cheat sheet
DESCRIPTION
Formulário para EletromagnetismoTRANSCRIPT
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