Formulaes Flashcards
Coloumb’s Law
F = 1/4πε0 qQ/ℛ^2 ℛ(hat)
Principle of superposition
F = F1 + F2 + F3 + …
E = E1 + E2 + E3 + …
Density of field lines
n/4πr^2 where n is the number of field lines
What are Maxwell’s equations?
∇ . E = p/ϵ Gauss’s law
∇ . B = 0 Gauss’s law
∇ x E = - ∂B/∂t Faraday’s Law
∇ x B = μ0 (J + ϵ ∂E/∂t) Ampere-Maxwell Law
What are Maxwell’s equations in Integral Form?
∫ E . da = qenc/ϵ Gauss’s law
∫ B . da = 0 Gauss’s law
∫ E . dl = -∂/∂t ∫ B .da Faraday’s Law
∫ B .dl = μ0 (Ienc + ϵ ∂/∂t ∫ E .da) Ampere-Maxwell Law
The Electric Field Strength
F = QE
where E = 1/4πε0 ∫ qi/ℛ^2 ℛ(hat) dr
and ℛ(hat) = ℛ(vector)/ℛ
General work done or energy
W = Q ΔV
where ΔV = - ∫ E .dl
The separation vector
where ℛ = r-r’
Electric field for a line charge
E = 1/4πε0 ∫ λ(r)/ℛ^2 ℛ(hat) dl
where λ(r) is the charge-per-unit-length
Electric field for a surface charge
E = 1/4πε0 ∫ σ(r)/ℛ^2 ℛ(hat) da
where σ(r) is the charge-per-unit-area
The electric dipole moment
p = qd
The electric dipole potential
V(r) ~ 1/4πϵ qdcosθ/r^2
and p = qd
V(r) ~ p/4πϵ cosθ/r^2
Electric field for a volume charge
E = 1/4πε0 ∫ p(r)/ℛ^2 ℛ(hat) dτ
where p(r) is the charge-per-unit volume
the flux of E through a surface
Φ = ∫ E.da
Starting from Gauss’s law in integral form derive Gauss’s law in differential form.
∫ E.da = Q(enc)/ε0
∫ E.da = ∫ ∇ . E dτ
Qenc = ∫ p dτ
∫ ∇ . E dτ = ∫ p/ε0 dτ
∇ . E = p/ε0
Electric potential derivation
V(r) = - (r ∫ O) E.dl
where O is some standard reference point.
V(b) - V(a) = (b ∫ a) ∇V . dl
The potential difference is therefore
E = -∇V
Show that V(r) = q/4πε0r
V(b) - V(a) = - (b ∫ a) E . dl
E = 1/4πε0 q/r^2
integrate
V = q/4πε0r
Derive poisson’s equation and therefore Laplace’s equation
∇. E = ∇. (-∇V)
∇. E = -∇^2V
from Gauss’s law
∇^2V = -p/ε0
Laplace’s equation when regions lacking charge p = 0
∇^2V = 0
Derive the energy density associated with the electric field.
W = (b ∫ a) F. dl and F = -QE
W = Q[V(b) - V(a)]
W = 1/2 (N Σ i=1) (N Σ j =1) qi Vij on formula sheet
W = 1/2 ∫ pV dτ
W = ε0/2 ∫ (∇ .E) V dτ
use vector identitiy
W = ε/2 ∫ E^2 dτ
The potential for a volume charge
V(r) = 1/4πε0 ∫ p(r)/ℛ dτ
The energy of a continuous charge distribution
W = ε/2 ∫ E^2 dτ
Area of : Circle, Sphere and Cylinder
Circle : πr^2
Sphere : 4πr^2
Cylinder : 2πrl
ratio of conductors
R = distance from the conductor surface/ size of the conductor
Integral limits for outer and inner
(r ∫ R) , where r is inner and R is outer.
The torque in an electric field
N = p x E
The work done during rotation by a dipole in an electric field
U = -p . E
Surface charge density
σ = Pcosθ = P . n
Bound volume charge
pb = - ∇. P
Electric Displacement.
D = ϵ0E + P
Derive Gauss’s law in terms of electric displacement.
∇ . E = pf+pb/ϵ0
∇ . (ϵ0E + P) = pf
pb = - ∇. P
and σb = P . n
thus D = ϵ0E + P
hence ∇. D = pf in differential form
and ∫ D . da = qf,enc in integral form.
Polarisation
P = ϵ0 Χe E
Show that D =ϵ0ϵr E
D = ϵ0E + P
D = ϵ0E + ϵ0ΧeE
D = ϵ0(1+ Χe)E
D = ϵ0ϵrE
Derive the normal and parallel components of electric field.
∫ E .da = Qenc/ε = σA/ε
E⊥above -E⊥below = σ/ε
∫ E . dl = 0
E∥above = E∥below
Derive the normal and parallel components of electric displacement and polarisation.
∇ x D = ε (∇ x E) + ∇ x P
∇ x D = ∇ x P
such that D⊥above - D⊥below = σf = 0 if no free charge
D∥above - D∥below = P∥above - P∥below
Derive the relationship of capacitance
V = V+ - V-
V = (+plate ∫ -plate) E.dl
C = Q/V
Derive the Energy stored in Capacitance.
dW = Vdq = q/c dq
since C = Q/V
W = 1/C (Q ∫ 0) qdq
W = Q^2/2C = 1/2 CV^2 = 1/2 QV.
What is the Energy Stored in a Capacitor with Dielectric
W = ε/2 ∫ εr E^2 dτ
W = 1/2 ∫ D . E dτ
Derive the Lorentz Force Law
dFmag = I(dl x B)
dFmag = I dt(v x B)
Fmag = Q(v x B)
F = Q(E + v x B)
Work done by magnetic force
dW = Fmag . dl
dW = Q(v x B). vdt
dW = 0
Derive the Continuity Equation
J = I/πa^2
I = ∫ J .da
∫ J . da = ∫ ∇ . J dτ
∫ ∇ . J dτ = -d/dt ∫ p dτ
Giving the continuity equation
Ampere’s Law
∫ B . dl = µ0 I
Derive the Time-Independent Ampere-Maxwell Law.
Ienc = ∫ J .da
1/μ ∫ B .dl = ∫ J .da
1/μ ∫ ∇ x B . da = ∫ J.da
∇ x B = μJ
Amperian Loops
∫ (∇ x B) . da = ∫ B . dl = µ ∫ J .da
where ∫ J .da = Ienc
Derive Magnetic flux
∇ . B = 0
apply divergence theorem
∫ ∇ . B dτ = ∫ B . ds = 0
hence Φ = ∫ B . dS
where Φ1 + Φ2 = ∫ B . dS + ∫ B . dS = 0
Magnetic Vector Potential
B = ∇ x A
Vector Magnetic Potential for a dipole
A(r) = µ0 (m x r)/4πr^3
Show that ∇^2 A = -µ0J
∇ x B = ∇ x ( ∇ x A)
∇ x B = ∇ ( ∇ . A) - ∇^2 A = µ0J
such that ∇^2A = -µ0J
Induced dipole moment and dipole moment
p = αE : induced dipole moment
where α is the atomic polarisability
P = Np : dipole moment
N atoms per unit volume and P polarisation
In a vacuum
free space such that
Xe = 0 and Xm = 0
ϵ = ϵ0 and μ = μ0
In matter
ϵ = ϵ0ϵr and μ = μ0μr
Torque in the Magnetic Field
N = m x B
The work done by a rotating dipole in a magnetic field
U = -m . B
Magnetic moment
m = nIA
Total current density
J = Jf + Jb
Bound surface current
Kb = M x n
Bound volume current
Jb = ∇ x M
for steady currents ∇ . Jb = 0
Derive the auxiliary field H
1/µ0 ( ∇ x B) = Jf + Jb
Jb = ∇ x M
1/µ0 ( ∇ x B) = Jf + ∇ x M
∇ x (B/µ0 - M) = Jf
hence H = B/µ0 - M
and in integral form, ∫ H . dl = If,enc
Magnetic Susceptibility
M = χm H
Show that B = µ0µr H
B = µ0(H + M)
B = µ0(1+ χm) H
hence B = µ0µrH
Relative permeability
µr = 1 + χm
Perpendicular component of magnetic field.
B⊥above = B⊥below
Derive the parallel component of magnetic field and hence the auxiliary parallel component.
H⊥above - H⊥below = -(M⊥above - M⊥below)
B∥above - B∥below = µ0(Kf + Kb)
=> H∥above - H∥below = Kf
Motional electromotive force (EMF)
ε = -d/dt ∫ (B .dS)
the RHS of the integral form of faradays law.
Magnetic induction
∫ E .dl = - ∫ ∂B/∂t . dS
∫ E .dl = - dΦm/dt
Φm = LI
Derive Ohm’s Law
J = σf
f = F/q
J = σ(E + v x B)
J = σE
Derive V = IR
I = JA = (σE)A
= (σ V/L)A = (σA/L)V
I = V/R
Show that emf = electric potential
ε = ∫ f . dl = ∫ fs . dl
V = - (b ∫ a) E. dl
V = ( b ∫ a) fs .dl = ∫ fs . dl
V = ε
Magnetic flux
Φ = ∫ B . dS
flux rule for EMF
ε = - dΦ/dt
Derive the differential form of Faraday’s law starting from the definition of EMF.
ε = -dΦ/dt = ∫ E . dl
- ∫ dB/dt . dS = ∫ E . dl apply stokes theorem to RHS
∫ - dB/dt . dS = ∫ ∇ x E . dS
=> -dB/dt = ∇ x E
derive the emf induced.
ε = - dΦ/dt
where Φ = LI
ε = -L dI/dt
Derive the work done (induced) per unit time
dW/dt = -εI = LI dI/dt
W = 1/2 LI^2
energy stored in the magnetic field
W = 1/2µ0 ∫ B^2 dτ
such that
Umag = 1/2 ∫ B. H dτ
Derive the Ampere-Maxwell Law
Start from the continuity equation
∇ . J = -∂p/∂t = -∂/∂t (ε0 ∇. E)
∇ . J = - ∇ . (ε0 ∂E/∂t)
such that
∇ x B = µ0J + µ0ε0 ∂E/∂t
Displacement current
jd = ε0 ∂E/∂t
such that ∂D/∂t = jd
Derive the polarisation current
pb = - ∇ . P
∂pb/∂t = - ∂/∂t ∇ . P
∂pb/∂t = - ∇ . ∂P/∂t
- ∇ . Jp = -∇ . ∂P/∂t
=> Jp = ∂P/∂t
The free current density
Jf
The bound current density, in magnetic materials
Jb
Maxwell’s fix for ∂E/∂t ≠ 0
jd
The polarisation current in dielectrics when ∂E/∂t ≠ 0
Jp
Completed Ampere-Maxwell law in matter
∇ × B = μ0 [Jf +Jb+jd+Jp]
Derive the auxiliary field H in terms of Jf, Jb, jd and Jp
∇ × B = μ0 [Jf +∇×M+jd+Jp]
∇ × ( B − μ0M ) = μ0 [Jf +jd +Jp]
μ0 ∇ × H = μ0 [Jf + jd + Jp]
∇ × H = Jf + jd + Jp.
where substitutions can be made for jd and Jp
such that
∇ × H = Jf + ∂D/∂t
Maxwell’s equations in static fields
no time-dependence
∇ · E = p/ε0
∇ · B = 0
∇ × E = 0
∇ × B = μ0J
Maxwell’s equations in quasi-static conditions
a good conductor
∇ · E = p/ε0
∇ · B = 0
∇ × E = −∂B/∂t
∇ × B = μ0J
Maxwell’s equations in free space.
No sources or sinks
∇ · E = 0
∇ · B = 0
∇ × E = −∂B/∂t
∇ × B = μ ε ∂E/∂t
speed in wave equation
c = 1/√µε
Derive the Wave equation.
Start from faraday’s law in free space
Take cross product of both sides
and use the vector identity.
To determine if solutions are of the wave equation.
Substitute solution in the plane wave solutions.
where they are related through c = ω/k
Concentric
Spherical - point charge
Coaxial
Cylindrical (line charge)
Pillbox
Plane (surface charge)
Out of the page and into the page
𝇇 out of the page
⨂ into the page
the total energy stored in electromagnetic fields
u = 1/2(ε0E^2 +1/μ0B^2) energy density
U = 1/2 ∫ (ε0E^2 +1/μ0B^2) dτ
Poynting vector and energy stored in fields Uem.
Energy crossing per area per unit time
S = 1/μ0 (E x B)
dW/dt = ∫ S . da
Power through a surface area.
P = ∫ S . da
Derive the displacement current Id
Start from the integral form of ampere-maxwell’s equation.
Replace dE/dt with Jd.
Id = ∫ Jd . da
Derive the time-averaged poynting vector. Given the plane wave solution E = E0 exp[i(kz-ωt)] and B = B0 exp[i(kz-ωt)].
take the plane wave solutions and substitute them into the Poynting vector.
looking at the real fields Re
B0 = E0/c where cos^2 -> 1/2 over one cycle.
giving <S> = E0^2/2cµ0</S>
refractive index
v = 1/√µε
c = 1/√µ0ε0
v = c/n
n = √(εµ/ε0µ0) = √εrµr
in general n = √εr
complex permittivity
εr = ε1 + iε2
relationship between v, k and
v ̃= ω/k ̃= 1/√ε0εrµ0µr
Skin depth
d = c/ωη = 2c√ε1 /ωε2
Reflection
R = (E0,r/E0,i)^2 which gives R on the formula sheet
Transmission
T = ε2v2/ε1v1(E0,t/E0,i)^2 which gives transmission on the formula sheet
What is the relation between transmission and reflection
R + T = 1
Current
I = dq/dt
Derive the relationship between pb and σb from polarisation
Use divergence theorem
∫ ∇ · P dτ + ∫ P . da = 0
P . da = P. n(hat) dS
and relationships on the formula sheet
r(hat) x dl
|r(hat)||dl|sin theta
where r(hat) = r(vector)/r
For a non magnetic media
μr = 1
Diamagnetism
I = e/T = ev/2πR
derive the change in magnetic moment
Δm = -e^2R^2/4me B
coloumb’s attraction = centripetal acceleration
1/4πϵ e^2/R^2 = me v^2/R
speed changes in presence of magnetic field
1/4πϵ e^2/R^2 + eṽB = me ṽ^2/R
eṽB = me/R (ṽ^2 -v^2)
Δv = ṽ -v = eRB/2me use binomial expansion
Δm = -1/2 e(Δv)Rz(hat)
Δm = -e^2R^2/4me B
Derive Poynting’s Theorem
dW = F.dl
v = dl/dt
dW = q(E+vxB).vdt
q = pdV and pv = J
dW = q(E.v)dt
dW/dt = ∫ E.J dV
E. J = 1/μ E. (∇ x B) – ϵE . ∂E/ ∂t
Curl identity
Then substitute for faradays law
B . ∂B/∂t = 1/2 ∂/∂t (B^2)
E . ∂E/∂t = 1/2 ∂/∂t (E^2)
Gather E^2 and B^2 terms
Arriving at pointing theorem
β
β = μ1v1/μ2v2 = μ1n2/μ2n1
cross product and dot product
cross product => sin(theta)
dot product => cos(theta)
parallel and perpendicular
parallel => sin(theta)
perpendicular => cos(theta)
Conduction current
Jc = σE
