Helmholtzs Theorem I s t h i s s o l u t i o n u n i - - PowerPoint PPT Presentation

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Helmholtzs Theorem I s t h i s s o l u t i o n u n i - - PowerPoint PPT Presentation

Jul 11, 2023 554 likes •750 views

Helmholtzs Theorem I s t h i s s o l u t i o n u n i q u e ? Y E S , a s l o n g a s t h e v e c t o r f e l d ) i t s e l f g o e s t o z e r o a t i n f n i t t F ( r I f

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SLIDE 1

I s t h i s s

  • l

u t i

  • n

u n i q u e ?

Helmholtz’s Theorem

Y E S , a s l

  • n

g a s t h e v e c t

  • r

f e l d F (r ) i t s e l f g

  • e

s t

  • z

e r

  • a

t i n f n i t t I f t h e d i v e r g e n c e D ( r ) a n d t h e c u r l C ( r )

  • f

a v e c t

  • r

f u n c t i

  • n

F ( r ) a r e s p e c i f e d , a n d i f t h e t b

  • t

h g

  • t
  • z

e r

  • f

a s t e r t h a n 1 / r

2

a s r g

  • e

s t

  • i

n f n i t t , a n d i f F ( r ) i t s e l f g

  • e

s t

  • z

e r

  • a

s r g

  • e

s t

  • i

n f n i t t , t h e n F ( r ) i s u n i q u e l t g i v e n b t F =−∇ U+∇ ×W

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SLIDE 2

Ma x w e l l ’ s E qu a t i

  • n

s

∇ . ⃗ E= ρ ϵ0 ∇ . ⃗ B=0 ∇× ⃗ E=−∂ ⃗ B ∂t ∇× ⃗ B=μ0 ⃗ J+μ0ϵ0 ∂ ⃗ E ∂t Maxwell’s equations specify the divergence and curl of the electric and magnetic fields. Using Helmholtz’s theorem, we can then determine the electric and magnetic fields from Maxwell’s equations.

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SLIDE 3

Ma x w e l l ’ s E qu a t i

  • n

s

  • E

L E C T R O S O S T A T I C S

∇ . E = ρ/ϵ0 ∇ × E=0 E(r)=−∇( 1 4 πϵ0∫ ρ(r ') ℜ dV ') = 1 4 πϵ0∫ ρ(r ') ℜ

2

^ ℜ dV ' ℜ=r−r ' Coulomb’s Law! For a continuous surface charge, E(r)= 1 4 πϵ0∫ σ(r ') ℜ

2

^ ℜ da' For a continuous line charge, E(r)= 1 4 π ϵ0∫ λ(r ') ℜ

2

^ ℜ dl' For a collection of discrete charges, E(r)= 1 4π ϵ0∑

i

qi ℜi

2 ^

ℜi

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SLIDE 4

E L E C T R O S T A T I C S

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SLIDE 5

E L E C T R O S T A T I C S

∇ . E = ρ/ϵ0 ∇ × E=0 E(r)= 1 4 πϵ0∫ ρ(r ') ℜ

2

^ ℜ dV ' E(r)= 1 4 πϵ0∑

i

qi ℜi

2 ^

ℜi Maxwell’s Equations for Electrostatics Coulomb’s Law Helmholtz Theorem Force on a test charge Q, F(r)=Q E(r) In principle, we’re done with Electrostatics!

σ z

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SLIDE 6

E L E C T R i c F i e l d L i n e s

  • Field lines begin on positive charges
  • Field lines end on negative charges, or they extend upto infinity
  • The strength of the field is indicated by the density of the field lines
  • Field lines can never cross
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SLIDE 7

E L E C T R i c F i e l d L i n e s – C l

  • s
  • s

e d S u r f a c e

Flux ΦE=∮

S E . d S

Flux ∝ Number of field lines Flux through a closed surface is a measure of the total charge inside the surface – Gauss’s Law

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SLIDE 8

G a u s s ’ s L a w

∮ E .d a=∫

1 4π ϵ0( q r

2 ^

r)(r

2sinθ dθ d ϕ ^

r)= q ϵ0

Point charge q at origin Collection of point charges qi

Principle of superposition

∮ E .d a=∑

i=1 N

(∮ E .d a)=∑

i=1 N

(

qi ϵ0) E=∑

i=1 N

Ei

∮

S

E .d a= 1 ϵ0 Qenc

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SLIDE 9

G a u s s ’ s L a w

∫

surface

E.d S = q 4 π ϵ0

∫

surface

^ r .d S r

2

= q 4 π ϵ0

∫

surface

|d S|cosθ r

2

= q 4 π ϵ0

∫

surface

d Ω = q 4 πϵ0 4 π = q ϵ0

If the point is located outside then the contributions exactly cancel Use superposition principle ---> Add contribution from each charge dΩ=|d ⃗ S|cosθ r

2

=sinθ dθ d ϕ

The solid angle subtended by a surface S is defined as the surface area of a unit sphere covered by the surface's projection onto the sphere. A measure of how large an object appears to an

  • bserver looking from that point
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SLIDE 10

G a u s s ’ s L a w – D i f f e r e n t i a l F

  • r

m

∫

surface

E.d S = ∫

vol

∇ . EdV 1 ϵ0 Qenc = ∫

vol

ρ(r) ϵ0 dV

∮

S

E .d a= 1 ϵ0 Qenc ⇒ ∇ . E = ρ(r) ϵ0

∇ . E = ρ(r) ϵ0 E = 1 4π ϵ0 q r

2 ^

r Helmholtz Theorem Gauss’s Law Only for static charges Valid for moving charges!

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SLIDE 11

C U R L OF T H E E L E C T R I C F I E L D E = 1 4 πϵ0 q r

2 ^

r ∇ × E = 0

q a b

∫

a b

E . d l= ∫

a b

(

1 4 πϵ0 q r

2 ^

r).(dr ^ r+r dθ ^ θ+rsinθd ϕ ^ ϕ)

∫

a b

E . d l= 1 4 π ϵ0∫

a b

q r

2 dr=

q 4π ϵ0( 1 ra − 1 rb)

∮ E .d l=0

Stokes Theorem

⇒ ∇ × E=0

True for any charge configuration due to superposition! Valid only for static charges

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SLIDE 12

G a u s s ’ s L a w + S Y MME T R Y

  • S

P H E R E

ρ(r)

Consider a spherically symmetric charge distribution ρ(r) Eϕ=0 Why? Rotate about the z-axis

∮ E . d l=0

⇒ Eϕ=0 Eθ=0 Why? Rotate about the x-axis

∮ E . d l=0

⇒ Eθ=0 Apply Gauss's Law: Er .4 π R

2 = 1

ϵ0∫

R

ρ(r)4π r

2.d r

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SLIDE 13

G a u s s ’ s L a w + S Y MME T R Y

  • C

Y L I N D E R

Eϕ=0 Why? Rotate about the z-axis

∮ E . d l=0

⇒ Eϕ=0 Consider a long, narrow wire with a charge per unit length λ Ez=0 Why? Flip about z-axis Nothing distinguishes z from -z ⇒ E z=0 Apply Gauss's Law: Eρ.2πρ = 1 ϵ0 λ

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SLIDE 14

G a u s s ’ s L a w + S Y MME T R Y

  • S

U R F A C E

Consider an infinite sheet of charge with a surface charge density σ E// (Ex , E y)=0 Why? Rotate the sheet about any point Translate by any in-plane vector Field cannot change ⇒ E//=0 Apply Gauss's Law:

∫ E .d a = 2 A|E|= 1

ϵ0 σ A E= σ 2ϵ0 ^ n

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SLIDE 15

G a u s s ’ s L a w + S Y MME T R Y

  • S

U R F A C E σ −σ

Consider two parallel plates with equal and opposite charge densities ±σ. What is the electric field? E={ σ 2ϵ0 ^ n+ σ 2ϵ0 ^ n= σ ϵ0 ^ n between the plates everywhere else C= ϵ0 A d

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SLIDE 16

G a u s s ’ s L a w + S y mme t r y

What is the flux of the electric field through the shaded face? q 24ϵ0

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SLIDE 17

T H E E L E C T R I C P O T E N T I A L

Gauss’s Law is always true. It may not always be useful! If we can take advantage of the symmetries of a problem, Gauss’s Law can be a very powerful tool.

∇ × E = 0 E = −∇ V V (r) ≡ Electric Potential ∇ . E=ρ/ϵ0 ⇒ ∇

2V=−ρ/ϵ0

Poisson’s Equation

In regions of no charge ∇

2V = 0

Laplace Equation