26

Dz2 = 2.48 x 10-9

D2 = -1.062 x 10-9ax + 1.107 x 10-9ay + 2.48 x 10

-9az D2 = -1.062ax + 1.107ay + 2.48az nC/m

2


a) DN2 = D2ay = (-1.062ax + 1.107ay + 2.48az) ay = 0 + 1.107 + 0 = 1.107 nC/m2

b) Dt2 = Et20R2 = (76.2)(8.854 x 10-12)(4) = 2.70 x 10-9

C/m2 = 270 nC/m2

c) D2 = -1.062ax + 1.107ay + 2.48az nC/m2

d)

P2 = -0.797ax + 0.830ay + 1.86az C/m2

e) ;

CAPACITANCE

Capacitance the ratio of the magnitudes of the total charge on either conductor to

the potential difference between

conductors. And it is measured in farads

(F), where a farad = 1 Coulomb/Volt

where: Q = surface integral over the

positive conductor.

0V

QC

CAPACITANCE

CAPACITANCE

Consider two conductors embedded in a homogenous dielectric. Conductor M2

carries a total positive charge Q, and M1

carries an equal negative charge. There

are no other charges present, and the total

charge of the system is zero.

CAPACITANCE

Let Vo, the potential difference between M2 and M1, and carrying a unit charge

from the negative to the positive surface,

and the capacitance is given by,

dLE

dSEC S

CAPACITANCE

CAPACITANCE

Consider a simple two-conductor system in which the conductors are identical,

infinite parallel planes with separation d.

Choosing the lower conducting plane at z

= 0 and the upper one at z = d, a uniform

sheet of surface charge S on each conductor leads to the uniform field.

CAPACITANCE

( = permittivity of the homogenous dielectric)

(charge on the lower plane)

(charge on the upper plane)

Potential difference between the lower and upper plane,

zS aE

zSaD

SZN DD

ZN DD

ddzdLEV Sd

Slower

upper

0

0

CAPACITANCE

where:

S = surface charge density

S = surface area

or

The total energy stored in the capacitor,

Therefore,

And the charge value,

SQ S

d

S

V

QC

0

2

222

0 0 2

22

2

1

2

1

2

1

2

1

d

d

SSddSdzdvEW

SSS d S

volE

C

QQVCVWE

2

0

2

02

1

2

1

2

1

SAMPLE PROBLEM 11

Find the relative permittivity of the

dielectric material used in a parallel plate

capacitor if : (a) C = 40 nF, d = 0.1 mm,

and S = 0.15 m2, (b) d = 0.2 mm, E = 500

kV/m, and S = 10 C/m2; (c) D = 50 C/m2 and the energy density is 20 J/m3.

Solution to Problem 11 (a)

(b)

Solution to Problem 11 (c)

Given:

D = 50 x 10-6 C/m2

W = 20 J/m3

SEVERAL CAPACITANCE

Consider a Coaxial cable or coaxial capacitor that has inner radius a, outer

radius b, and length L. The value of the

capacitance is given by,

)/ln(

2

ab

LC

SEVERAL CAPACITANCE

Consider a spherical capacitor formed of two concentric spherical conducting shells of radius a and b, b > a. The expression for the electric field,

( = dielectric permittivity

between the region of the

spheres)

24 r

QEr

SEVERAL CAPACITANCE

Potential difference using the line integral,

(Q =represents the total charge o the inner sphere)

ba

QVab

11

4

SEVERAL CAPACITANCE

The Capacitance is given by,

or

(allowing the outer sphere to become infinitely large, b = .)

ba

V

QC

ab11

4

aC 4

SEVERAL CAPACITANCE

Coating the sphere with a different dielectric layer, for which = 1, extending from r = a to r = r1,

)(4

)(4

4

12

0

12

1

2

rrr

QE

rrar

QE

r

QD

r

r

r

SEVERAL CAPACITANCE

The potential difference is,

1011

2

0

2

1

1111

4

44

1

1

rra

QVV

r

Qdr

r

QdrVV

a

ra

ra

SEVERAL CAPACITANCE

Therefore,

1011

1111

4

rra

C

SEVERAL CAPACITANCE

Consider a parallel-plate capacitor of area S and spacing d, where d is small

compared to the linear dimensions of the

plate. The capacitance is given by 1S/d, using a dielectric permittivity 1. Suppose another part of the capacitor dielectric has

a permittivity of 2, then the potential difference given by Vo is,

SEVERAL CAPACITANCE

(E1 and E2 are

both uniform)

(E is normal)

or

22110 dEdEV

21 NN DD

2211 EE

SEVERAL CAPACITANCE

Eliminating E2;

The surface charge density,

Since D1 = D2, the capacitance is given by,

21210

1/dd

VE

2

2

1

1

01111

dd

VEDS

212

2

1

110011

11

CCS

d

S

dV

S

V

QC S

SAMPLE PROBLEM 12

Find the capacitance of: (a) 20 cm of 58C/U coaxial cable having an inner conductor 0.0295 in. in diameter, an outer conductor having an inside diameter of 0.116 in., and a polyethylene dielectric; (b) a conducting sphere 1 cm in diameter, covered with a layer of polyethelene 1 cm thick, in free space; (c) a conducting sphere 1 cm in diameter, covered with a layer of polyethelene 1 cm thick, and surrounded by a concentric conducting sphere 1.5 cm in radius.


a)

b)

c)


b)

SAMPLE PROBLEM 13

A parallel-plate capacitor contains three

dielectric layers, Let R1 = 1, d1 = 0.2 mm, R2 = 2, d2 = 0.3 mm, R3 = 3, d3 = 0.4 mm, and S = 20 cm2. (a) Find C. (b) Find the

percentage of the total stored energy

located in each of the three regions.


a)

b)

CAPACITANCE OF A TWO-WIRE

LINE


LINE

Consider the potential field of two infinite line charges. A positive line charge in the

xz plane at x = a and a negative line

charge at x = -a. The potential of a single

line charge with zero reference at a radius

of Ro is,

R

RV L 0ln

2


LINE

Combining the potential field,

Letting R10 = R20 (placing zero reference at equal distance from each line) and R1 = x and R2 =y,

120

210

2

20

1

10 ln2

lnln2 RR

RR

R

R

R

RV LL

22

22

22

22

ln4

ln2 yax

yax

yax

yaxV LL


LINE

Choosing an equipotential surface V = V1 and letting K1 = dimensionless parameter of potential V1

LVeK /4

11


LINE

Therefore, 22

22

1yax

yaxK

2

1

12

2

1

1

1

2

1

1

K

Kay

K

Kax

01

12 22

1

12

ay

K

Kaxx


LINE

In equipotential surface where V = V1, the surface is independent value of z ( a cylinder) and intersects the xy plane in a circle of radius b,

Centered at x = h, y = 0;

where: ( x = 0)

1

2

1

1

K

Kab

1

1

1

1

K

Kah

22 bha

b

bhhK

22

1


LINE

But

Therefore,

The value of the capacitance is given by,

or

LVeK /2

10

1

0

ln

4

K

VL

110 ln

2

ln

4

K

L

K

L

V

LC L

)/(cosh2

/ln

2122 bh

L

bbhh

LC

SAMPLE PROBLEM 14

Find the capacitance between a circular

conducting cylinder in air, 2 mm in radius,

and: (a) a conducting plane, 1 cm distant

from the cylinder axis; (b) a conducting

plane, 1 cm distant from the cylinder axis;

(c) a similar cylinder, axes separated by 1

cm.


a)

b)

c)

L = 1

= 0 = 8.854 x 10-12

r = b = 2mm = 2 x 10-3 m

d = h = 0.01 m


0.024 = 1cm + 1cm + 2mm + 2mm

CONDUCTORS, DIELECTRICS, AND CAPACITANCE.pdf

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