Electrostatics - Arraytool · 03/10/2012 · SI UnitsCoulomb’s LawGauss LawVoltage &...

SI Units Coulomb’s Law Gauss Law Voltage & Energy Poisson’s Equation Capacitance Boundary Conditions MoI Summary Problems

Electrostatics

S. R. [email protected]

School of Electronics EngineeringVellore Institute of Technology

October 18, 2012

Electrostatics EE208, School of Electronics Engineering, VIT

mailto:[email protected]


Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy

5 Poisson’s Equation

6 Capacitance

7 Boundary Conditions

8 MoI

9 Summary

10 Problems



SI Base Units

Unit Name Unit Symbol Quantity Name Quantity Symbol

meter m length l

kilogram kg mass m

second s time t

ampere A electric current I

kelvin K thermodynamic temperature T

candela cd luminous intensity Iv

mole mol amount of substance n



A Few Named Units Derived from SI Base Units

Name Symbol Quantity Expression in terms of SI base units

couloumb C electric charge s.A

farad F electric capacitance kg−1.m−2.s4.A2

henry H inductance kg.m2.s−2.A−2

hertz Hz frequency s−1

joule J energy, work, heat kg.m2.s−2

newton N force, weight kg.m.s−2

ohm Ω electric resistance, impedance, reactance kg.m2.s−3.A−2

radian rad angle dimensionless

siemens S electrical conductance, susceptance kg−1.m−2.s3.A2

steradian sr solid angle dimensionless

tesla T magnetic field, magnetic flux density kg.s−2.A−1

volt V voltage, electric potential, electromotive force kg.m2.s−3.A−1

watt W power, radiant flux kg.m2.s−3

weber Wb magnetic flux kg.m2.s−2.A−1



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



First of all ... One Important Vector Calculus Identity

Let’s find out the gradient for the following equation:

∇(

1‖~r‖

)= ∇

(1‖rr‖

)From the definition of gradient in spherical coordinate system,

∇w = ∑i

(1hi

∂w∂qi

qi

)

we get,

∇(

1‖~r‖

)=

∂

∂r

(1r

)r = − 1

r2 r = − 1r2

~r‖~r‖ = − ~r

‖~r‖3 . (1)



Coulomb’s Law - Statement

The force~Fir,stat (ir stands for irrotational) acting on the charge q located at~r, due to the presence ofthe charge q1 located at ~r1 in an otherwise empty space, is given as

~Fir,stat (~r) =qq1

4πε0

~r− ~r1

‖~r− ~r1‖3 = − qq1

4πε0∇(

1‖~r− ~r1‖

)(2)

where vacuum primittivity, ε0 = 107/(4πc2

)≈ 8.8542× 10−12 Fm−1(SI unit).



Electric Field (Intensity) - Definition

For many purposes it is useful to introduce the concept of electric field, where electric field~Eir,stat isdefined by the limiting process

~Eir,stat ≡ limq→0

~Fir,stat

q=

q1

4πε0

~r− ~r1

‖~r− ~r1‖3 (3)

where~Fir,stat is the electric force, as defined in equation (2), due to a net electric charge q1 on the testparticle with a small electric net electric charge q .

Note that~Fir,stat and thus~Eir,stat are singular at~r = ~r′ .



Electric Field Due to a Point Source - I

...............Positive Source .................................... Negative Source



Electric Field Due to a Point Source - II

...............Positive Source .................................... Negative Source



Electric Field Due to Several Discrete Charges

In the presence of several discrete electric charges qi, located at the points ~ri, i = 1, 2, 3, . . . , respec-tively, in an otherwise empty space, the assumption of linearity of vacuum allows us to superimposetheir individual electrostatic fields into a total electrostatic field

~Eir,stattotal ≡∑

ilimq→0

~Fiir,stat

q=

14πε0

∑i

(qi

~r−~ri

‖~r−~ri‖3

). (4)



Electric Field Due to a Pair of Charges

...............Electric Field ............................. Magnitude of Electric Field



Electric Field Due to a Pair of Charges

...............Opposite polarities ............................. Same polarities



Electric Field Due to Continuous Charge Distribution

~dEir,stat

=dq′

4πε0

~r−~r′∥∥∥~r−~r′∥∥∥3 =

ρ′e,vol

(~r′)

dv′

4πε0

~r−~r′∥∥∥~r−~r′∥∥∥3

⇒ ~Eir,stat =

˚V′

~dEir,stat

=1

4πε0

˚V′

~r−~r′∥∥∥~r−~r′∥∥∥3 ρ′e,vol

(~r′)

dv′ (5)

= −∇

14πε0

˚V′

ρ′e,vol

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

(6)



Why~E produced due to electric charges is always irrotational?



Since ∇× [∇α (~r)] ≡~0 for any R3 scalar field, we immediately find that in electrostatics

∇×~Eir,stat = −∇×

14πε0

∇

ˆV′

ρ′e,vol

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

= − 1

4πε0∇×

∇ˆ

V′

ρ′e,vol

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

= ~0 (7)

i.e., that~Eir,stat is an irrotational (also known as curl-free / lamellar / conservative / gradient) field.



Irrotational & Rotational Fields - How do they look like?



- 4 - 2 0 2 4

- 4

- 2

0

2

4

x

y

...............Rotational Field .................................. Corresponding Curl



Now ... What About These Fields?

.................Point charge .................................... A pair of charges



Dirac Delta Function - Heuristic Description

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-2 -1 0 1 2

The Dirac delta can be loosely thought of as a function on the real line which is zeroeverywhere except at the origin, where it is infinite,

δ (x) =

+∞, x = 00, x 6= 0

and which is also constrained to satisfy the identityˆ +∞

−∞δ (x) dx = 1.



Dirac Delta Function - A Few Properties

• δ (−x) = δ (x) (Symmetry Property)

• ´ +∞−∞ δ (αx) dx =

´ +∞−∞ δ (u) du

|α| =1|α| (Scaling Property)

• ´ +∞−∞ f (x) δ (x− x0) dx = f (x0) (Translation or Sifting Property)

• δ (x)⇔ 1



Volume Charge Densities of Point, Line, and SheetCharges

X′ , Y′ , Z′ ⇒ Source coordinates

X, Y, Z ⇒ Test charge coordinates



Electric Field Due to Discrete Charges - Revisited

Point Charge:

Volume charge density corresponding to a point charge of strength q1 at ~r1 = x1x + y1y + z1 z can berepresented using the Dirac delta function as,

ρe,vol

(~r′)= q1δ

(~r′ − ~r1

)= q1δ (x′ − x1) δ (y′ − y1) δ (z′ − z1) .

Substituting the above volume charge density into (6) gives

⇒ ~Eir,stat =q1

4πε0

~r− ~r1

‖~r− ~r1‖3 , (8)

which is nothing but (3).



Electric Field Due to Discrete Charges - Revisited

Multiple Point Charges:

Volume charge density corresponding to a set of point charges can be represented as,

ρe,vol

(~r′)=

N

∑i=1

qiδ(~r′ −~ri

).

Once again, substituting the above volume charge density into (6) gives

⇒ ~Eir,stat =N

∑i=1

(qi

4πε0

~r−~ri

‖~r−~ri‖3

), (9)

which is nothing but (4).



Now ... Some Special Cases



Electric Field Intensity Due to a Line Charge

Electric field due to a small differential charge densityat ~r′ = z′ z is given as

~dEir,stat

=ρL (z′) δ (x′) δ (y′) dx′dy′dz′

4πε0

~r−~r′∥∥∥~r−~r′∥∥∥3


4πε0

xx + yy + (z− z′) z[x2 + y2 + (z− z′)2

] 32

.

Converting the above equation into cylindrical coordi-nate system gives

~dEir,stat


4πε0

ρρ + (z− z′) z[ρ2 + (z− z′)2

] 32

.



Electric Field Intensity Due to a Line ChargeTo get the total electric filed, let’s evaluate the volume integral given below:

~Eir,stat =

˚V′

~dEir,stat

=

˚V′

ρL (z′) δ (x′) δ (y′)4πε0

ρρ + (z− z′) z[ρ2 + (z− z′)2

] 32

dx′dy′dz

=

ˆ z′=+∞

z′=−∞

ρL (z′)4πε0

ρρ + (z− z′) z[ρ2 + (z− z′)2

] 32

dz′

For the simplest case, assuming ρL (z′) = ρL (where ρL is a constant),

~Eir,stat =ρL

4πε0

ˆ z′=+∞

z′=−∞

ρρ[ρ2 + (z− z′)2

] 32

dz′ +ˆ z′=+∞

z′=−∞

(z− z′) z[ρ2 + (z− z′)2

] 32

dz′

︸︷︷︸this term is zero

=ρL

4πε0

ˆ z′=+∞

z′=−∞

ρ[ρ2 + (z− z′)2

] 32

dz′

ρ (10)



Electric Field Intensity Due to a Line ChargeThe previous integral (10) can be calculated as shown below:

ˆ z′=+∞

z′=−∞

ρ[ρ2 + (z− z′)2

] 32

dz′ = ρ

ˆ ξ=−∞

ξ=+∞

1

[ρ2 + ξ2]32(−dξ) , where ξ = z− z′

= ρ

ˆ ξ=+∞

ξ=−∞

1

[ρ2 + ξ2]32

dξ

= ρ

ˆ τ=0

τ=π

1

[ρ2 + ρ2cot2τ]32

(−ρ csc2 τdτ

), where ξ = ρcotτ

= −ρ2ˆ τ=0

τ=π

1

[ρ2 csc2 τ]32

(csc2 τdτ

)= − 1

ρ

ˆ τ=0

τ=πsin τdτ =

2ρ

(11)

So, finally, from (10) and (11),~Eir,stattotal is given as

~Eir,stat =ρL

2πε0ρρ (12)



Electric Field Intensity Due to a Sheet Charge - I ***

Electric field due to a small differential charge density at ~r′ = x′ x + z′ z is given as

~dEir,stat

=ρS (x′ , z′) δ (y′) dx′dy′dz′

4πε0

~r−~r′∥∥∥~r−~r′∥∥∥3

=ρS (x′ , z′) δ (y′) dx′dy′dz′

4πε0

(x− x′) x + yy + (z− z′) z[(x− x′)2 + y2 + (z− z′)2

] 32

.



Electric Field Intensity Due to a Sheet Charge - I ***To get the total electric filed, let’s evaluate the volume integral given below (also, let’s assume thesimplest case, ρS (x′ , z′) = ρS, where ρS is a constant):

~Eir,stat =

˚V′

~dEir,stat

=

˚V′

ρS (x′ , z′) δ (y′) dx′dy′dz′

4πε0

(x− x′) x + yy + (z− z′) z[(x− x′)2 + y2 + (z− z′)2

] 32

=

ˆ z′=+∞

z′=−∞

ˆ x′=+∞

x′=−∞

ρS

4πε0

(x− x′) x + yy + (z− z′) z[(x− x′)2 + y2 + (z− z′)2

] 32

dx′dz′

=

ˆ ˆρS

4πε0

(x− x′) x[(x− x′)2 + y2 + (z− z′)2

] 32

dx′dz′


+

ˆ ˆρS

4πε0

yy[(x− x′)2 + y2 + (z− z′)2

] 32

dx′dz′

+

ˆ ˆρS

4πε0

(z− z′) z[(x− x′)2 + y2 + (z− z′)2

] 32

dx′dz′




Electric Field Intensity Due to a Sheet Charge - I ***

So,~Eir,stat is given as

~Eir,stat =

ˆ ˆ1[

(x− x′)2 + y2 + (z− z′)2] 3

2dx′dz′

(ρSy

4πε0

)y

=

ˆˆ x′=+∞

x′=−∞

1[(x− x′)2 + Ω2

] 32

dx′

dz′(

ρSy4πε0

)y, where Ω2 = y2 + (z− z′)2

=

(ˆ z′=+∞

z′=−∞

(2

Ω2

)dz′)(

ρSy4πε0

)y, from (10)

=

(ˆ z′=+∞

z′=−∞

1

y2 + (z− z′)2 dz′)(

ρSy2πε0

)y

=

(ˆ ξ=+∞

ξ=−∞

1y2 + ξ2 dξ

)(ρSy

2πε0

)y, where ξ = z− z′

=

[1y

tan−1(

ξ

y

)∣∣∣∣ξ=+∞

ξ=−∞

](ρSy

2πε0

)y =

(π

2+

π

2

)( ρS

2πε0

)y =

ρS

2ε0y. (13)



Electric Field Intensity Due to a Sheet Charge - II

Electric field at point~r = xx + yy due to the strip located at at ~r′ = x′ x is given as

~dEir,statstrip =

ρSdx′

2πε0

∥∥∥~r−~r′∥∥∥ ~r−~r′∥∥∥~r−~r′

∥∥∥=

(ρS

2πε0

)dx′(−x′ x + yy

x′2 + y2

).



Electric Field Intensity Due to a Sheet Charge - II

To calculate the total electric filed, let’s integrate along x′ as shown below:

~Eir,stat =

ˆ x′=∞

x′=−∞

(ρS

2πε0

)(−x′ x + yy

x′2 + y2

)dx′

=

(ρS

2πε0

)ˆ x′=∞

x′=−∞

yyx′2 + y2 dx′ −

ˆ x′=∞

x′=−∞

x′ xx′2 + y2 dx′︸︷︷︸

this term is zero

=

(ρSy

2πε0

)(ˆ x′=∞

x′=−∞

1x′2 + y2 dx′

)y

=

(ρSy

2πε0

)[1y

tan−1(

x′

y

)∣∣∣∣x′=+∞

x′=−∞

]y

=

(ρS

2πε0

)× πy =

ρS

2ε0y. (14)

We can see that both (14) and (13) are the same.



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



Faraday’s Experiment & Electric Flux Density

What Faraday did was, he enclosed the inner sphere of charge +Q with another bigger outer

sphere.




Then, he discharged the outer sphere by connecting it to the ground momentarily.




When measured, outer sphere has a total charge of −Q. Now, what does that mean ?? ...




Flux:

If we assume that there exists a concept called electricflux which represents the transfer of charge from onesphere to another, then the total flux (

‚) is equal to

Ψ = +Q.

Flux Density:

And electric flux density at any point~r = rr is definedas

~Dir,stat =Q

4πr2 r.




Flux Density:

A more general definition of electric flux density at anypoint~r = rr due to a continuous source is given as

~Dir,stat =1

4π

˚V′

ρ′e,vol

(~r′) ~r−~r′∥∥∥~r−~r′

∥∥∥3 dv′ (15)



Gauss’ Law - Statement & Physical Interpretation

Gauss’ Law:The electric flux passing through any closed surface is equal to the total charge enclosed by thatsurface. ˚

V(ρ′e,vol) dv′ =

‹S~Dir,stat · ~ds (16)

Divergence Theorem: ˚V

(∇ ·~F

)dv =

‹S~F · ~ds

Combining Gauss’ law and divergence theorem gives

∇ · ~Dir,stat = ρ′e,vol. (17)



Let’s Go Back to Line Charge



Let’s Go Back to Sheet Charge



One More Important Vector Calculus Identity ***

∇2(

1‖~r‖

)= −4πδ (~r)

∇2(

1‖~r‖

)= ∇ · ∇

(1‖~r‖

)= ∇ ·

(− 1

r2 r)

= − 1r2 sin θ

[∂

∂r

(r2 sin θ × 1

r2

)]= 0, when~r 6=~0

Since the the term represented in red color has a singularity at~r =~0, let us solve this problem fromthe definition of divergence itself ...

∇2(

1‖~r‖

)= ∇ ·

(− 1

r2 r)

= −4πr2 ×

(1/r2

)dv

= −[

limr→0

(4πr2

r2

)] [limr→0

dv−1]

= −4πδ (~r) (18)



Gauss Law Derivation - From the Definition of ElectricFlux Density ***

According to Helmholtz’s theorem, a well-behaved vector field is completely known if one knowsits divergence and curl. Taking the divergence of (5) gives

∇ · ~Dir,stat = −∇ ·

14π∇

˚V′

ρ′e,vol

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

= − 1

4π∇2

˚V′

ρ′e,vol

(~r′)

∥∥∥~r−~r′∥∥∥ dv′

= − 1

4π

˚V′∇2

1∥∥∥~r−~r′∥∥∥ ρ′e,vol

(~r′)

dv′

=

˚V′

δ(~r−~r′

)ρ′e,vol

(~r′)

dv′

=

˚V′

δ(~r′ −~r

)ρ′e,vol

(~r′)

dv′ = ρ′e,vol (~r) . (19)



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



Electrostatic Scalar Potential - Definition

Since ∇×~Eir,stat =~0,~Eir,stat can be represented as

~Eir,stat = −∇Vstat (20)

where Vstat is usually defined as electrostatic scalar potential.Now, comparing (21) and (6) gives

Vstat =1

4πε0

ˆV′

ρ′e,vol

(~r′)

∥∥∥~r−~r′∥∥∥ dv′ (21)



The Gradient Theorem

A line integral through a gradient or conservative vector field can be evaluated by evaluating theoriginal scalar field at the endpoints of the curve:

ˆγ[~p,~q]∇φ (~r) · ~dr = φ (~q)− φ (~p) (22)

The gradient theorem implies that line integrals through irrotational vector fields are path indepen-

dent.



Electrostatic Scalar Potential - Physical Interpretation

Energy needed to bring a unit charge from ∞ to~r in the presence of charge q at~0 is given as

WE = −ˆ

C1

~Eir,stat · ~dr = −ˆ

C2

~Eir,stat · ~dr

Evaluating the above expression gives

−ˆ

C2

~Eir,stat · ~dr = −ˆ r

∞

(q

4πε0r2

)dr =

−q4πε0

(−1r

)∣∣∣∣r∞=

q4πε0r

= Vstat (23)



Electrostatic Scalar Potential - Physical Interpretation

So, finally notations that we will use in this course:

VstatAB = Vstat

A −VstatB

VstatAB = −

ˆ A

B~Eir,stat · ~dl

VstatA = Vstat

A∞ = −ˆ A

∞

~Eir,stat · ~dl



Analysis of a DipoleVoltage due to the charges +q and −q at point P is,

Vstat =q

4πε0

[1‖~r1‖

− 1‖~r2‖

]=

q4πε0

[‖~r2‖ − ‖~r1‖‖~r1‖ ‖~r2‖

]. (24)

For far-field approximation ~r1 ‖ ~r2 ‖~r, we get

‖~r2‖ − ‖~r1‖ = d cos θ

‖~r1‖ = ‖~r2‖ = ‖~r‖ . (25)

Substituting the above equations in (24) gives

Vstat =qd cos θ

4πε0 ‖~r‖2 . (26)

If we define dipole momentum as~P = q~d,

Vstat =~P · r

4πε0 ‖~r‖2 . (27)



Analysis of a Dipole

Now, we can obtain electric field at point P by using the formula~Eir,stat = −∇Vstat as shown below:

~Eir,stat = −(

3

∑i=1

1hi

∂Vstat

∂qiqi

)

= −(

∂Vstat

∂rr +

1r

∂Vstat

∂θθ

)= −

(− 2qd cos θ

4πε0r3 r− qd sin θ

4πε0r3 θ

)=

qd4πε0r3

(2 cos θr + sin θθ

).



Energy of an Electrostatic System

Energy spent in bringing all these charges from ∞ is

WstatE = 0 + q2Vstat

21 + q3Vstat31 + q3Vstat

32 + · · · . (28)

Also, it is easy to prove that q2Vstat21 = q1Vstat

12 , i.e.,

q2Vstat21 = q2

q1

4πε0r21= q1

q2

4πε0r12= q1Vstat

12 .



Energy of an Electrostatic System

So, (28) can be rewritten as

WstatE = q1Vstat


23 + · · · . (29)

Now, combining (28) and (29) gives

2WstatE =

(q2Vstat

21 + q1Vstat12)+(q1Vstat

13 + q3Vstat31)+(q2Vstat

23 + q3Vstat32)+ · · ·

= q1(Vstat

12 + Vstat13 + · · ·

)+ q2

(Vstat

21 + Vstat23 + · · ·

)+ q3

(Vstat

31 + Vstat32 + · · ·

)+ · · · .

Each sum of potentials in parentheses is the combined potential due to all the charges except for thecharge at the point where this combined potential is being found. In other words,

2WstatE = q1Vstat


3 + · · · .

So, from the above equation, the total energy of the electrostatic system is given as

WstatE =

12

N

∑i=0

qiVstati =

12

˚V

ρe,volVstatdv. (30)



Energy Density of an Electrostatic SystemFrom the previous equation,

WstatE =

12

˚V

ρe,volVstatdv

=12

˚V

(∇ · ~Dir,stat

)Vdv.

From the vector identity

∇ · (wA) = w∇ ·A + A · ∇w,

we get

WE =12

˚V

(∇ · ~Dir,stat

)Vstatdv

=12

˚V

[∇ ·

(Vstat~Dir,stat

)− ~Dir,stat · ∇Vstat

]dv

=12

˚V

[∇ ·

(Vstat~Dir,stat

)]dv− 1

2

˚V

[~Dir,stat · ∇Vstat

]dv

=12

‹S

(Vstat~Dir,stat

)dv︸︷︷︸

= 0, if S is ∞ sphere

− 12

˚V

[~Dir,stat · ∇Vstat

]dv =

˚V

[12

ε0

∥∥∥~Eir,stat∥∥∥2]

dv. (31)



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



Poisson’s Equation - Derivation

From (21) and (19), we get

∇ ·~Eir,stat =ρ′e,vol

ε0

⇒ ∇ ·(−∇Vstat) = ρ′e,vol

ε0

⇒ ∇2Vstat = −ρ′e,vol

ε0(32)

where (32) is known as Poisson’s equation.



Poisson’s Equation - Uniqueness



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



Capacitor - Basics

A Few Basic Equations:

• Q =´ t

0 idt = Cv (assuming that current doesn’t exist for t < 0)

• i = dQdt = C dv

dt

• Energy =´ t

0 (power)dt =´ t

0 vidt =´ t

0 vC dvdt dt = C

´ v0 vdv = 1

2 Cv2

• I (jω) = jωCV (jω) =⇒ V(jω)I(jω) = 1

jωC = Z (jω) (using Fourier transformation)



Evaluation of Capacitance

Starting with the Assumption of Charge



Parallel Plate Capacitor

STEP1: Using the Gauss law and assuming that fring-ing fields are negligible, we can prove that

~D (x) = ρSx.

STEP2: From the definition of voltage, we have

V0d = V0 −Vd = −ˆ x=0

x=d~E · ~dl

= −ˆ x=0

x=d

(ρS

ε0

)dx

=ρSdε0

=QdAε0

.

From the above equation

Q =

(Aε0

d

)V0d. (33)



Cylindrical Capacitor

STEP1: Using the Gauss law and assuming that fring-ing fields are negligible, we can prove that

~D (ρ) =ρSaρ

ρ.


Vab = Va −Vb = −ˆ ρ=a

ρ=b~E · ~dl

= −ˆ ρ=a

ρ=b

ρSaε0ρ

dρ

=ρSaε0

ln(

ba

)=

Q2πaL

(aε0

)ln(

ba

)

= Q

(2πLε0

ln( b

a

) )−1

(34)



Spherical Capacitor

STEP1: Using the Gauss law, we can prove that

~D (r) =ρSa2

r2 r.


Vab = Va −Vb = −ˆ r=a

r=b~E · ~dl

= −ˆ r=a

r=b

ρSa2

ε0r2 dr

=ρSa2

ε0

1r

∣∣∣∣r=a

r=b=

(Q

4πa2

)a2

ε0

(1a− 1

b

)

= Q

(4πε01a −

1b

)−1

. (35)



Evaluation of Capacitance

Starting with the Assumption of Voltage

(i.e., using the Poisson’s Equation)



Parallel Plate Capacitor

STEP1: Using Poisson’s equation and boundary condi-tions, one gets

∇2V =∂2V∂x2 = 0

⇒ V = Ax + B

⇒ V = V0

(1− x

d

).

STEP2: From the the relation between electric field andvoltage, we have

~E = −∇V = − ∂V∂x

x =V0

dx.

On the surface of a metal plate (PEC), charge density isgiven by Dnorm. So,

Q = AρS = AV0

ε0d=

(A

ε0d

)V0. (36)



Cylindrical CapacitorSTEP1: Using Poisson’s equation and boundary condi-tions, one gets

∇2V =1ρ

∂

∂ρ

(ρ

∂V∂ρ

)= 0

⇒ ρ∂V∂ρ

= A

⇒ V = A ln ρ + B = V0ln (b/ρ)

ln (b/a).

STEP2: From the the relation between~E and V,

~E = −∇V = − ∂V∂ρ

ρ =V0

ρ

1ln (b/a)

ρ.

On the surface of a metal plate (PEC), charge density isgiven by Dnorm. So,

Q = AρS =

(2πLε0

ln (b/a)

)V0. (37)



Conical Capacitor (!)STEP1: Using Poisson’s equation and boundary condi-tions, one gets

∇2V =1

r2 sin θ

∂

∂θ

(sin θ

∂V∂θ

)= 0

⇒ sin θ∂V∂θ

= A

⇒ V = A ln(

tanθ

2

)+ B = V0

ln(tan θ

2

)ln(tan α

2

) .

STEP2: From the the relation between~E and V,

~E = −∇V = − 1r

∂V∂θ

θ = −V01

r sin θ ln(tan α

2

) θ.

On the surface of a metal plate (PEC), charge density is given by Dnorm. So,

Q =

¨ρSds = −

ˆ 2π

0

ˆ r

0

(ε0V0

r sin α ln(tan α

2

) ) r sin αdφdr =2πrε0

ln(cot α

2

)V0 (38)



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



Boundary Conditions - Tangential Components

12

34

5

6

7 + + + + +

Using the Stokes’ Theorem,

(ˆ1+

ˆ3+

ˆ2+

ˆ4

)(~E · ~dl

)=

¨ (∇×~E

)· ~ds(ˆ

1+

ˆ3

)(~E · ~dl

)= 0, (∵ ∇×~E =~0)

⇒ (E1tangential − E2

tangential)4l = 0

⇒ E1tangential = E2

tangential

D1tangential

εr1=

D2tangential

εr2.



Boundary Conditions - Normal Components

12

34

5

6

7 + + + + +

Using the divergence theorem (and Gauss law),

(¨5+

¨6+

¨7

)(~D · ~ds

)= Qe,cylinder

⇒(D2

normal −D1normal

)ds = ρeds

⇒ D2normal −D1

normal= ρe

⇒ εr2E2normal − εr1E1

normal= ρe.



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



Method of Images



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



Electrostatics - Summary



Outline

1 SI Units

2 Coulomb’s Law

3 Gauss Law

4 Voltage & Energy


6 Capacitance


8 MoI

9 Summary

10 Problems



Coulomb’s Law

1 Find~E at P (1, 1, 1) caused by four identical 3 nC charges located at P1 (1, 1, 0), P2 (−1, 1, 0),P3 (−1,−1, 0), and P4 (1,−1, 0).Ans: 6.82x + 6.28y + 32.8z V/m [H1, E2.2, P35]

2 An infinitely long uniform line charge is located at y = 3, z = 5. If ρL = 30 nC/m, findelectric field intensity at the origin.Ans: −47.582y− 79.3z V/m

3 In a rectangular region −2 ≤ x ≤ 2, −3 ≤ y ≤ 3, z = 0, a surface charge density is given byρS =

(x2 + y2 + 1

)3/2 C/m2. If no other charge is present, find~E at P(0, 0, 1).

Ans: 0.2157× 1012 z V/m



Gauss’ Law

1 Calculate the total charge within the universe for the given volume charge density,

ρv = e−2r

r2 C/m3. Also, calculate the electric flux density ~D at~r = rr.

Ans: 6.28 C, 1−e−2r

r2 C/m2 [H1, D2.4, P38]

2 Evaluate both sides of the Gauss Law for the field~D = 2ρ (z + 1) cos φρ− ρ (z + 1) sin φφ + ρ2 cos φz µC/m2, and the region 0 ≤ ρ ≤ 2,0 ≤ φ ≤ 0.5φ, and 0 ≤ z ≤ 4.

3 A non-uniform volume charge density, ρv = 120r C/m3, lies within the spherical surface

r = 1m, and ρv = 0 elsewhere. (a) Find Dr everywhere. (b) What surface charge density ρs2

should be on the surface r = 2m so that Dr|r=2− = Dr|r=2+ ?



Voltage - I

1 given the nonuniform field~E = yx + xy + 2z, determine the work expended in carrying 2Cfrom A (1, 0, 1) to B (0.8, 0.6, 1) along the arc of the circle x2 + y2 = 1, z = 1.Ans: -0.96 J [H1, E4.1, P87]

2 Again find the work required to carry 2 C from A to B in the same field, but this time use thestraight-line path from B to A.Ans: -0.96 J [H1, E4.2, P88]

3 Let~E = yx at a certain instant of time, and calculate the work required to move a 3C chargefrom (1, 3, 5) to (2, 0, 3) along the straight line segments joining:

• (1, 3, 5) to (2, 3, 5) to (2, 0, 5) to (2, 0, 3)• (1, 3, 5) to (1, 3, 3) to (1, 0, 3) to (2, 0, 3)

Ans: -9 J; 0 J [H1, D4.3, P90]



Voltage - II

1 An electric field is expressed in Cartesian coordinates by~E = 6x2x + 6yy + 4z V/m. Find:

• VMN if points M and N are specified by M (2, 6,−1) and N (−3,−3, 2)• VM if V = 0 at Q (4,−2,−35)• VN if V = 2 at P (1, 2,−4)

Ans: -139 V; -120 V; 19 V [H1, D4.4, P93]

2 If V = x2y (z + 3) V, then find the electric field intensity~E at point PCart (3, 4,−6).Ans: 72x + 27y− 36z V/m

3 If~E = 5x + 2y + 4z kV/m in a medium with εr = 4, then find the corresponding volumeenergy density.Ans: 769.878 µJ/m3

4 Given the potential V = 4xy + 3z2 V, what is the volume charge density ρv at pointPCart (−4, 3, 6)?

Ans: −6ε0 C/m3



Capacitance

1 Two parallel plates are separated by a dielectric of thickness 2mm with a dielectric constantof 6. If the area of each plate is 40 cm2 and the potential difference between them is 1.5 kV,then find the magnitude of the electric field inside the dielectric medium.

2 A parallel-plate capacitor with area 0.30m2 and separation 5.5 mm contains three dielectricswith interfaces normal to~E and ~D as follows: εr1 = 3, εr2 = 4, εr3 = 6d1 = 1 mm, d2 = 2 mm,and d3r1 = 2.5 mm. Find the capacitance.

Ans: 2.12nF



Miscellaneous

1 Determine the value of~E in a material for which the electric susceptibility is 2.5 and

P = 2.3× 10−7pC/m2.


Electrostatics - Arraytool · 03/10/2012 · SI UnitsCoulomb’s LawGauss LawVoltage &...

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Transcript of Electrostatics - Arraytool · 03/10/2012 · SI UnitsCoulomb’s LawGauss LawVoltage &...