Basics∇ ∙ D= ρ=∂ q /∂ Vn̂( DII−DI )=σF

∇× E=0∇ ∙ J=0∆ φ (r )=−ρ (~r ) /ε (Poisson)

∆ φ (r )=0 (Laplace)

DefinitionsD=ε EE=−∇ φJ=κ E=∂ I /∂ SF=Q Eε=ε o εr

Solutions

φab=W /q=−∫a

b

E ∙ d r

φ (r )= 14 πε

∑i=1

n q i

‖r−~r i‖= 1

4 πε∫ ρ(~r )

‖r−~r‖d~V

Parametrizationd ~r=‖d ~r /d ~¿‖d

~¿

d~A=‖ d ~r

d ~¿1

×d ~rd ~¿2

‖d~¿1 d

~¿2

Special operations∇× (f F )=f ∇× F+∇ f × F∇ ∙ ( f F )=f ∇ ∙ F+F∇ ∙ f

∇ (ln‖r‖)= r

‖r‖2

lnxy=−ln

yx=ln x−ln y

ln xa=a ln xln ( x )−ln (−x )=ln (−x2 ) (complex number)

Electrostatics∆ φ (r )=−ρ (~r ) /ε+RB (Poisson – Kirchhoff, Multipole, Images)

Multipole Images Images

∆ φ (r )=0 (Laplace – Separation, Images)

Separation

Stationary current field ρ̇=0 Separation method

∇× E=0∇ ∙ J=0

Kirchhoff’s integral limr → ∞

φ (r )=0

1) Parametrize2) Calculate the norm ||r - r’||3) Calculate the potential using the given charge density

φ (r )= 14 πε

∫ ρ(~r )‖r−~r‖

d~V

Multipole method r >> r’

1) Parametrize2) Calculate the Monopole moment Q and/or the Dipole moment P

using the given line (λ(r)) or surface (σ(r)) charge density3) Calculate the potential using Q and P

Q=∫ ρ(~r )d~VP=∫~r ρ (~r )d~V

φ (r )= 14 πε ( Q

‖r‖+r ∙ P

‖r‖3 )Separation method

1) Derive the separation approach2) Determine the constraints (the

last constraint should lead to Fourier integral)

3) Take care with the period of the integral (T or T/2).

∆ φ (r )=0 φ (r )=X ( x )Y ( y ) Z ( z )X ' '

X⏟¿ k1

2

+ Y ' '

Y⏟¿k2

2

+ Z ' '

Z⏟¿k3

2

=0

k 12+k2

2+k32=0

+k2⟶C1 ekx+C2 e−kx

−k 2⟶C1 cos (kx )+C2 sin(kx )

φ (r )={ (C1 ekx+C2 e−kx ) . (C3 e jkx+C4 e− jkx )…(C1 ekx+C2 e−kx ) . (C3cos ( jkx )+C4 sin( jkx))

…

…

Method of Images (finite RB)

1) Draw and locate the image charges2) Determine the position of the charges3) Calculate the potential φ(r). Take care if one has Q, λ

or extensive loads. The latter can be replaced by a compensation charge

4) If asked, calculate the surface charge density σ(r) using its normal vector ñ using the electric field E

φ (r )= 14 πε

∑i=1

n q i

‖r−~ri‖

*) for a ball, Q'=−R‖x‖

Q ,‖x '‖= R2

‖x‖, x '=R2 x

‖x‖2

Capacitance coefficients1) Calculate the potential φ(r) using Separation or

Images2) Determine the charges λi on conductors depending on

φ(r) – Inversion of matrix3) Use Maxwell’s coefficients of potential to determine

the capacitances

C=QU

= Qφ1−φ2

⇒one charge

C ij={ g ij⇒ i≠ j , mutual (Gegen)

−∑k=1

N

g ik⇒ i= j , self (Eigen)

[ φ1 (r )⋮

φN (r )]⏟Φ

= 14 πε [ p11 ⋯ p1 N

⋮ ⋱ ⋮pN 1 ⋯ pNN

]⏟P

∙[ λ1 (~r )⋮

λN (~r )]⏟Λ

Φ=P ∙ Λ → Λ=P−1 ∙ Φ

P−1=G=[ g11 ⋯ g1 N

⋮ ⋱ ⋮gN 1 ⋯ gNN

]Trigonometricssin2 x+cos2 x=1sin (−x )=−sin xcos (−x )=cos xtan (−x )=−tan x1+ tan2 x=sec2 x1+cot2 x=cosec 2 xcosec x=1 /sin xsec x=1 /cos xtan x=sin x /cos xcot x=1/ tg x=cos x /sin xsin (a ± b )=sin a .cos b± cos a . sin bcos (a±b )=cos a .cosb∓sin a .sin b

tan (a+b )=tan a+tan b1−tan a . tan b

tan (a−b )=tan a− tan b1+tan a . tan b

cos2 x=12

(1+cos 2 x )

sin2 x=12

(1−cos 2 x )

sin 2 x=2 sin x . cos x

tan2 x= 2 tan x

1− tan2 x

|sinx2|=√ 1−cos x

2

|cosx2|=√ 1+cos x

2

tanx2=1−cos x

sin x= sin x

1+cos x

Function Derivative Integral

un n .un−1u 'un+1

n+1+C , for n ≠−1

ln|u|+C , for n=1

u . v u' . v+u . v ' ---

uv

u' . v−u . v 'v2 ---

au au . ln (a ) . u '(a>0 , a ≠1 )

au

ln (a )+C , (a>0 , a ≠1 )

eu eu .u ' eu+C

ln u1u

u ' u . ln (u )−u+C

sin (u ) u' .cos (u ) −cos (u )+C

cos (u ) −u' .sin (u ) sin (u )+C

tan (u ) u' . sec2 (u ) ln|sec (u )|+C

‖r−~r‖ −r−~r‖r−~r‖ ---

1‖r−~r‖

−r−~r‖r−~r‖3 ---

Recurrence integrals

∫sin2au du=u2−

sin (2au )4 a

+C

∫cos2 au du=u2+

sin (2au )4 a

+C

System coordinates

sin x .cos y=12

[sin ( x− y )+sin ( x+ y ) ]

sin x .sin y=12

[cos ( x− y )−cos ( x+ y ) ]

cos x .cos y=12

[cos (x− y )+cos ( x+ y ) ]

cos x . sin y=12

[sin ( x+ y )−sin ( x− y ) ]

sin x .cos x=12

sin 2 x

sin x−sin y=2sin ( x− y2 ) .cos ( x+ y

2 )1−cos x=2 sin2 x

2

1+cos x=2cos2 x2

cos x= e ix+e−ix

2

sin x= e ix−e−ix

2 i

cosh x= ex+e−x

2

sinh x= ex−e−x

2