Part 1:Electrostatics

PART 1

ELECTROSTATICS

2.1 COULUMB’S LAW

2.2 ELECTRIC FIELD INTENSITY

2.3 LINE, SURFACE & VOLUME CHARGES

2.4 ELECTRIC FLUX DENSITY

2.5 GAUSS’S LAW

2.6 ELECTRIC POTENTIAL

2.7 BOUNDARY CONDITIONS

2.8 CAPACITANCE

INTRODUCTION

Electromagnetics is the study of the effect of charges at rest and charges in motion.

Some special cases of electromagnetics:

Electrostatics: charges at rest

Magnetostatics: charges in steady motion (DC)

Electromagnetic waves: waves excited by charges in time-varying motion

Maxwell’sequations

Fundamental laws of classical electromagnetics

Special cases

Electro-statics

Magneto-statics

Electro-magnetic

Kirchoff’s Laws

Statics: 0t

Geometric Optics

TransmissionLine

TheoryCircuitTheory

Input from other

disciplines

INTRODUCTION (Cont’d)

• Electrical phenomena caused by friction

are part of our everyday lives, and can be

understood in terms of electrical charge.

• The effects of electrical charge can be

observed in the attraction/repulsion of

various objects when “charged.”

• Charge comes in two varieties called “positive” and “negative.”

• Objects carrying a net positive charge attract those carrying a net negative charge and repel those carrying a net positive charge.

• Objects carrying a net negative charge attract those carrying a net positive charge and repel those carrying a net negative charge.

• On an atomic scale, electrons are negatively charged and nuclei are positively charged.

• Electric charge is inherently quantized such that the charge on any object is an integer multiple of the smallest unit of charge which is the magnitude of the electron charge e = 1.602 10-19 C.

• On the macroscopic level, we can assume that charge is “continuous.”

COULUMB’S LAW

In the late 18th century, Colonel Charles

Augustus Coulomb invented a sensitive

torsion balance that he used to

experimentally determine the force

exerted in one charge by another.

COULUMB’S LAW (Cont’d)

He found that the force is proportional to the

product of two charges, inversely proportional

to the square of the distance between the

charges and acts in a line containing the two

charges.

The proportional constant, k is:

36101085.8

Where the free space permittivity with a value given by:

041 rk ,

Charge Q1 exerts a vector force F12 in Newton's (N) on charge Q2,

122120

If more than two charges, use the principle ofsuperposition to determine the force on aparticular charge.

If there are N charges, Q1,Q2...QN locatedrespectively at point with position vectorsr1,r2...rN the resultant force F on a charge Qlocated at point r is the vector sum of the forcesexerted on Q by each charges Q1,Q2...QN

rrrrrr

304 rr

Or generally,

322320

122120

QQTOTAL

For example,

2112 FF

EXAMPLE 1

Suppose 10nC charge Q1 located at (0.0, 0.0, 4.0m)and a 10nC charge Q2 located at (0.0, 4.0m, 0.0). Findthe force acting on Q2 from Q1.

To employ Coulomb’s Law, first find vector 12R

aa rrR

441212

Magnitude of 12R

44 2212

SOLUTION TO EXAMPLE 1

And zyzy aa

0.40.4

2410854.84

10101010

122120

SOLUTION TO EXAMPLE 1 (Cont’d)

2.2 ELECTRIC FIELD INTENSITY

If Q1 is fixed to be at origin, a second charge

Q2 will have force acting on Q1 and can be

calculated using Coulomb’s Law. We also

could calculate the force vector that would

act on Q2 at every point in space to generate

a field of such predicted force values.

It becomes convenient to define electric field intensity E1 or force per unit charge as:

This field from charge Q1 fixed at origin results

from the force vector F12 for any arbitrarily

chosen value of Q2

ELECTRIC FIELD INTENSITY (Cont’d)

Coulomb’s law can be rewritten as

to find the electric field intensity at any point in space resulting from a fixed charge Q.

Let a point charge Q1 = 25nC be located

at P1 (4,-2,7). If ε = ε0, find electric field

intensity at P2 (1,2,3).

EXAMPLE 2

By using the electric field intensity,

This field will be:

41025 aR

zyx aaarrR 4431212

4112 R

Where,

4434110854.84

123120

122120

If there are N charges, Q1,Q2...QN locatedrespectively at point with position vectorsr1,r2...rN the electric field intensity at point r is:

NQQrrrr

rrrrrr

FIELD LINES

The behavior of the fields can be visualized usingfield lines:

Field vectors plotted within a regular grid in 2D space surrounding a point charge.

Some of these fieldvectors can easily bejoined by field lines thatemanate from thepositive point charge.

The direction of the arrow indicates the direction of electric fields

The magnitude is given by density of the lines

FIELD LINES (Cont’d)

The field lines terminated

at a negative point charge

The field lines for a pair

of opposite charges

FIELD LINES (Cont’d)

2.3 LINE,SURFACE & VOLUME CHARGES

Electric fields due to continuous charge distributions:

To determine the charge for each distributions:

Line charge:

Surface charge:

Volume charge:

LINE,SURFACE & VOLUME CHARGES (Cont’d)

Infinite Length of Line Charge:

To derive the electric field intensity at any

point in space resulting from an infinite

length line of charge placed conveniently

along the z-axis

LINE CHARGE

Place an amount of charge in coulombs along the z axis.

The linear charge density is coulombs of charge per meter length,

Choose an arbitrary point P where we want to find the electric field intensity.

LINE CHARGE (Cont’d)

The electric field intensity is:

zzEEE aaaE

But, the field is only vary with the radial distance from the line.

There is no segment of charge dQ anywhere on the z-axis that will give us . So,E

zzEE aaE

Consider a dQ segment a distance z above radial axis, which will add the field components for the second charge element dQ.

The components cancel each other (by symmetry) , and the adds, will give:

Recall for point charge,

For continuous charge distribution, thesummation of vector field for each chargesbecomes an integral,

The differential charge,

dzdldQ

The vector from source to test point P,

Which has magnitude, and a

unit vector,

22 z R

So, the equation for integral of continuous charge distribution becomes:

2222204 z

Since there is no component,

Hence, the electric field intensity at any point ρaway from an infinite length is:

For any finite length, use the limits on the integral.

EXAMPLE 3

Use Coulomb’s Law to

find electric field

intensity at (0,0,h) for

the ring of charge, of

charge density,

centered at the origin in

the x-y plane.

By inspection, the ring

charges delivers only

and contribution

to the field.

component will be

cancelled by symmetry.

Each term need to be determined:

The differential charge,

addldQ

The vector from source to test point,

unit vector,

22 ha R

The integral of continuous charge distribution becomes:

2222204 ha

ad zL aaE

ad aE2

Rearranging,

Easily solved,

ah aE2

EXAMPLE 4

An infinite length line of charge

exists at x = 2m and z = 4m. Find the

electric field intensity at the origin.

Sketch in three dimensions and the cross section:

The vector from line charge to the origin:

zx aaaR 42

Which has magnitude, and a unit

vector,

zxR aaaa204

Inserting into the infinite line charge equation:

4.142.7

2010854.82104

SURFACE CHARGE

Infinite Sheet of Surface Charge:

To derive the electric field intensity at point

P at a height h above a charge sheet of

infinite area (x-y plane).

The charge distribution, is inS 2mC

SURFACE CHARGE (Cont’d)

Consider a differential charge,

dddSdQ

The vector from surface charge to the origin:

unit vector,

22 h R

Where, for continuous charge distribution:

The equation becomes:

2222204 ha

dd zS aaE

dd aE2

Since only z components exists,

0 0 23

A general expression for the field from a sheet

charge is:

NS aE02

Where is the unit vector normal from the

sheet to the test point.Na

EXAMPLE 5

An infinite extent sheet of charge

exists at the plane y = -2m. Find the electric

field intensity at point P (0, 2m, 1m).

210mnC

Sketch the figure:

The unit vector directed away from the sheet and toward the point P is ya

10854.821010

VOLUME CHARGE

A volume charge is distributed over a

volume and is characterized by its volume

charge density, inV 3mC

The total charge in a volume containing a

charge distribution, is found by

integrating over the volume:V

EXAMPLE 6

Find the total charge

over the volume with

volume charge density,

V dVQ The total charge, with volume:

dzdddV Thus,

10854.7

VOLUME CHARGE (Cont’d)

To find the electric field intensity resulting from a volume charge, we use:

RdVdQ a

Since the vector R and will vary over the volume, this triple integral can be difficult. It can be much simpler to determine E using Gauss’s Law.

2.4 ELECTRIC FLUX DENSITY

Consider an amount of charge

+Q is applied to a metallic

sphere of radius a.

Enclosed this charged sphere

using a pair of connecting

hemispheres with bigger radius.

ELECTRIC FLUX DENSITY (Cont’d)

The outer shell is grounded. Remove the ground

then we could find that –Q of charge has

accumulated on the outer sphere, meaning the

+Q charge of the inner sphere has induced the –Qcharge on the outer sphere.

Electric flux, extends from the positivecharge and casts about for a negative charge. Itbegins at the +Q charge and terminates at the–Q charge.

The electric flux density, D in is:2mC

where ED

This is the relation between D and E, where is the material permittivity. The advantage of using electric flux density rather than using electric field intensity is that the number of flux lines emanating from one set of charge and terminating on the other, independent from the media.

We can find the total flux over a surface as:

We could also find the electric flux density, Dfor:

Infinite line of charge:

02L So,

So, NS aD

NS aE02

Infinite sheet of charge:

Volume charge distribution:

RV dV a

04 So, R

V dV aR

EXAMPLE 7

Find the amount of electric flux through the surface at z = 0 with

mymx 3050 ,

243 mCxxy zx aaD

The differential surface vector is

zdxdyd aS

We could have chosen but the

positive differential surface vector is pointing in

the same direction as the flux, which give us a

positive answer.

zdxdyd aS

Therefore,

dxdyxxy

EXAMPLE 8

Determine D at (4,0,3) if there is a point

charge at (4,0,0) and a line

charge along the y axis.

How to visualize ?!

Let total flux,

LQTOTAL DDDWhere DQ is flux densities due to point charge

and DL is flux densities due to line charge.

Where,

3,0,00,0,43,0,4

Which has magnitude, and a unit vector,

R aaa 3

And aD

Where, 5

340,0,03,0,40,0,03,0,4 zx aaa

218.024.0

Therefore, total flux:

242240

18.024.0318.0

LQTOTAL

aaa DDD

2.5 GAUSS’S LAW

If a charge is enclosed, the net flux passing through the enclosing surface must be equal to the charge enclosed, Qenc.

Gauss’s Law states that:

The net electric flux through any closed surface isequal to the total charge enclosed by that surface

encQd SD

It can be rearranged so that we have relation

between the Gauss’s Law and the electric flux.

dVdQ SD

GAUSS’S LAW (Cont’d)

Gauss’s Law is useful in finding the fields for

problems that have high degree of symmetry.

• Determine variables influence D and what

components D present

• Select an enclosing surface, called Gaussian

Surface, whose differential surface is directed

outward from the enclosed volume and is

everywhere (either tangent or normal to D)

GAUSS’S LAW APPLICATION (Cont’d)

Use Gauss’s Law to determine electricfield intensity for each cases below:

Point Charge

Infinite length of Line Charge

Infinite extent Sheet of Charge

POINT CHARGE

• Point Charge:

It has spherical coordinate

symmetry, where the field

is everywhere directed

radially away from the

origin. Thus,

rrD aD

For a gaussian surface, we could find the differential surface vector is:

rddrd aS sin2So,

POINT CHARGE (Cont’d)

Since the gaussian surface has a fixed radius,

Dr will be constant and can be taken from

integration to yield

POINT CHARGE (Cont’d)

By using Gauss’s Law, where:

So, which leads to expected result:24 r

rrQ aE 2

04 See page 16!

INFINITE LENGTH LINE OF CHARGE

• Infinite length line of charge:

Find D and then E at any point zP ,,

A Gaussian surface containing the point P is placed around a section of an infinite length line of charge density L

occupying the z-axis.

An element of charge dQ along the line will give Dρ and Dz. But second element of dQ will result in cancellation of Dz. Thus,

aD DThe flux through the closed surface is:

bottomtop

SDSDSD

Where,

aS , aS

dzddddd

sidezbottom

Then, we know that Dρ is constant on the side of gaussian surface

DhdzdD

dzdDdh

The charge enclosed by the gaussian surface:

hdzQ L

Lenc 0

encL QhDhd 2SD

We know that,

2LD Thus, as

expected: aE

INFINITE EXTENT SHEET OF CHARGE

• Infinite extent sheet of charge:

Determine the field everywhere resulting from an infinite extent sheet of charge ρS

placed on the x-y plane at z = 0.

Locate a point at which we want to find the field along the z axis at height h.

Gaussian surface must contain this point and surround some portion of the charged sheet.

A rectangular box is employed as the Gaussian surface surrounding a section of sheet charge with sides 2x, 2y and 2z

SHEET OF CHARGE (Cont’d)

Only a DZ component will be present, and the

charge enclosed is simply:

dydxdSQ

No flux through the side of the box, so find

the flux through the top and bottom surface

bottomzzz

topzzz

bottomtop

SDSDSD

Notice that the answer is independent of the height of the box.

Then we have:

And electric field intensity, as expected:

NS aE02

Related to Gauss’s Law, where net flux isevaluated exiting a closed surface, is theconcept of divergence.

Expression for divergence by applying Gauss’sLaw might be too lengthy to derive, but it can bedescribed as:

The expression is also called the point form of Gauss’s Law, since it occurs at some particular point in space. For instance,

Plunger stationary – no net movement of molecules

Plunger moves up – net movement where air molecules diverging air is expanding

Plunger pushes in – net flux is negative and molecules diverging air is compressing

EXAMPLE 9

Suppose:

Find the flux through the surface of a cylinder

with and by evaluating the

left side and the right side of the divergence

theorem.

hz 0 a

Remember the divergence theorem?

dVdS DD

We can first evaluate the left side of the divergence theorem by considering:

bottomtop

SDSDSD

A sketch of this cylinder is shown withdifferential vectors.

The integrals over the top and bottom surfaces are each zero, since:

For evaluation of the right side of the divergence

theorem, first find the divergence in cylindrical

coordinate:

Performing a volume integration on this divergence,

This is the same!

2.6 ELECTRIC POTENTIAL

To develop the concept of electric potential and show its relationship to electric field intensity.

In moving the object from point a to b, the work can be expressed by:

adW LF

dL is differential length vector along some portion of the path between a and b

ELECTRIC POTENTIAL (Cont’d)

The work done by the field in moving the charge from a to b is

afieldE dQW LE

If an external force moves the charge against the field, the work done is negative:

adQW LE

We can defined the electric potential difference, Vba

as the work done by an external source to move a charge from point a to point b as:

QWV LE

Where,abba VVV

Consider the potential difference between twopoints in space resulting from the field of apoint charge located at origin, where theelectric field intensity is radially directed, thenmove from point a to b to have:

aba dr

rQdV aaLE 2

The absolute potential at some finite radius from a point charge fixed at the origin:

If the collection of charges becomes a continuous distribution, we could find:

Where,

Line charge

Surface charge

Volume charge

kQV104

Or generally,

The principle of superposition, where applied to electric field also applies to potential difference.

Based on figure, if a closed path is chosen, the integral will return zero potential:

Three different paths to calculate work moving from the origin to point P against an electric field.

0 LE d

EXAMPLE 10

Two point charges -4 μC and 5 μC are

located at (2,1-,3) and (0,4,-2)

respectively. Find the potential at

(1,0,1).

Let and CQ 41 CQ 52

144 rrrr

Where,

262,4,12,4,01,0,1

62,1,13,1,21,0,1

264105

441,0,1

rrrrQQV

Therefore,

kVV 872.51,0,1

The electrostatic potentialcontours from a point chargeform equipotential surfacessurrounding the point charge.The surfaces are alwaysorthogonal to the field lines.The electric field can bedetermined by finding the max.rate and direction of spatialchange of the potential field.

Therefore,

VEThe negative sign indicates that the field is pointing in the direction of decreasing potential.

By applying to the potential field:

rV aaE 2

IMPORTANT!!Three ways to calculate E:

If sufficient symmetry, employ Gauss’s Law.

Use the Coulomb’s Law approach.

Use the gradient equation.

EXAMPLE 11

Consider a disk of charge ρS, find the

potential at point (0,0,h) on the z-axis and

then find E at that point.

Find that,

dddSdQ

and 22 hr

With rdQV

04then,

Let and leads to

integral then,

How to calculate the integral?

duu 21

To find E, need to know how V is changing with position. In this case E varies along the z-axis, so simply replace h with z in the answer for V, then proceed with the gradient equation.

2202201

BOUNDARY CONDITIONS

So far we have considered the existence of electric

field in a region consisting of two different media,

the condition that the field must satisfy at the

interfacing separating the media called “boundary

condition”. Thus, we could see how the fields

behave at the boundary between a pair of

dielectrics or between a dielectric and a conductor.

First boundary condition can be determined by performing a line integral of E around a closed rectangular path,

BOUNDARY CONDITIONS

Fields are shown in each medium along with normal and tangential components. For static fields,

0 LE d

Integrate in the loop clockwise starting from a,

adddd LELELELE

BOUNDARY CONDITIONS

Evaluate each segment,

dLEdLEd

wEdLEd

aaaaLE

BOUNDARY CONDITIONS

dLEdLEd

wEdLEd

aaaaLE

Summing for each segment, then we have the first boundary condition:

21 TT EE

BOUNDARY CONDITIONS

Second boundary condition can be determined by applying Gauss’s Law over a small pillbox shaped Gaussian surface,

BOUNDARY CONDITIONS

encQd SD

BOUNDARY CONDITIONS

The Gauss’s Law,

Where,

sidebottomtop

dddd SDSDSDSD

The pillbox is short enough, so the flux passes through the side is negligible.

So, only top and bottom where:

SDdSDd

NNNNbottom

NNNNtop

Which sums to:

encNN QSDD 21

BOUNDARY CONDITIONS

And the right side of Gauss’s Law,

SdSQ SSenc

Thus, it leads to the second boundary condition:

SNN DD 21

This is when the normal direction from medium 2 to medium 1.

BOUNDARY CONDITIONS

If the normal direction is from medium 1 to medium 2,

SNN DD 12

BOUNDARY CONDITIONS

Generally,

S 2121 DDa

For a boundary conditions between a dielectric and a good conductor,

0TEBecause in a good conductor, E = 0. And since the electric flux density is zero inside the conductor,

BOUNDARY CONDITIONS

EXAMPLE 12

Consider that the field E1 is known as:

Find the field E2 in the other dielectrics.

mVzyx aaaE 5431

BOUNDARY CONDITIONS

We can employ Poisson’s and Laplace’sequations to help find the potential functionwhen conditions at the boundaries are specified.

From divergence theorem expression,

V DBy considering ,ED

From the gradient expression,

VEWhich gives us Poisson’s equation,

In charge free medium in which , it becomes Laplace’s equation:

BOUNDARY CONDITIONS

EXAMPLE 13

Determine the electric potential in the dielectric region between a pair of concentric spheres that have a potential difference Vab.

The charge distribution is: 30

This employs Poisson’s equation and the potential is only a function of r,

Multiply with r2 and integrate to obtain,

ArrrVr

Dividing both sides with r2 and integrate again,

BrArrV

Assume that the Vab consists of a voltage Va on the inner conductor and the outer conductor is grounded. So,

BbAbVbrV

BaAaVarV

ababAabV

aAaVVV

rrbaab

From (3) you can get:0

a abba

abVbaB a

Then, the potential between the spheres is

given by:

rarV a

CAPACITANCE

The amount of charge that accumulates as a function of potential difference is called the capacitance.

The unit is the farad (F) or coulomb per volt.

CAPACITANCE (Cont’d)

Two methods for determining capacitance:

Q Method

• Assume a charge +Q on plate a and a charge –Q on plate b.

• Solve for E using the appropriate method (Coulomb’s Law, Gauss’s Law, boundary conditions)

• Solve for the potential difference Vab between the plates (The assumed Q will divide out)

V Method

• Assume Vab between the plates.

• Find E , then D using Laplace’s equation.

• Find ρS, and then Q at each plate using conductor dielectric boundary condition (DN = ρS )

• C = Q/Vab (the assumed Vab will divide out)

CAPACITANCE (Cont’d)

EXAMPLE 14

Use Q method to find the capacitance for

the parallel plate capacitor as shown.

Place charge +Q on the inner surface of the top plate, and –Q charge on the upper surface of the bottom plate, where the charge density,

Use conductor dielectric boundary, to obtain:

dSQ SSQ

S from

zSQ aD from SN D

We could find the electric field intensity, E

zrSQ aE0

The potential difference across the plates is:

Finally, to get the capacitance:

dSC r0

EXAMPLE 15

Use V method to find the capacitance for a length Lof coaxial line of inner conductor radius a and

outer radius b, filled with dielectric permittivity as

shown.

Employ Laplace’s equation to find the potential

field everywhere in the dielectric. Assume that

fringing fields are neglected and the field is only

in function of ρ.

Laplace’s equation becomes:

Integrating twice to obtain:

BAV ln

Apply boundary condition to determine A and B. Let V(b) = 0 and V(a) = Vab to get:

VAbAB abln

,ln So,

abbVV ab

Apply the gradient to obtain E:

and also..

abVabr

We can find Q on the inner conductor with:

abVLaL

abaVSQ abrabr

S ln22

At inner conductor, the flux is directed outward indicating a positive surface charge density,

abaVabr

Thus, we can find the capacitance:

EXAMPLE 16

Conducting spherical shells with radius a =10cmand b = 30cm are maintained at a potentialdifference of 100V, such that V(r = a) = 100V andV(r = b) = 0V. Determine:

• The V and E in the region between the shells.

• If εr = 2.5 in the shells, determine the totalcharges induced on the shells and the capacitanceof the capacitor.

5.23.01.0

EXAMPLE 16 (Cont’d)

Employ Laplace’s equation to find the potential

field and the field is only in function of r.

Laplace’s equation becomes:

Multiply by r2,02

Integrating once gives

Integrating again gives

To obtain the value of constant A and B, use boundary conditions, where:

BbAVbr

rbAV 11

Therefore, the potential difference V

rbVV11

So, the electric field intensity E

Apply the gradient to obtain E:

rVV aaE 2

The total charges Q is:

r sin111 2

This yields to:

Thus, substituting the values of a, b and V0

to get the total charges, Q

1723.41.0

11005.210854.84 12

And for the potential difference V = V0.

So, the capacitance, C

723.41

1001723.4

PART 1

PRACTICAL APPLICATION

Laser Printer

Electret Microphone

Electrolytic Capacitors

LASER PRINTER

• OPC drum : Organic Photoconductive Cartridge –has a special coating will hold electrostatic charge.

• The surface is photoconductive – will discharge if surface hit by light.

i. A portion of drum passes under a negativecharged wire large negative charges induces apositive charge on the drum.

ii. Image to be printed is delivered to this chargedregion by laser and spinning mirror combination.

iii.Wherever the laser light strikes the drum, thephotoconductive material is discharged.

LASER PRINTER (Cont’d)

iv. The drum rolls past a toner The toner is blackpowder and positive charge drawn to thoseportions of the charge that have been dischargedby the laser.

v. Paper is fed through same speed as drum itpasses over positively charged wire that gives thepaper a strong negative charge.

vi. The positively charged toner drum is transferred tothe stronger negative charged on the paper thenit passes near a negatively charged wire thatremoves the negative charge from the paper,prevents it from statically clinging to the drum

vii.The paper and loose toner powder passed throughheated fuser rollers powder melts into the paperfiber the warm paper exits the printer.

• The drum continues rolling, passing through highintensity light discharges all the photoconductorsto erase the image from the drum, and ready forapplication of positive charge again from coronawire.

SUMMARY (1)

•The force exerted on a charge Q1 on charge Q2 in a medium of permittivity ε is given by Coulomb’s Law:

Where is a vector from charge Q1 to Q2121212 aR R

•Electric field intensity E1 is related to force F12 by:

The Coulomb’s Law can be rewritten as:

For a continuous charge distribution:

RRdQ aE 2

04•For a point charge at origin:

rrQ aE 2

SUMMARY (2)

•For an infinite length line charge ρL on the z axis

•For an infinite extent sheet of charge ρS

•Electric flux density related to field intensity by:

ED 0 r

SUMMARY (3)

Where εr is the relative permittivity in a linear, isotropic and homogeneous material.

SUMMARY (4)

• Electric flux passing through a surface is given by:

• Gauss’s Law states that the net electric flux through any closed surface is equal to the total charge enclosed by that surface:

encQd SD

Point form of Gauss’s Law is

SUMMARY (5)

• The electric potential difference Vab between a pair of points a and b in an electric field is given by:

aba VVdV LE

Where Va and Vb are the electrostatics potentials at aand b respectively.

For a distribution of charge in the vicinity of the origin, where a zero reference voltage is taken at infinite radius:

SUMMARY (6)

• E is related to V by the gradient equation:

VEWhich for Cartesian coordinates is:

zyx zV

xVV aaa

• The conditions for the fields at the boundary between a pair of dielectrics is given by:

S 2121 DDa21 TT EE and

Where ET1 and ET2 are the electric field components tangential to the boundary, a21 is a unit vector from medium 2 to 1 and ρS is the surface charge at the boundary. If no surface charge is present, the components of D normal to the boundary are equal:

21 NN DD

SUMMARY (7)

At the boundary between a conductor and a dielectric, the conditions are:

0TE and SN D

• Poisson’s equation is:

Where the Laplacian of V in Cartesian coordinates is given by:

In a charge free medium, Poisson’s equation reduces to Laplace’s equation

SUMMARY (8)

• Capacitance is a measure of charge storage capability and is given by:

SUMMARY (9)

For coaxial cable:

For two concentric spheres:

abV Lab ln

baQVab

Part 1:Electrostatics

Education

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ContentsChapter 1 Introduction to Electrostatics Electrostatics is the study of time-independent distributions of charges and elds. 1.1 Coulomb’s Law The foundation of electrostatics

UNIT 9 Electrostatics and Currents 1. Friday March 9 th 2 Electrostatics and Currents.

ELECTROSTATICS and CAPACITOR - Tiwari Academy … · Chapter 1 and 2 ELECTROSTATICS and CAPACITOR Electrostatics : study of electric charges at rest. Experiment : A glass rod rubbed

CP: Electrostatics February 1, 2010. Electrostatics – Chap. 32.

ELECTROSTATICS KIT - Nuffield type€¦ · INSTRUCTION SHEET 1 ELECTROSTATICS KIT - Nuffield type EM1771-001 electrostatics – ‘Nuffield’ DESCRIPTION: This is a larger sized

5. ELECTROSTATICS - Sakshi EducationS(egkyyr452nm03xqjt3sqo5mp))/EAMCET/QR...ELECTROSTATICS Synopsis : 1. Study of stationary electric charges at rest is known as electrostatics. 2.

Chapter 1 Introduction to Electrostatics

Electrostatics-1 atc

Chapter 1. Electrostatics I

Chapter 1 Introduction to Electrostatics - Babanski · Chapter 1 Introduction to Electrostatics Electrostatics is the study of time-independent distributions of charges and ﬁelds.

Chapter 1 Electrostatics - CAPE PHYSICSizifundo.weebly.com/.../batce_cape_physics_unit_2_lesson_1_notes.pdfChapter 1 Electrostatics ... Thus, if you hold and rub rods of polythene

Physics 1220/1320 Electromagnetism – part one: electrostatics Lecture Electricity, chapter 21-26.

Electrostatics Class 12- Part 3