Electricity & Magnetism I

Northeastern Illinois University

Electricity & Magnetism I Electrostatics: Basic Principles I

Greg Anderson Department of Physics & Astronomy Northeastern Illinois University

Spring 2017

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Northeastern Illinois Overview University

Electric Charge Coulomb’s Law The Electric Field Examples Curl, Divergence & Potential Gauss’s Law Dirac Delta Green Funct. Potential Multipole Torque and Potential

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Northeastern Illinois University

Electric Charge Electromagnetism Quantization Quantization II Quantization III Conservation Conservation II

Coulomb’s Law The Electric Field Electric Charge Examples Curl, Divergence & Potential

Gauss’s Law

Dirac Delta

Green Funct.

Potential

Multipole Torque and Potential

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Northeastern Illinois Electromagnetism University

With the exception gravity, almost every force that you experience in everyday life is electro-magnetic in origin. ◆ EM forces bind electrons and nuclei into atoms.

◆ EM forces bind atoms into molecules. O H H ◆ EM forces bind atoms & molecules into solids.

Electric forces are produced by electric charges.

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Northeastern Illinois Electric Charge Quantization University

Electric charge is quantized: A charged object has a surplus or deﬁcit in the number of electrons relative to protons. ◆ e = fundamental unit of charge

Q =+e, Q = e proton electron − ◆ For any charge:

Q = ne, n =0, 1, 2,... ± ± SI Units: [Charge] = Coulomb = C

e =1.6 10−19 C. ×

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Northeastern Illinois Electric Charge Quantization II University

Electric charge is quantized: A charged object has a surplus or deﬁcit in the number of electrons relative to protons. For any charged object:

Q = ne, n =0, 1, 2,... ± ±

◆ Let Ne be the number of electrons in an object

◆ Let Np be the number of protons in an object. n = N N p − e

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Northeastern Illinois Charge & Charged Constituents University

The total charge of any composite object is the sum of the charges of its charged constituents: electrons and protons. ◆ N electrons, each with charge e contribute: e − Q = N ( e)= N e electrons e − − e

◆ Np protons, each with charge +e contribute:

Qprotons = Npe Example + − N = 3 Together they give a total charge: + p +− − Ne = 4

Q =(Np Ne)e − Q = 1e − −

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Northeastern Illinois Electric Charged is Conserved University

Electric Charge is Conserved: The total charge of an isolated system never changes.

For an Isolated System:

Qinitial = Qﬁnal

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Northeastern Illinois Charge Conservation Examples University

Examples of electric charge conservation:

Pair production − − γ + γ e+ + e + −→ Pair annihilation +

e+ + e− γ + γ −→ −

Ionization H − − H + γ p+ + e + −→

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Northeastern Illinois University

Electric Charge

Coulomb’s Law Electrostatics Selected E&M Heros Coulomb’s Law Gravitational Analogy Coulomb’s Law II Coulomb’s Law & Superposition Coulomb’s Law The Electric Field

Examples Curl, Divergence & Potential

Gauss’s Law

Dirac Delta

Green Funct.

Potential

Multipole Torque and Potential

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Northeastern Illinois Electrostatics University

Electrostatics: The science of the interactions between electric ﬁelds and electric charges in static conﬁgurations.

All of electrostatics can be reduced to: ◮ Coulomb’s Law ◮ Principle of superposition.

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Northeastern Illinois Selected E&M Heros University

Volta Amp´ere Oersted Coulomb Stokes Laplace Maxwell Galvani Thomson (Kelvin) Franklin Faraday DuFey Gauss 1700 1750 1800 1850 1900 1950

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Northeastern Illinois Coulomb’s Law University

Charles Augustin de Coulomb (1785) y q Empirical force of Q on q: qQ qQ F r r q = K 2 ˆ = 2 ˆ r r 4πǫ0r ˆr = unit vector from Q to q ˆr r ˆr = x r Q | | 1 K = =8.99 109 N m2/C2 4πǫ0 × ·

[q] = Coulomb = C, ǫ0 = permittivity of the vacuum.

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Northeastern Illinois Gravitational Analogy University

m q

r r

ˆr ˆr M Q

Mm Qq F = G ˆr F = K ˆr − r2 r2

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Northeastern Illinois Coulomb’s Law II University

A more general notation: Electrostatic force of q2 on q1

q1 q q r F = K 1 2 ˆr, ˆr = 1 r2 r r | | x1 ˆr x + r = x , r = x x , 2 1 ⇒ 1 − 2 q2 q1q2 x1 x2 x2 F = − on q 1 4πǫ x x 3 1 0 | 1 − 2 | 1 What is the force on q2? K = 4πǫ0 F2 =

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Northeastern Illinois Coulomb’s Law II University

A more general notation: Electrostatic force of q2 on q1 q1 q q r F = K 1 2 ˆr, ˆr = 1 r2 r r | | x1 ˆr x + r = x , r = x x , 2 1 ⇒ 1 − 2 q2 q1q2 x1 x2 x2 F = − on q 1 4πǫ x x 3 1 0 | 1 − 2 | 1 q1q2 x2 x1 K = F2 = − on q2 4πǫ0 4πǫ x x 3 0 | 1 − 2 |

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Northeastern Illinois Coulomb’s Law & Superposition University

Force on q from q q i j r1 q1 q q x x F = i j i − j r2 i 4πǫ x x 3 0 | i − j | r x1 3 q2 x Force on q from q1, q2, . . . , qN x2 N q qqk x xk x3 3 F q = − 3 4πǫ0 x xk Xk=1 | − | Electric ﬁeld, force per unit charge:

F N q x x E x q k k ( ) = lim = − 3 q→0 q 4πǫ0 x xk Xk=1 | − |

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Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field Action at a Distance Field Theory Electric Monopole Field + Electric Monopole Field The Electric Field − The Electric Field Example: Hexagon Example: Hexagon II Electric Field & Charge Continuum Charge Densities

Examples Curl, Divergence & Potential

Gauss’s Law

Dirac Delta

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Northeastern Illinois Action at a Distance? University

Coulomb’s law appears to be a theory of “action at a distance.” F + Q Q F = k 1 2 r2 + F

Q: What is the mediator of the interaction between these spacially separated objects? A: Electromagnetic Fields.

c 2004-2017 G. Anderson Electricity & Magnetism – slide 18 / 71 x Northeastern Illinois Field Theory University

Action at a Distance: q q x x F = i j i − j j on i 4πǫ x x 3 0 | i − j | Field Theory (Maxwell and Faraday): ◆ Charges produce ﬁelds: E

F N q x x E x q k k ( ) = lim = − 3 q→0 q 4πǫ0 x xk Xk=1 | − |

◆ Forces are local interactions between charges and ﬁelds:

F = qE

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Northeastern Illinois Electric Monopole Field + University

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Northeastern Illinois Electric Monopole Field + University

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Northeastern Illinois Electric Monopole Field + University

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Northeastern Illinois Electric Monopole Field University −

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Northeastern Illinois Electric Monopole Field University −

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Northeastern Illinois Electric Monopole Field University −

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Northeastern Illinois The Electric Field University

Electric ﬁeld from q P 1 r q1 1 q1 x x1 E1(x)= − r2 4πǫ x x 3 0 | − 1 | r x1 3 q2 x Electric ﬁeld from q1, q2, . . . , qN x2 N q qk x xk x3 3 E x ( )= − 3 4πǫ0 x xk Xk=1 | − | Electric force on a test charge q:

F = qE

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Northeastern Illinois Charges on a Regular Hexagon University

Electric ﬁeld at x y 6 ′ 2 1 q x x E x k k ( )= − ′ 3 4πǫ0 x x a k=1 | − k | π X 3 Position of charge k 3 6 x kπ kπ x′ = ˆia cos + ˆja sin k 3 3

4 5 1 √3 1 √3 ( , ):( 1, 0) : ( , ) ± 2 2 ± ± 2 − 2

q 6 (x a cos kπ/3)ˆi +(y a sin kπ/3)ˆj E x ( )= − − 3/2 4πǫ0 [(x a cos kπ/3)2 +(y a sin kπ/3)2] Xk=1 − −

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Northeastern Illinois Charges on a Regular Hexagon II University y

2 1 6 ′ a qk x xk π E(x)= − ′ 3 x x 3 3 6 x 4πǫ0 k Xk=1 | − | 1 √3 1 √3 ( , ):( 1, 0) : ( , ) ± 2 2 ± ± 2 − 2 4 5

q 6 (x a cos kπ/3)ˆi +(y a sin kπ/3)ˆj E x ( )= − − 3/2 4πǫ0 [(x a cos kπ/3)2 +(y a sin kπ/3)2] Xk=1 − − On the x axis (y = 0):

q 6 (x−a cos kπ/3) Ex(x, 0) = 3/2 4πǫ0 k=1 [(x−a cos kπ/3)2+(a sin kπ/3)2] −aq 6 (sin kπ/3) Ey(x, 0) = P 3/2 =0 4πǫ0 k=1 [(x−a cos kπ/3)2+(a sin kπ/3)2] P c 2004-2017 G. Anderson Electricity & Magnetism – slide 24 / 71

Northeastern Illinois Charges on a Regular Hexagon II University y

2 1 6 ′ a qk x xk π E(x)= − ′ 3 x x 3 3 6 x 4πǫ0 k Xk=1 | − | 1 √3 1 √3 ( , ):( 1, 0) : ( , ) ± 2 2 ± ± 2 − 2 4 5 On the x axis (y = 0):

q 6 (x−a cos kπ/3) Ex(x, 0) = 3/2 4πǫ0 k=1 [(x−a cos kπ/3)2+(a sin kπ/3)2] −aq 6 (sin kπ/3) Ey(x, 0) = P 3/2 =0 4πǫ0 k=1 [(x−a cos kπ/3)2+(a sin kπ/3)2] P Far from the origin (x a) (1+ δ)−3/2 = 1 3 δ + 15 δ2 + ≫ − 2 8 ··· 1 6q 9qa2 E (x, 0) + x ≈ 4πǫ x2 2x4 0 c 2004-2017 G. Anderson Electricity & Magnetism – slide 24 / 71

Northeastern Illinois Electric Field & Charge Continuum University

′ x x P r = − x Electric ﬁeld from dq x′ dq x x′ dE(x)= − 4πǫ x x′ 3 0 | − | Charge density: ρ(x)

dq = ρ(x′)d3x′

1 x x′ E(x)= − ρ(x′)d3x′ 4πǫ x x′ 3 0 Z | − |

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Northeastern Illinois Charge Densities University

◆ Volume charge density: ρ dV dq = ρ(x)d3x, Q = ρ(x)d3x ρ Z ◆ Surface charge density: σ

dA dq = σ(x)dA, Q = σ(x)dA σ Z ◆ Linear charge density: λ dℓ

dq = λ(x)dℓ, Q = λ(x)dℓ λ Z

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Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field

Examples Charged Wire Charged Loop Spherical Shell Spherical Shell II Curl, Divergence Electrostatic Examples & Potential

Gauss’s Law

Dirac Delta

Green Funct.

Potential

Multipole Torque and Potential

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Northeastern Illinois Example: Uniformly Charged Wire University

z Electric ﬁeld from dq = λdz′ ℓ 2 dq ˆr ′ r = x2 + z′ E x dz d ( )= 2 4πǫ0 r p θ x x cos θ = dE √x2 + z′2 q = 2λℓ dq x ℓ x − dEx( )= 2 ′2 3/2 4πǫ0 (x + z ) 1 ℓ xλdz′ λℓ E (x, 0, 0) = = x 2 ′2 3/2 4πǫ0 − (x + z ) 2πǫ x√x2 + ℓ2 Z ℓ 0 Consider the limits ℓ x, x ℓ. ≫ ≫

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Northeastern Illinois Example: Uniformly Charged Wire University

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Northeastern Illinois Example: Uniformly Charged Wire University

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Northeastern Illinois Example: Uniformly Charged Loop University

z Electric ﬁeld from dq = λadθ dE dq ˆr E x d ( )= 2 4πǫ0 r r = √a2 + z2 z φ cos φ = √a2 + z2 x a dq z q = λ2πa dq x dEz( )= 2 2 3/2 4πǫ0 (a + z ) Along the z axis:

0 z 0 1 qz → Ez(0, 0,z)= 2 2 3/2 4πǫ0 (a + z ) −→  1 q 2 z a  4πǫ0 z ≫ 

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Northeastern Illinois Example: Spherical Shell University z a = R sin θ P Charge of inﬁntesimal ring

dθ dq = σ(2πR sin θ)Rdθ

Recall, for a charged ring: θ R x O 1 qh σ Ez(0, 0,h)= 2 2 3/2 4πǫ0 (a + h )

Height above ring: h = r R cos θ − σR2 π (r R cos θ)sin θdθ E (r)= r 2 2− 2 3/2 2ǫ0 [R sin θ +(r R cos θ) ] Z0 − Let u = cos θ,

σR2 1 (r Ru)du Er(r)= 2 −2 3/2 2ǫ0 −1 [R + r 2rRu] c 2004-2017 G. Anderson Z − Electricity & Magnetism – slide 30 / 71

Northeastern Illinois Example: Spherical Shell II University z Charge of sphere

Q =4πR2σ θ R x O Electric ﬁeld at a distance r σ σR2 r R E (r)= − +1 r 2ǫ r2 r R 0 | − | Inside and outside the sphere ) r

( Qˆr 2 r>R E 4πǫ0r Er(r)= ( 0 r

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Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field

Examples Curl, Divergence & Potential Curl & Divergence Curl, Divergence & Toward E = V −∇ Vanishing Curl Tangential E Potential Gauss’s Law

Dirac Delta

Green Funct.

Potential

Multipole Torque and Potential

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Northeastern Illinois Curl & Divergence of E University

Electric ﬁeld from a static distribution of charge:

1 x x′ E(x)= d3x′ − ρ(x′) 4πǫ x x′ 3 0 Z | − | As a consequence of this relation:

E = 0 (staticconﬁgurations) ∇× E = ρ/ǫ (Gauss’s Law) ∇· 0 Electric Potential

E =0, E = V ∇× ⇒ −∇

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Northeastern Illinois Toward E = V University −∇

First an identity:

1 1 x x′ = − x x′ = − ∇ x x′ x x′ 2 ∇| − | − x x′ 3 | − | | − | | − | The electric ﬁeld: 1 x x′ E(x)= d3x′ ρ(x′) − 4πǫ x x′ 3 0 Z | − | Using the identity above: 1 1 E(x)= d3x′ ρ(x′) −4πǫ ∇ x x′ 0 Z | − |

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Northeastern Illinois Toward E = V University −∇

First an identity:

1 1 x x′ = − x x′ = − ∇ x x′ x x′ 2 ∇| − | − x x′ 3 | − | | − | | − | The electric ﬁeld: 1 1 E(x)= d3x′ ρ(x′) −4πǫ ∇ x x′ 0 Z | − | does not act on primed coordinates: ∇ 1 ρ(x′) E(x)= d3x′ −∇4πǫ x x′ 0 Z | − |

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Northeastern Illinois Toward E = V University −∇

First an identity:

1 1 x x′ = − x x′ = − ∇ x x′ x x′ 2 ∇| − | − x x′ 3 | − | | − | | − | The electric ﬁeld can be written: 1 ρ(x′) E(x)= d3x′ V −∇4πǫ x x′ ≡ −∇ 0 Z | − | The Electric Potential, V

1 ρ(x′) V (x)= d3x′ 4πǫ x x′ 0 Z | − |

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Northeastern Illinois Electrostatics: E =0 University ∇×

Electric Potential 1 ρ(x′) V (x)= d3x′ 4πǫ x x′ 0 Z | − | The Electric ﬁeld is the gradient of an electric potential

E(x)= V (x) −∇ Curl of the electric ﬁeld vanishes in electrostatics:

E = V =0 ∇× −∇×∇

E =0 ∇×

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Northeastern Illinois Tangential Component of E University

E =0 E dℓ =0 ∇× ⇒ · IC For static conﬁgurations, the tangential component of the electric ﬁeld must be continuous across a surface.

dℓ1

dℓ2

E dℓ + E dℓ =0 E = E t1 · 1 t2 · 2 ⇒ t1 t2

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Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field

Examples Curl, Divergence & Potential

Gauss’s Law Integral form of Gauss’s Law Gauss’s Law Gauss’s Law: Uniformly Charged Sphere Gauss’s Law: Uniformly Charged Sphere Gauss’s Law: Long straight wire Gauss’s Law: Uniformly Charged Plane

Dirac Delta

Green Funct.

Potential

Multipole Torque and Potentialc 2004-2017 G. Anderson Electricity & Magnetism – slide 37 / 71

Northeastern Illinois Integral form of Gauss’s Law University

E = ρ/ǫ (Gauss’s Law, diﬀerential form) ∇· 0 Integral form:

3 E dA = d x E (Gauss’s Theorem) S · V ∇· H R = 1 d3x ρ(x) = 1 Q ǫ0 V ǫ0 enclosed Charge produces a divergenceR of the electric ﬁeld.

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Northeastern Illinois Gauss’s Law: Uniformly Charged Sphere University

1 Q E dA = d3xρ(x)= ρ · ǫ ǫ IS 0 ZV 0 R Electric ﬂux outside sphere:

r Φ = E dA = 4πr2E(r) E · By IS symmetry: Q E(r)= (r>R) 4πǫ r2 E = E(r)ˆr 0

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Northeastern Illinois Gauss’s Law: Uniformly Charged Sphere University

Charge inside sphere: r

3 4 3 ρ Q(r) = V d xρ(x)= ρ 3 πr = Q(R)(r/R)3 R r R Φ = E dA = 4πr2E(r)= Q(r)/ǫ E · 0 IS

By symmetry: ρr Qr E(r)= = (r

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Northeastern Illinois Gauss’s Law: Long straight wire University

r 1 E dA = d3xρ(x) h · ǫ λ I 0 Z E dA = E(r)h(2πr) · I d3xρ(x)= λh Z Cylindrical symmetry λ E = ˆr 2πrǫ0 E = E(r)ˆr

cylindrical coordinates

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Northeastern Illinois Gauss’s Law: Uniformly Charged Plane University

1 A E dA = d3xρ(x) · ǫ I 0 Z σ E dA =2E(z)A · I d3xρ(x)= σA Z σ/2ǫ kˆ for z > 0 E(x)= 0 Planar symmetry: σ/2ǫ kˆ for z < 0 − 0 E = E(z)kˆ

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Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field

Examples Curl, Divergence & Potential

Gauss’s Law Dirac Delta Dirac’s Delta Function The Dirac Delta Function 3D dirac Delta Function Active Learning

Green Funct.

Potential

Multipole Torque and Potential

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Northeastern Illinois The Dirac Delta Function University

Recall the Kronecker delta tensor δij

δij =1, Aiδij = Aj i i X X Dirac introduced a continuum analog δ δ(x a) ij → −

∞ −∞ dxδ(x a) =1 − width 0 → R ∞ height → ∞ −∞ dxf(x)δ(x a) = f(a) − area= 1 R 0 (x = a) 6 δ(x a)= −  (x = a)  ∞ a 

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Northeastern Illinois 3D dirac Delta Function University

In three dimensions:

δ3(x a)= δ(x a )δ(y a )δ(z a ) − − x − y − z The three dimentional delta function satisﬁes: d3xδ3(x a) =1 − R d3xδ3(x a)f(x) = f(a) − R x = a δ3(x a)= ∞ − 0 x = a 6

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Northeastern Illinois Active Learning University

For each of the charge distributions: ◆ A point charge, Q, at x = a. ◆ A line charge, λ, on the z axis ◆ A surface charge, σ, in the z = 0 plane Express the charge density ρ(x) in terms of the charges given above and appropriate delta functions.

Hint: What is the dimension of δ(x a)? −

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Northeastern Illinois Active Learning University

◆ ρ(x)= Qδ3(x a)= Qδ(x a )δ(y a )δ(z a ) − − x − y − z ◆ ρ(x)= λδ(x)δ(y) ◆ ρ(x)= σδ(z)

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Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field

Examples Curl, Divergence & Potential Gauss’s Law Green Functions & Dirac Delta

Green Funct. Green functions Poisson’s Equation Green’s Function 2 for −∇ Green’s Function 2 for II −∇ Poisson’s Equation

Potential

Multipole Torque and Potential

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Northeastern Illinois Green functions University

Consider the diﬀerential equation:

OˆΦ(x)= f(x), e.g., Oˆ = 2, f = ρ/ǫ −∇ 0 The Green function of the operator Oˆ is deﬁned by:

3 OGˆ O(x, x′)= δ (x x′) − The formal solution to the diﬀerential equation is then:

3 Φ(x)= GO(x, x′)f(x′)d x′ Z

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Northeastern Illinois Green functions University

Consider the diﬀerential equation:

OˆΦ(x)= f(x), e.g., Oˆ = 2, f = ρ/ǫ −∇ 0 The Green function of the operator Oˆ is deﬁned by:

3 OGˆ O(x, x′)= δ (x x′) − The formal solution to the diﬀerential equation is then:

3 Φ(x)= GO(x, x′)f(x′)d x′ Z Proof: 3 OˆΦ(x)= Oˆ GO(x, x′)f(x′)d x′ c 2004-2017 G. Anderson Electricity & Magnetism – slide 48 / 71 Z Northeastern Illinois Green functions University

Consider the diﬀerential equation:

OˆΦ(x)= f(x), e.g., Oˆ = 2, f = ρ/ǫ −∇ 0 The Green function of the operator Oˆ is deﬁned by:

3 OGˆ O(x, x′)= δ (x x′) − The formal solution to the diﬀerential equation is then:

3 Φ(x)= GO(x, x′)f(x′)d x′ Z Proof: 3 OˆΦ(x)= OGˆ O(x, x′)f(x′)d x′ c 2004-2017 G. Anderson Electricity & Magnetism – slide 48 / 71 Z Northeastern Illinois Green functions University

Consider the diﬀerential equation:

OˆΦ(x)= f(x), e.g., Oˆ = 2, f = ρ/ǫ −∇ 0 The Green function of the operator Oˆ is deﬁned by:

3 OGˆ O(x, x′)= δ (x x′) − The formal solution to the diﬀerential equation is then:

3 Φ(x)= GO(x, x′)f(x′)d x′ Z Proof: 3 3 OˆΦ(x)= δ (x x′)f(x′)d x′ − c 2004-2017 G. Anderson Electricity & Magnetism – slide 48 / 71 Z Northeastern Illinois Green functions University

Consider the diﬀerential equation:

OˆΦ(x)= f(x), e.g., Oˆ = 2, f = ρ/ǫ −∇ 0 The Green function of the operator Oˆ is deﬁned by:

3 OGˆ O(x, x′)= δ (x x′) − The formal solution to the diﬀerential equation is then:

3 Φ(x)= GO(x, x′)f(x′)d x′ Z Proof: OˆΦ(x)= f(x) c 2004-2017 G. Anderson Electricity & Magnetism – slide 48 / 71

Northeastern 2 Illinois Green’s Function for University −∇

The Green function of an operator Oˆ is deﬁned by:

3 OGˆ (x x′)= δ (x x′) O − − The Green function of 2 is then: −∇ 2 3 G 2(x x′)= δ (x x′) −∇ −∇ − − We will show the solution is: 1 G 2(x x′)= −∇ − 4π x x | − ′ |

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Northeastern 2 Illinois Green’s Function for II University −∇

Proof: 2(1/r)= 4πδ3(r). Recall (1/r)= ˆr/r2 ∇ − ∇ − 1 1 ∂ 1 2 = (1/r)= r2 − =0 (r = 0) ∇ r ∇·∇ r2 ∂r r2 6 Yet the integral over the volume is non-vanishing:

2 1 d3x = d3x 1 V ∇ r V ∇· ∇ r = 1 ˆrdA R RS ∇ r · = 1 (r2dΩ) HS − r2 = 4π −H Compare to: 1 2 d3x = 4πδ3(x x′)d3x = 4π ∇ r − − − ZV ZV

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Northeastern Illinois Poisson’s Equation University

2V = E = ρ/ǫ (Poisson’s Equation) −∇ ∇· 0 Green’s solution to Poisson’s equation:

1 3 1 3 ρ(x′) V (x)= d x′ G 2 (x x′)ρ(x′)= d x′ ǫ −∇ − 4πǫ x x 0 Z 0 Z | − ′ | Proof: ′ 2 1 2 3 ρ(x ) V (x) = d x′ ′ ∇ 4πǫ0 ∇ x x | − ′ | 1 3 2 ρ(x ) R ′ = 4πǫ0 d x′ x x ∇ | − | 1 3 3 = R d x′ ρ(x′)δ (x x′) −ǫ0 − = ρ(xR)/ǫ − 0 c 2004-2017 G. Anderson Electricity & Magnetism – slide 51 / 71

Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field

Examples Curl, Divergence & Potential

Gauss’s Law Dirac Delta Energy & Potential Green Funct.

Potential Gradients and Line Integrals I Gradients and Line Integrals II Potential Energy: Two Point Charges Potential Energy Energy & E-ﬁeld

Multipole Torque and Potential

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Northeastern Illinois Gradients and Line Integrals I University

f Gradient and inﬁntesimal change: F dU = U dℓ ∇ · Fundamental theorem for gradients

f U dℓ = U U ∇ · f − i i Zi Electric Force and Potential Energy

f ∆U = Uf Ui = F dℓ F = U − − i · −∇ Z

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Northeastern Illinois Gradients and Line Integrals II University

f Gradient and inﬁntesimal change: E dV = V dℓ ∇ · Fundamental theorem for gradients

f V dℓ = V V ∇ · f − i i Zi Electric Field and Potential

f ∆V = Vf Vi = E dℓ E = V − − i · −∇ Z

c 2004-2017 G. Anderson Electricity & Magnetism – slide 54 / 71

Northeastern Illinois Potential Energy: Two Point Charges University

Start with one charge q1 at x1 q1 To move a second charge q2 from x1 ∞ q2 to x2 in the presence of a charge q1: x q q 2 U = q V (x ) = 1 2 12 2 1 2 4πǫ x x 0 | 1 − 2 | Note, order of assembly is irrelevant, and:

1 1 qiqj U12 = [q2V1(x2)+ q1V2(x1)] = 2 2 4πǫ0 xi xj i=j X6 | − | double sum over i, j, excluding i = j

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Northeastern Illinois Energy of Charges & Distributions University

Interaction energy of N point-charges:

N 1 q q U = i j 2 4πǫ0 xi xj j=1 i=j X X6 | − | Generalizing to total energy of a continuous distribution of charge1: q dq = ρd3x, → →

1 3 P3 ρR(x)ρ(x′) U = d xd x′ 2 4πǫ x x Z 0 | − ′ | Goal: Recast U from an expression in terms of sources to an expression in terms of ﬁelds. 1 This adds in the self energy: Utotal = Uinteraction + Uself

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Northeastern Illinois Energy & E-ﬁeld University

For any charge distribution:

′ 1 3 3 ρ(x )ρ(x) ′ U = 2 d x d x′ 4πǫ0 x x | − | 1 3 = 2 R d x V (x)ρ(x) = Rǫ0 d3xV 2V = ǫ0 d3xV V − 2 ∇ − 2 ∇·∇ = ǫ0 R d3x (V V ) R ( V )2 − 2 ∇· ∇ − ∇ = ǫ0 Rd3x E2h surface term i 2 − R Thus, the energy stored in an electric ﬁeld is

ǫ0 3 2 U = 2 d x E c 2004-2017 G. Anderson Electricity & Magnetism – slide 57 / 71 R Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field

Examples Curl, Divergence & Potential

Gauss’s Law Dirac Delta Multipole Expansion Green Funct.

Potential

Multipole Multipole Expansion Two Charges Two Charges II N Point Charges Moments Electric Monopole Electric Diplole Electric Dipole Field Quadrupole Octupole

Torquec 2004-2017 and G. Anderson Electricity & Magnetism – slide 58 / 71 Potential Northeastern Illinois Electrostatic Multipole Expansions University

The electric ﬁeld and electric potential of a general charge distribution can be organized as an expansion in powers of 1/r. Multipole expansions of E and V : P x 1 Q ˆr p ˆr 2 ˆr x ·2 3 V ( ) = 4πǫ0 r + r + ·Qr · + ... n o ′ ρ(x ) 1 Qˆr 3ˆr(ˆr p) p E x 2 · 3 − ( ) = 4πǫ0 r + r + ... n o monopole + dipole + quadrupole + octupole + ···

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Northeastern Illinois Expansion for Two Charges University

P 1 q q V (x)= 1 + 2 4πǫ x x x x x 0 1 2 x1 | − | | − | q1 For r r and r r , we will show: ≫ 1 ≫ 2 x2 2 2 1 1 ˆr xi 3(ˆr xi) ri 4 q2 = + · 2 + · 3 − + r− x xi r r 2r O | − |

First, note that:

2 2 2 x xi = x xi = x + x 2x xi | − | | − | i − · 2 2 = rp 1+ r /r p2ˆr xi/r r√1+ ǫ i − · ≡ c 2004-2017 G. Anderson Electricity & Magnetism – slide 60 / 71 p Northeastern Illinois Two Charges Continued University

For small ǫ: P 1 = 1 1 x xi r √1+ǫ | − | x 1 1 3 2 3 x1 = 1 ǫ + ǫ + ǫ q1 r − 2 8 O

x2 2 2 ǫ = ri /r 2ˆr xi/r q2 − ·

Organize this expansion as a series in ri/r 1: ≪ 2 1 1 1 ri xi 3 xi 2 3 2 ˆr ˆr x xi = r 1 2 r 2 r + 8 2 r + r− | − | − − · · O 2 2 1 n ˆr xi 3(ˆr xi) ri 3 o = 1+ · + · 2 − + r− r r 2r O c 2004-2017 G. Anderson n Electricityo & Magnetism – slide 61 / 71

Northeastern Illinois For N Point Charges University

x 1 N qi V ( ) = 4πǫ0 i=1 x xi | − | 2 2 1 N qi ˆr xi 3(ˆr xi) ri 3 = P 1+ · + · 2 − + r− 4πǫ0 i=1 r r 2r O n o Deﬁne moments: P

Q = i qi monopole moment

p = Pi qixi dipole moment qi 2 ab = (3xi xi r δab) quadrupole moment Q2 Pi 2 a b − i Multipole expansionP for V in terms of moments:

1 Q ˆr p ˆr 2 ˆr 4 V (x) = + ·2 + ·Q3 · + (r− ) 4πǫ0 r r r O c 2004-2017 G. Anderson Electricity & Magnetism – slide 62 / 71 n o Northeastern Illinois Moments of a General Distribution University P x

1 Q ˆr·p ˆr·Q2·ˆr −4 V (x) = + 2 + 3 + (r ) x′ 4πǫ0 r r r O For a continuous charge distribution:

q dq = ρ(x′)d3x′ i −→ i X Z Z Deﬁne moments integrals:

3 ′ ′ Q = d x ρ(x ) monopole moment 3 ′ ′ ′ p = R d x x ρ(x ) dipole moment 1 3 ′ ′ ′ ′2 ′ 2ab = R d x (3xax r δab)ρ(x ) quadrupole moment Q 2 b − c 2004-2017 G. Anderson Electricity & Magnetism – slide 63 / 71 R Northeastern Illinois Electric Monopole University

Outside of a spherically symmetric distribution of charge centered at the origin:

x V (x) = 1 Q 4πǫ0 r

Q 1 Qˆr E(x) = 2 4πǫ0 r

◆ Equipotential lines: Lines of constant V ◆ E-ﬁeld lines: Lines tangent to E

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Northeastern Illinois Pure Electric Dipole University

◆ Two equal but opposite charges:

q = q = q , Q =0 1 − 2 ⇒

◆ Dipole moment: p = q(x1 x2)= qd −

q1 = q P Point like dipole: d 0, with p ﬁxed: x → p ˆr θ V (x) = · 2 p 4πǫ0r p cos θ 2 V (r, θ) = 4πǫ0r q2 = q 3ˆr(p ˆr) p − E = V = · 3− −∇ 4πǫ0r c 2004-2017 G. Anderson Electricity & Magnetism – slide 65 / 71

Northeastern Illinois Electric Dipole Field University

E-ﬁeld and equipotential lines for a point dipole: p cos θ V (r, θ) = 2 4πǫ0r

p E r θˆ = 3 2cos θˆ + sin θ 4πǫ0r

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Northeastern Illinois Electric Quadrupole Field University

2 Linear electric quadrupole: Q2 =2qd q P x d θ 2q − d q

2 Q2 (3cos θ 1) V (r, θ) = 3 − 4πǫ0 2r d 0, q , Q ﬁxed → → ∞ 2

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Northeastern Illinois Electric Octupole Field University

Pure Electric Octupole:

1 (5cos2 θ 3)cos θ V (r, θ) −4 ∝ 4πǫ0 2r

c 2004-2017 G. Anderson Electricity & Magnetism – slide 68 / 71

Northeastern Illinois University

Electric Charge

Coulomb’s Law The Electric Field

Examples Curl, Divergence & Potential

Gauss’s Law Dirac Delta Torque and Potential Green Funct.

Potential

Multipole Torque and Potential Torque on a Dipole Potential of a Dipole

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Northeastern Illinois Torque on a Dipole University

For an external ﬁeld E, the torque on p is:

τ = x dF = x E(x)ρ(x)d3x × × Z Z For E = constant over ρ: E τ = ( xρ(x)d3x) E p × R = p E = τ ×

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Northeastern Illinois Potential of a Dipole University

For an external ﬁeld E:

3 U = ρ(x′)V (x′)d x′ Z Expand V around dipole center:

V (x′) V (x)+(x′ x) V (x) E p ≃ − ·∇ x

3 U = d x′ ρ(x′)x′ V (x) ·∇ = p E R− ·

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