Introduction to Conductors & Dielectrics

CONDUCTORS & DIELECTRICS

Presented by

R.Lenin Raja, B.E.,M.E.,MIEEE.,(PhD)

Assistant Professor/ECE,

Sri Vidya College of Engineering & Technology.

1FDTP on Electromagnetic Fields [EC6403]4 July 2015

Electricity & Magnetism – Current & Resistance


Definition of Current• Current is the flow of electrical

charge, i.e. amount of charge persecond moving through a wire, i =dq/dt.

• It is a scalar, not a vector, but it has adirection—positive in the direction offlow of positive charge carriers.

• Any way that you can get charges tomove will create a current, but atypical way is to attach a battery to awire loop.

• Charges will flow from the + terminalto the – terminal (again, it is reallyelectrons that flow in the oppositedirection, but current is defined asthe direction of positive chargecarriers).

Units: ampere 1 A = 1 C/s

Total current in

11 A

Total current out

- 3 A = 8 A

8 A

4 July 2015

Current in a Circuit

• What is the current in the wire marked i in the figure below?

Total current in

11 A

Total current out

- 3 A = 8 A

8 A


Current At Junctions

• What is the current in all of the wire sections that are not marked?

8 A

2 A

5 A 3 A

6 A


Current Density

• When we care only about the total current i in a conductor, we do not have to worry about its shape.

• However, sometimes we want to look in more detail at the current flow inside the conductor. Similar to what we did with Gauss’ Law (electric flux through a surface), we can consider the flow of charge through a surface. To do this, we consider (charge per unit time) per unit area, i.e. current per unit area, or current density. The units are amps/square meter (A/m2).

• Current density is a vector (since it has a flow magnitude and direction). We use the symbol . The relationship between current and current density is

J

AdJi

Small current densityin this region

High current densityin this region


Resistance• Resistance is defined to be . That is, we apply a voltage V, and ask how

much current i results . This is called Ohm’s Law.

• If we apply the voltage to a conducting wire, the current will be very large so R is small.

i

VR

+

_

1.5 Vbatteryi

i large, soR small

wire

If we apply the voltage to a less conducting material, such as glass, the current will be tiny so R is very large.

i

i small, soR large glass

filament

The unit of resistance is the ohm, W. (Greek letter omega.)

1 ohm = 1 W = 1 volt per ampere = 1 V/A

RVCircuitDiagram

Resistor


Current Through a Resistor

3. What is the current through the resistor in the following circuit, if V = 20 V and R = 100 W?

A. 20 mA.

B. 5 mA.

C. 0.2 A.

D. 200 A.

E. 5 A.

RVCircuitDiagram

4 July 2015

Current Through a Resistor

4. If the current is doubled, what changes?

A. The voltage across the resistor doubles.

B. The resistance of the resistor doubles.

C. The voltage in the wire between the battery and the resistor doubles.

D. The voltage across the resistor drops by a factor of 2.

E. The resistance of the resistor drops by a factor of 2.

RVCircuitDiagram

4 July 2015

Resistivity and Conductivity• Rather than consider the overall resistance of an object, we can discuss the

property of a material to resist the flow of electric current.

• This is called the resistivity. The text uses (re-uses) the symbol r for resistivity. Note that this IS NOT related to the charge density, which we discussed earlier.

• The resistivity is related not to potential difference V and current i, but to electric field E and current density J.

J

Er

Units V/m over A/m2 = Vm/A = ohm-meter =W m

Definition of resistivity

Note that the ability for current to flow in a material depends not only on the material, but on the electrical connection to it.

High resistance

Low resistance

r

1 Definition of conductivity

Note use (re-use) of for conductivity. NOT surface charge density.


More on Resistivity• Since resistivity has units of ohm-meter, you might think that you can just divide by

the length of a material to find its resistance in ohms.

?/ LR r

AiJLVE / and / since

LRAAi

LV

J

E/

/

/rresistivity is

A

LR r Resistance from resistivity

Dependence on temperature: you can imagine that a higher temperature of a material causes greater thermal agitation, and impedes the orderly flow of electricity. We consider a temperature coefficient a: )( 000 TT -- arrr


Resistivity of a Resistor

5. Three resistors are made of the same material, with sizes in mm shown below. Rank them in order of total resistance, greatest first.

A. I, II, III.

B. I, III, II.

C. II, III, I.

D. II, I, III.

E. III, II, I.

4

4

52

6

3

I.

II.

III.

Each hassquarecross-section

4 July 2015

Electric Power

• Recall that power is energy per unit time, (watts). Recall also that for an arrangement of charge, dq, there is an associated potential energy dU = dqV.

• Thus,

• In a resistor that obeys Ohm’s Law, we can use the relation between R and i, or Rand V, to obtain two equivalent expressions:

• In this case, the power is dissipated as heat in the resistor.

dt

dUP

iVVdt

dqP Rate of electrical energy transfer

Units: 1 VA = (1 J/C)(1 C/s) = 1 J/s = 1 W

RiP 2

R

VP

2

Resistive dissipation


Superconductivity

• In normal materials, there is always some resistance, even if low, to current flow. This seems to make sense—start current flowing in a loop (using a battery, say), and if you remove the battery the current will eventually slow and stop.

• Remarkably, at very low temperatures (~4 K) some conductors lose all resistance. Such materials are said to be superconductors. In such a material, once you start current flowing, it will continue to flow “forever,” like some sort of perpetual motion machine.

• Nowadays, “high-temperature” superconductors have been discovered that work at up to 150 K, which is high enough to be interesting for technological applications such as giant magnets that take no power, perhaps for levitating trains and so on.


Electrostatics

(Free Space With Charges & Conductors)

Outline

Maxwell’s Equations (In Free Space)

Gauss’ Law & Faraday’s Law

Applications of Gauss’ Law

Electrostatic Boundary Conditions

Electrostatic Energy Storage


E-Gauss:

Faraday:

H-Gauss:

Ampere:

Maxwell’s Equations (in Free Space with Electric Charges present)

DIFFERENTIAL FORM INTEGRAL FORM

Static arise when , and Maxwell’s Equations split into decoupled

electrostatic and magnetostatic eqns.

Electro-quasistatic and magneto-quasitatic systems arise when one (but not

both) time derivative becomes important.

Note that the Differential and Integral forms of Maxwell’s Equations are related through

Stoke’s Theorem and Gauss’ Theorem 16FDTP on Electromagnetic Fields [EC6403]4 July 2015

Charges and Currents

There can be a nonzero charge density in the absence of a current density .

There can be a nonzero current density in the absence of a charge density .

Charge

conservation

and KCL for

ideal nodes


Flux of through closed surface S = net charge inside V

Gauss’ Law


Point Charge Example

Apply Gauss’ Law in integral form

making use of symmetry to find

• Assume that the image charge is

uniformly distributed at .

Why is this important ?

• Symmetry


Gauss’ Law Tells Us …

… the electric charge can reside only on the surface of the conductor.

[If charge was present inside a conductor, we can draw a Gaussian surface around

that charge and the electric field in vicinity of that charge would be non-zero !

A non-zero field implies current flow through the conductor, which will transport the

charge to the surface.]

… there is no charge at all on the inner surface of a hollow conductor.

… that, if a charge carrying body has a sharp point, then the electric field at that

point is much stronger than the electric field over the smoother part of the body.

Lets show this by considering

two spheres of different size,

connected by a long, thin wire …20FDTP on Electromagnetic Fields [EC6403]4 July 2015

ba

SPHERE B

SPHERE A

WIRE

Because the two spheres are far apart, we can

assume that charges are uniformly distributed

across the surfaces of the two spheres,

with charge qa on the surface of sphere A

and qb on the surface of sphere B

… and the E-field on the surface of the spheres is:

Note that Ea >> Eb if b >> a 21FDTP on Electromagnetic Fields [EC6403]4 July 2015

Lighting Rods are connected to the ground. When a cloud

carrying electric charges approaches, the rod attracts opposite

charges from the ground. The Electric field at the tip of the

rod is much stronger than anywhere else. When the E-field

exceeds the

air breakdown strength (of 33 kV/cm), charges start to travel

to ground.

Lightning Rod

When a conductive body

contains sharp points,

the electric field on

these points is much

stronger than that on the

smooth part of the

conducting body.


Faraday’s Law

Path I Path II

A unique path-independent potential

may be defined if and only if

Path I Path II

and

Dynamic form:

Static form:


Normal is discontinuous at a surface charge.

Tangential is continuous at a surface.

A static field terminates

perpendicularly on a conductor

Boundary Conditions


Point Charges Near Perfect Conductors

+ + + + + + ++ + + + + + ++ + + + + + +

-

- - - -

---

--

-

-

--

----

-

-

-

--

Time t = 0

Time t >> 0

+ + + + + + ++ + + + + + ++ + + + + + +

--

- - -

---

--

-- -

--

---- -

-- -


Point Charges Near Perfect Conductors

- --- -- -- --- -- - -- + + + ++ + + ++ + + ++ + + +

+ + + +

Negative charge on top

surface of conductorPositive charge on

top and bottom

surface of

conductor

+


Uniqueness and Equivalent Image Charges

- --- -- -- --- -- - -- + + + ++ + + +

Equivalent Image Charge


Electrostatic Boundary Conditions

Tangential field is continuous

There is a jump in the normal electric field as

one passes through a surface charge


Energy Stored in

Electric Fields

1. Begin with a neutral

reference conductor, the

charge reservoir. Its

potential is zero, by

definition.

2. Move charges from the

reference conductor into

free space, thereby creating

an electric field and doing

work in the process. The

work is stored as potential

energy in the electric fields.

3. Account for all the work

done, and thereby derive

the energy stored in the

electric fields.

4. The argument directly

extends to systems with

multiple conductors

(and dielectrics).

• The work done by moving charge to a location

with potential is . More generally, the work

done to make an incremental charge change to a

charge density is

• Gauss’ Law

ZERO ! WHY ? ENERGY DENSITY [J/m3]


Energy Stored in Electric Fields

The energy stored in an electric field is ½ ε0 E2.

The maximum achievable field strength is

typically limited by electric breakdown

Note: Dielectric constant is


Energ

y [

eV]

+

_

+

_Remember this

unit of energy:

1 eV = 1.6 x 10-19 J

0 V

ENERGY OF AN ELECTRON IN AN

ALKANEMOLECULE IN

GASOLINE

ENERGY OF AN ELECTRON IN H2O or CO2

ENERGY GIVEN OFF AS HEAT IN THE PROCESS OF GASOLINE COMBUSTION

O = C = OO

H H

H H

H - C – C – H

H H

--

--

1.16 Å1.54 Å

1.10 Å

Hydrogen ground state

energy is -13.6 eV

If the hydrogen radius was

twice as long, what would be

the ground state energy ?

Modeling Atoms and Molecules

as Capacitors that Store Energy

These product molecules have shorter bond lengths than the initial reactant molecule, hence the charge in them

sits closer together. This can be modeled as a higher capacitance.

Since voltage V = Q/C = E d is reduced,stored energy W/Volume = ½ εoE

2

is reduced in these molecules

Burn this molecule by reacting it with oxygen


Leyden Jar


Franklin’s motor (1748)


KEY TAKEAWAYS

• Maxwell’s Equations (in Free Space with Electric Charges present):

E-Gauss:

Faraday:

H-Gauss:

Ampere:

DIFFERENTIAL FORM INTEGRAL FORM

AIR BREAKDOWN STRENGTH is 33 kV/cm

• Boundary conditions for E-field:

. Normal E-field – discontinuous

. Tangential E-Field - continuous

• Energy stored in the

electric field per unit volume is:

• Dielectric constant in free space is

NEW UNIT OF ENERGY:

• Energy released when fuel molecules are oxidized since the charges in the products

are positioned closer together than in reactants (hence in a lower energy state)

GOOD FACTS TO REMEMBER:


Polarization


Propagation and polarizationIn isotropic media(e.g. free space, amorphous glass, etc.)

More generally,

(reminder :Anisotropic in media, e.g. crystals, one could E not have parallel to D)

planar wavefront

electric field vector E

wave-vector k


Linear polarization (frozen time)


Linear polarization (fixed space)


Circular polarization (frozen time)


Circular polarization:linear components


Circular polarization (fixed space)


/4 plate

Linear polarization

birefringentl/4 plate

Circular polarization


λ/2 plate

Linear (90o-rotated)

polarization

Linear polarization

birefringentλ/2 plate


Think about that

mirror

birefringent

λ/4 plate

Linear polarization


Relationship between E and B

Vectors k, E, B form a

right-handed triad.

Note: free space or isotropic media only


Reflection & transmission@ dielectric interface

I. Polarization normal to plane of incidence




Continuity of tangential electric fieldat the interface:

Since the exponents must be equalfor all x, we obtain




Continuity of tangential electric fieldat the interface:

law of reflection

Snell’s law of refraction

so wave description is equivalent to Fermat’s principle!! ☺




Incident electric field:

Reflected electric field:

Transmitted electric field:

Need to calculate the reflected and transmitted amplitudes E0r, E0t

i.e. need two equations




Continuity of tangential electric fieldat the interface gives us one equation:

which after satisfying Snell’s lawbecomes




The second equation comes from continuity of tangential magnetic field

at the interface:




So continuity of tangential magnetic field Bx at the interface y=0 becomes:



II. Polarization parallel to plane of incidence

Following a similar procedure ...


1-D Finite-Element Methods

with Poisson’s Equation


Outcomes Based Learning

Objectives

By the end of this laboratory, you will:

– Understand how to approximate the heat-

conduction/diffusion and wave equations in

two and three dimensions

– You will understand the differences between

insulated and Dirichlet boundary conditions


The Target Equation

Recall the first of Maxwell’s equations (Gauss’s equation):

If we are attempting to solve for the underlying potential function, under the assumption that it is a conservative field, we have

This is in the form of Poisson’s equation

0

r

E

2

0

ur


The Target Equation

In one dimension, this simplifies to:

Define:

and thus we are solving for

2

2

0

xdu x

dx

r

2

2

0

def xdV x u x

dx

r

-

0V x


The Integral

If , it follows that

for any test function f(x) and therefore

Substituting the alterative definition of V(x)into this equation, we get

0x V xf

0V x

0

b

a

x V x dxf

2

2

0

0

b

a

xdx u x dx

dx

rf

-


The Integral

Consider the first test function f1(x) :

3

1

2 2

2 210

0

xb

a x

xd dx u x dx u x dx

dx dx

rf

-


Integration by Parts

Again, take

but before we apply integration by parts, expand the integral:

Everything in the second integral is known: bring it to the right:

2

2

0

0

b

a

xdx u x dx

dx

rf

-

2

2

0

0

b b

a a

xdx u x dx x dx

dx

rf f

-

2

2

0

b b

a a

xdx u x dx x dx

dx

rf f



The left-hand integral is no different from before,

and performing integration by parts, we have

0

0

b b b

a aa

b b

a ax b x a

xd d dx u x x u x dx x dx

dx dx dx

xd d d db u x a u x x u x dx x dx

dx dx dx dx

rf f f

rf f f f

-

- -

2

2

0

b b

a a

xdx u x dx x dx

dx

rf f



First, substituting in the first test function:

which yields

3 3

1 13 1

2 3 2 1 2 2

0

x x

x xx x x x

xd d d dx u x x u x x u x dx x dx

dx dx dx dx

rf f f f

- -

2

2 22

0

b b

a a

xdx u x dx x dx

d x

rf f

1 1 1

1 3x x x x - - -


The System of Linear Equations

Again, recall that we approximated the solution by unknown piecewise linear functions:

where we define on

2 11 2 1 2

1 2 2 1

3 22 3 2 3

2 3 3 2

11 1

1 1

n nn n n n

n n n n

x x x xu u x x x

x x x x

x x x xu u x x x

x x x xu x

x x x xu u x x x

x x x x

-- -

- -

- - - -

- -

- -

- - - -

11

1 1

defk k

k k k

k k k k

x x x xu x u u

x x x x

- -

- -

1k kx x x


Unequally Spaced Points

Recall that we approximated the left-hand integral by substituting the piecewise linear functions to get:

1 1

1 1

1

1

1

2

2

0

2

2

0

1 1

1 1 0

1 1 1

1 1 1 1 0

2 2

1 1 1 12 2 2

k k

k k

k

k

k

b b

k k

a a

x x

x x

x

k k k k

k k k k x

k k k

k k k k k k k k x

xdx u x dx x dx

d x

xdu x dx dx

d x

xu u u udx

x x x x

xu u u dx

x x x x x x x x

rf f

r

r

r

- -

-

-

-

-

- -

- -

- --

- -

-

- - - -

1kx


67

LAPLACE’S EQUATION, POISSON’S EQUATION AND UNIQUENESS THEOREM

1 LAPLACE’S AND POISSON’S EQUATIONS

2 UNIQUENESS THEOREM

3 SOLUTION OF LAPLACE’S EQUATION IN ONE VARIABLE

4 SOLUTION FOR POISSON’S EQUATION

FDTP on Electromagnetic Fields [EC6403]4 July 2015

LAPLACE’S AND POISSON’S EQUATIONS AND UNIQUENESS THEOREM

- In realistic electrostatic problems, one seldom knows the charge distribution –thus all the solution methods introduced up to this point have a limited use.

- These solution methods will not require the knowledge of the distribution of charge.


1 LAPLACE’S AND POISSON’S EQUATIONS

To derive Laplace’s and Poisson’s equations , we start with Gauss’s law in point form :

vED r

VE -Use gradient concept :

r

r

v

v

V

V

-

-

2Operator :

Hence :

(1)

(2)

(3)

(4)

(5) => Poisson’s equation

is called Poisson’s equation applies to a homogeneous media.

22 / mVV v

r-


0vrWhen the free charge density

=> Laplace’s equation(6)22 / 0 mVV

2

2

2

2

2

22

z

V

y

V

x

VV

In rectangular coordinate :


2 UNIQUENESS THEOREM

Uniqueness theorem states that for a V solution of a particular electrostatic problem to be unique, it must satisfy two criterion :

(i) Laplace’s equation

(ii) Potential on the boundaries

Example : In a problem containing two infinite and parallel conductors, oneconductor in z = 0 plane at V = 0 Volt and the other in the z = d plane at V = V0 Volt,we will see later that the V field solution between the conductors is V = V0z / dVolt.

This solution will satisfy Laplace’s equation and the known boundary potentials at z= 0 and z = d.

Now, the V field solution V = V0(z + 1) / d will satisfy Laplace’s equation but will notgive the known boundary potentials and thus is not a solution of our particularelectrostatic problem.

Thus, V = V0z / d Volt is the only solution (UNIQUE SOLUTION) of our particularproblem. 71FDTP on Electromagnetic Fields [EC6403]4 July 2015

3 SOLUTION OF LAPLACE’S EQUATION IN ONE VARIABLE

Ex.1: Two infinite and parallel conducting planes are separated d meter, with one of the conductor in the z = 0 plane at V = 0 Volt and the other in the z = d plane at V = V0 Volt. Assume and between the conductors.0

2 0v

r

Find : (a) V in the range 0 < z < d ; (b) between the conductors ;

(c) between the conductors ; (d) Dn on the conductors ; (e) on the conductors ; (f) capacitance per square meter.

E

D sr

Solution :

0v

rSince and the problem is in rectangular form, thus

02

2

2

2

2

22

z

V

y

V

x

VV (1)

(a)


0

0

2

2

2

2

22

dz

V

dz

d

dz

Vd

z

VV

We note that V will be a function of z only V = V(z) ; thus :

BAzV

Adz

dV

Integrating twice :

where A and B are constants and must be evaluated using given potential values at the boundaries :

00

BVz

dVA

VAdVdz

/0

0

(2)

(3)

(4)

(5)

(6)

(7)


)(0 Vzd

VV

Substitute (6) and (7) into general equation (5) :

dz 0

)/(ˆˆ

ˆˆˆ

0 mVd

Vz

z

Vz

z

Vz

y

Vy

x

VxVE

-

-

--

(b)

)/(2ˆ

200 mCd

VzED

-

(c)


)/(2

)ˆ(2

ˆˆ

2

ˆ2

ˆˆ

200

00

00

00

0

mCd

V

zd

VznD

d

V

zd

VznD

dzs

zs

r

r

--

-

-

(d) Surface charge :

0

/

V

ds

VQC

s

ab

r

)/(/2

/

2

0

00

0

2

mFdV

dV

VmC s

r

(e) Capacitance :

z = 0

z = d

V = 0 V

V = V0 V


Ex.2: Two infinite length, concentric and conducting cylinders of radii a and b are located on the z axis. If the region between cylinders are charged free and , V = V0 (V) at a, V = 0 (V) at b and b > a. Find the capacitance per meter length.

03

Solution : Use Laplace’s equation in cylindrical coordinate :

and V = f(r) only :

011

2

2

2

2

2

2

z

VV

rr

Vr

rrV

f


BrAV

r

A

r

V

Ar

Vr

r

Vr

r

r

Vr

rrV

ln

0

012

and V = f(r) only :

(1)


BrAV ln

BbAV

BaAVV

br

ar

ln0

ln0

Boundary condition :

ba

bVB

ba

VA

/ln

ln;

/ln

00 -

Solving for A and B :

ab

rbVV

/ln

/ln0

Substitute A and B in (1) :

(1)

bra ;


r

abr

VED

rabr

V

r

VrVE

ˆ/ln

ˆ/ln

ˆ

0

0

--

ab

rbVV

/ln

/ln0

abb

VrD

aba

VrD

brs

ars

/lnˆ

/lnˆ

0

0

r

r

--

Surface charge densities:

ab

Vb

ab

Va

brsbr

arsar

/ln

22

/ln

22

0

0

rr

rr

-

Line charge densities :


oab V

d

V

QC

r

Capacitance per unit length:

)/(

/ln

2/

0

mFabV

mCr


6/ and 0 ff 0f

VV 100 6/f

EV and

Ex.3: Two infinite conductors form a wedge located at

is as shown in the figure below. If this region is characterized by charged free. Find . Assume V = 0 V at

and at .

z

x f = 0

f = /6

V = 100V


Solution : V = f ( f ) in cylindrical coordinate :

01

2

2

2

2 fd

Vd

rV

BAV

Ad

dV

d

Vd

f

f

f0

2

2

f

f

/600

)6/(100

0

6/

0

A

AV

BV


Hence :

f

ff

ˆ600

ˆ1

r

d

dV

rVE

-

--

f

600V

6/0 f

for region :


BAV

A

d

dV

Ad

dV

d

dV

d

d

d

dV

d

d

rV

2/tanln

sin

sin

0sin

0sinsin

12

2

= /10

= /6

V = 50 V

x

y

z

6/ and 10/

E

Ex.4: Two infinite concentric conducting cone located at

10/ . The potential V = 0 V at

6/ and V = 50 V at . Find V and between the two conductors.

Solution : V = f ( ) in spherical coordinate :

2/tanlnsin

dUsing : 83FDTP on Electromagnetic Fields [EC6403]4 July 2015

BAV 2/tanln

BAV

BAV

12/tanln50

20/tanln0

6/

10/


Solving for A and B :

-

20/tan

12/tanln

20/tanln50 ;

20/tan

12/tanln

50

BA

1584.0

2/tanln1.95

20/tan

2/tanln

20/tan

12/tanln

50

V

ˆsin

1.95

ˆ1

r

d

dV

rVE

-

-

6/10/ Hence at region :

and


4 SOLUTION FOR POISSON’S EQUATION

0vrWhen the free charge density

Ex.5: Two infinite and parallel conducting planes are separated d meter, with one of the conductor in the x = 0 plane at V = 0 Volt and the other in the x = d plane at V = V0 Volt. Assume and between the conductors.04 0

vr

Find : (a) V in the range 0 < x < d ; (b) between the conductors E

Solution :

BAxx

V

Axdx

dV

dx

Vd

V v

-

-

-

-

2

2

0

0

0

2

2

2

r

r

r

rV = f(x) :


2

2

0

00

2

0

0

0

d

d

VA

Add

VV

BV

dx

x

r

r

-


BAxx

V -2

2

0

r

dx 0In region :

xd

Vxd

xV 00

2-

r

xxd

d

V

xdx

dVE

ˆ2

ˆ

00

--

-

r

;


xrv 1 and 0 rEx.6: Repeat Ex.6.5 with

BxAV

x

A

dx

dV

Adx

dVx

Vxdx

d

Exdx

d

E

D v

-

-

-

-

)1ln(

1

1

01

01

0

0

rSolution :


)1ln(

)1ln(

0

0

0

0

d

VA

dAVV

BV

dx

x

-

-


xdx

Vx

dx

dVE

d

xVV

ˆ)1ln()1(

ˆ

)1ln(

)1ln(

0

0

--

dx 0In region :BxAV - )1ln(


?Contact : R.Lenin raja,

Contact: 088700-82081 ; 08807082081Email: [email protected] ; [email protected]


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mailto:[email protected]

THANK YOU


Introduction to Conductors & Dielectrics

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