Flashcards
Electromagnetism
Cards in this deck (77)
- 1Coulomb's LawF=q(1)q(2)/4πε(0)r^2
Where F = The force on the charges(N)
q(1) or q(2) = the charge on the particle(C)
ε(0) = permitivity of free space(Farads/m)
r = distance between the two charges(m) - 2Coulomb's Law for multiple chargesF(tot) = qΣ(i) q(i)/ 4πε(0)r(i)^2
Where F(tot) = The total force on the charge(N)
q(i) = the charge on the particle(C)
ε(0) = permitivity of free space(Farads/m)
r = distance between the two charges(m) - 3Electric fluxdΦ(e) = E dA cosθ or [E].d[A]
Where dΦ(e) = infintesimal electric flux(Vm)
E = electric field(V/m)
[E] = The electric field vector(V/m)
d[A] = The vector of the infintesimal area(m^2)
dA = infintesima area(m^2)
θ = angle between electric field vector and area vector(rads) - 4Gauss's lawSurf int [E].d[A] = Q(enc)/ε(0)
Where Surf int = surface integral
[E] = The electric field vector(V/m)
d[A] = The vector of the infintesimal area(m^2)
Q(enc) = Charge enclosed by surface
ε(0) = permitivity of free space(Farads/m) - 5Divergence theoremSurf int [A].d[S] = vol int [∇] . [A] dV
- 6Stokes Theoremline int [A].d[L] = surf int[∇] X [A].dS
- 7Gauss's law differential form[∇].[E] = ρ(free)/ε(0)
Where [E] = Electric Field vector(V/m)
ρ(free) = free charge density(C m^-3)
ε(0) = permitivity of free space(Farads/m) - 8Divergence of a magnetic field[∇].[B] = 0
Where [B] = magnetic field vector(T) - 9Electric potential energyU(e) = qq(0)/ 4πε(0)r
Where U(e) = Electric potential energy(j)
q = Charge of electric field(C)
q(0) = Charge of test charge(C)
ε(0) = permitivity of free space(Farads/m)
r = distance between the two charges(m) - 10Electric potentialV(e) = q/4πε(0)r
Where V(e) = electric potential (V)
q = Charge of electric field(C)
ε(0) = permitivity of free space(Farads/m)
r = distance between the two charges(m) - 11Electric field from electric potential[E]=-∇V
Where [E] = electric field vector(V/m)
V = electric potential(V) - 12Electric potential from electric fieldΔV = path int [E] . d[L]
ΔV = Change in electric potential(V)
[E] = Electric field (V/m) - 13Faradays law integral formline int [E].d[L] = -∂/∂t surf int[B].d[A]
Where [E] = electric field vector(V/m)
[B] = Magnetic field vector(T) - 14Faradays law diff form[∇]X[E] = -∂[B]/∂t
Where[E] = electric field vector(V/m)
[B] = magnetic field vector(T) - 15Lorentz force[F] = q[E] + q([v]X[B])
Where [F] = Force on a moving charge in an electromagnetic field(N)
q = charge on particle(C)
[E] = electric field vector(V/m)
[v] = velocity of charge vector(m/s)
[B] = Magnetic field vector(T) - 16Force on a charge due to electric field[F] =q[E]
Where [F] = Force vector(N)
q = charge of particle(C)
[E] = Electric field Vector(v/m) - 17Force on a chrage due to a magnetic field[F] = q([v]X[B])
Where [F] = Force Vector(N)
q = charge on particle(C)
v = velocity of particle vector(m/s)
[B] = magetic field vector(T) - 18Biot Savart law[B] = μ(o)/4π q[v]X[r]/r^2
Where [B] = Magnetic field vector(T)
μ(o) = permeability of free space(Henry/m)
q = charge of particle(C)
[v] = velocity of particle vector(m/s)
[r] = distance from magnetic field vector(m)
r = distance from magnetic field(m) - 19What is the direction of the field lines in an electric fieldThey go from the posititve charge to the negative charge
- 20Define the electric fluxThe number of field lines crossing a surface
- 21What are the possible simplifications when using gauss's law on a point charge and why-[E] is in the same direction as d[A] at all point of the surface and therefore become scalars as cosθ = 1
-[E] is constant at all points on the surface ad=nd can therefore be taken out the integral - 22Out of the two gauss's laws which is local and which isn't and whyThe differential form is local as it relates the electric field at a point to the charge density at that point
The intergral from is not local as it relates the charge contained within a volume to the eelctric flux through the surface bounding that colume - 23explain the divergence of a magnetic fieldMagnetic monopoles do not exist therefore magnetic field lines are continous
- 24Define Faraday's lawA change in magnetic flux induces a voltage
- 25How are magnetic fields generatedMoving charges e.g. current in a wire
- 26What happens to stationairy charges in a magnetic fieldThey're unaffected
- 27How do you determine the direction of a magnetic field?right hand grid rule
- 28When a capacitor has plates that are very large relative to their separation What simplifications could we apply to Poisson's equation?-(x,y) >> z therefore the electric potential only depends on the z coordinate which means the eletricfield is only in the z direction. ∇^2 V = ∂^2V/∂z^2
-The charge density between the plates is zero ∇^2 V = 0 - 29Describe the method of imagesA conducting surface can be replaced by one or more image or mirror charges in electrostatic problems
- 30Describe the uniqueness theoremThere is only one electric potential that satisfies the equations and the boundary conditions
- 31Why do electric fields enter conductors normal to the surfaceThe electric potential in or on a conductor is constant
- 32Why is the electric potential in or on a conductor constantThe electric field inside a conductor is always zero
- 33Why is the electric field in a conductor always zero[F] = -e[E]
The electric field causes a force that is opposite in direction to the electric field
This causes a build up in chargeon the side closest to the electric field.
This build up in charge induces an electric field that exactly cancels out the applied field in the conductor - 34Ampere's law static fields int formint line [B].d[L] = μ(0)I(enc)
Where [B] = the magnetic field vector(T)
μ(0) = permeability of free space(Henry/m)
I(enc) = current enclosed in integration path(A) - 35Ampere's law static fields diff form[∇] x[B] = μ(0)[J]
Where [B] = magnetic field vector(T)
μ(0) = permeability of free space(Henry/m)
[J] = current density vector(Am^-2) - 36equation of continuity int form-d/dt vol int ρ = surf int [J].d[A]
ρ = charge density(C m^-3)
[J] = current density vector[A m^-2] - 37Equation of continuity diff form[∇].[J] = -∂ρ/∂t
Where [J] = Current density vector
ρ = charge density(C m^-3)
t= time(s) - 38What happens to the motion a charged particle in a non uniform magnetic fieldThe charged particle moves in a circular orbit. When the particle enters an area of lower magnetic field strength the radius incraeses. This leads to the particle going to areas of even weaker magnetic field strength and the process continues
- 39Ampere law(non static fields)line int [B] .d[L] = μ(0)I(enc) + μ(0)ε(0)dΦ(E)/dt
Where [B] = magnetic field vector(T)
μ(0) = permeability of free space(Henry/m)
I(enc) = current enclosed in integration path(A)
ε(0) = permativity of free space(Farads/m)
dΦ(E)/dt = the rate of change of flux(V m s^-1) - 40Ampere-Maxwell law diff form[∇] X [B] = μ(0)[J] + μ(0)ε(0)∂[E]/∂t
Where [B] = magnetic field vector(T)
μ(0) = permeability of free space(Henry/m)
[J]= current density (A m^-2)
ε(0) = permativity of free space(Farads/m)
∂[E]/∂t = change in the elctric filed strength per unit time(V m s^1) - 41Ampere-Maxwell law int formline int [B] . d[L] = μ(0)(surf int([J] + μ(0)ε(0)∂[E]/∂t).d[A])
Where [B] = magnetic field vector(T)
μ(0) = permeability of free space(Henry/m)
[J]= current density (A m^-2)
ε(0) = permativity of free space(Farads/m)
∂[E]/∂t = change in the electric filed strength per unit time(V m s^1) - 42What happens to an atom when an electric field is appliedThe nuclues moves with the field whilst the electrons move against it. This shifts the centres of charge slightly creating an electric dipole. This decreases the overall [E] field
- 43electric dipole moment[p] = q[d] or |p| = q|d|
Where
[p] = electric dipole moment vector(C m)
q = Charge of an electron(C)
[d] = speration of charge vector(m)(points from -ve to +ve)
|p| =magnitude of electric dipole moment(C m)
|d|= magnitude of seperation betwenn two charges(m) - 44What causes a permanent dipole moment?Asymmetry in molecules
- 45Which dipole arrangement has more potential energy alligned or anti allignedanti alligned
- 46difference in potential energy between alligned and anti alligned equationU = -[p].[E]
Where U = potential energy(j)
[p] = dipole moment vector(C m)
[E] = electric field vector(V m^-1) - 47Polarisation Vector[P] = n<[p]>
Where [P] = poalrisation vector(C m^-2)
n = number density of molecules(m^-3)
<[p]> = average polarisation vector(C m^-2) - 48Define electric susceptibilityHown susceptible a material is to being polarised
- 49PolarisationP = χ(E)ε(0)[E]
Where P = polarisation(C m^-2)
χ(E) = electric susceptibility
ε(0) = permativity of free space(Farads/m)
[E] = electric field vector(V m^-1) - 50Surface charge densityσ(b+-) = +-Pcosθ
Where σ(b+-) = surface charge density
P = poalrisation(C m^-2)
θ = angle between polarisation and vector normal to face - 51Volume Polarisation charges int formvol int ρ(b) dV = surf int [P].d[A]
Where ρ(b) = bound charge density(C m^-3)
[P] = polarisation vector(C m^-2) - 52Volume polarisation Charges diff formρ(b) = -[∇].[P]
Where ρ(b) = bound charge density(C m^-3)
[P] = polarisation vector(C m^-2) - 53D-Field int form[D]=ε(0)[E] + [P] = ε(0)ε(r)[E]
[D] = D field vector(V m^-1)
ε(0) = permativity of free space(Farads/m)
[E] = electric field vector(V m^-1)
[P] = polarisation vector(C m^-2)
ε(r) = (1 + χ(E)) = relative permittivity(Farads/m) - 54D-field diff form[∇] .[D] = ρ(f)
[D] = D-field vector(V m^-1)
ρ(f) = free charge density(C m^-3) - 55Prove that the perpendicular components of the D field going through a dielectric boundary are equalConstruct gausian pillbox
Assume there are only bound charges
int [D].d[A] = 0
Pillbox height,h, tends to 0
therefore ignore sides
int [D].d[A] = [D(2)].{n}(2)ΔS +
[D(1)].{n}(1)ΔS
{n(1)} = -{n(2)}
int [D].d[A] = [D(2)].{n}(2)ΔS -
[D(1)].{n}(2)ΔS = 0
[D(2)].{n}(2)ΔS =[D(1)].{n}(2)ΔS
[D(2)].{n}(2) =[D(1)].{n}(2)
[D(2)](perp) =[D(1)](perp) - 56Derive the electric field wave equation[∇]X[E] = -∂[B]/∂t
[∇]X[∇]X[E] = -∂([∇]X[B])/∂t
[∇]X[B] = μ(0)ε(0)∂[E]/∂t
[∇]X[∇]X[E] = μ(0)ε(0)∂^2[E]/∂t^2
[∇]X[∇]X[E] = [∇]([∇].[E]) - [∇^2][E]
since in free space [∇].[E] = ρ/ε(0) = 0
[∇^2][E] = μ(0)ε(0)∂^2[E]/∂t^2 - 57Derive the magnetic field wave equation[∇]X[B] = μ(0)ε(0)∂[E]/∂t
[∇]X[∇]X[B] = -∂([∇]X[E])/∂t
[∇]X[E] = -∂[B]/∂t
[∇]X[∇]X[B] = -μ(0)ε(0)∂^2[B]/∂t^2
[∇]X[∇]X[B] = [∇]([∇].[B]) - [∇^2][B]
[∇].[B] = 0 always true
[∇^2][B] = μ(0)ε(0)∂^2[B]/∂t^2 - 58Link between magnetic and electric field wave equations[B] = 1/ω ([K]X[E])
Where[B] = magnetic field vector(T)
ω = angular frequency(rad/s)
[K] = wave number(m^-1)
[E] = electric field vector(V/m) - 59general solution for wave equation[E] = E(0)cos(ωt-[k].[r]){n}
Where [E] = electric field vector(V/m)
E(o) = ampitude of wave
ω = angular frequency(rad/s)
t = time
[K] = wavenumber(m^-1)
[r] = position vector
{n} = unit vector in a direction - 60Poynting vector[S] = [E] X [H]
Where [S] = poynting vector(W m^-2)
[E] = electric field vector(V/m)
[H] = h-field vector(A/m) - 61Circularly polarised waves[E] = E(0)cos(ωt+[k].[r]){n} + E(0)sin(ωt+[k].[r]){n}
Where [E] = electric field vector(V/m)
E(o) = ampitude of wave
ω = angular frequency(rad/s)
t = time
[K] = wavenumber(m^-1)
[r] = position vector
{n} = unit vector in a direction - 62Magnetic moment proportionality[μ] ∝ [r] X [I]
Where [μ] = magnetic moment(J/Tesla)
[r] = radius of current loop(m)
[I] = current(A) - 63torque due to magnetic field[τ] = [μ] X [B]
Where [τ] = torque on atoms[N m]
[μ] = magnetic moment(J/Tesla)
[B] = Magnetic field(Tesla) - 64define DiamamgnetismA material that when there is an applied magnetic field produces a magnetic field that opposes and weakens the original magnetic field
- 65define ParamagnetismA material that when a magnetic field is applied it produces a magnetic field in the same direction as the original magnetic field and strengthens it
- 66Magnetisation[B] = [H] + 4π[M]
Where [B] = magnetic field(T)
[H] = H field(A/m)
[M] = Magnetisation(A/m) - 67linear magnetisation[M] = χ(m)[H]
Where [M] = magnetisation(A/m)
χ(m) = magnetic susceptibility(m^3/kg)
[H] = H-field(A/m) - 68Magnetic field H field link[B] = μ[H] = (1 + 4πχ(m))[H]
Where [B] = magnetic field(T)
μ = permeability(m kg s^-2 A^-2)
χ(m) = magnetic susceptibility(m^3/kg)
[H] = H-field(A/m) - 69Curl of H field[∇]X[H] = [J] + ∂[D]/∂t
Where [H] = H-field(A/m)
[J] = current density(A/m^2)
[D] = D field(C m^-2) - 70divergence of H field[∇].[H] = 0
- 71E field to H field[∇]X[E] = -μ(0)∂[H]/∂t
Where [E] = electric field(V/m)
μ(0) = permeability(m kg s^-2 A^-2)
[H] = H field(A/m^2) - 72Current density-conductivity[J] = σ[E]
Where [J] = current density(A/m^-2)
σ = conductivity(Siemens/m)
[E] = electric field(V/m) - 73Ampere-Maxwell for conductors[∇]X[B] = μ(0)σ[E]+ μ(0)ε(0)ε(r)∂[E]/∂t
Where [B] = magnetic field(T)
μ(0) = permeability of free space(m kg s^-2 A^-2)
σ = conductivity(Siemens/m)
[E] = electric field(V/m)
ε(0) = permitivity of free space(Farads/m)
ε(r) = relative permativity(Farads/m) - 74Ampere-Maxwell for conductors H field[∇]X[H] = σ[E] + ∂[D]/∂t
Where [H] = H field(A/m^2)
σ = conductivity(Siemens/m)
[E] = electric field(v/m)
[D] = D field(C/m^2) - 75Poission's equation∇^2V=-ρ/ε(0)
Where V = electric potential(V)
ρ = charge density(C m^-3)
ε(0) = permativity of free space - 76Laplaces equation and when it occurs∇^2V = 0
V = electric potential(V)
Used when charge density is 0 - 77Derive an equation for the velocity of an E.M wavesub equation of wave into
[∇^2][E] = μ(0)ε(0)∂^2[E]/∂t^2
until
K^2 = ω^2μ(0)ε(0)
K = 2π/λ
ω = 2πf
μ(0)ε(0) = K^2 /ω^2
(μ(0)ε(0))^1/2 = K/ω = 2π/2πλf = 1/λf = 1/c
c= 1/(μ(0)ε(0))^1/2
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- Flashcards on Electromagnetism, created by daniel upton on 05/04/2021.
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