Wright State University Lake Campus/2018-9/Phy2410/Equation sheet

Also available as File:Wsul file pdf 04.pdf

T1

$\varepsilon _{0}=$ 8.85×10⁻¹² F/m = vacuum permittivity.

e = 1.602×10⁻¹⁹C: negative (positive) charge for electrons (protons)

$k_{e}={\tfrac {1}{4\pi \varepsilon _{0}}}=$ = 8.99×10⁹ m/F

${\vec {F}}=Q{\vec {E}}$ where ${\vec {E}}={\tfrac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\tfrac {q_{i}}{r_{Pi}^{2}}}{\hat {r}}_{Pi}$

${\vec {E}}=\int {{\tfrac {dq}{{r}^{2}}}{\hat {r}}}$ where $dq=\lambda d\ell =\sigma da=\rho dV$

$E={\tfrac {\sigma }{2\varepsilon _{0}}}$ = field above an infinite plane of charge.

Find E

${\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{2}}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\vec {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{3}}}$ is the electric field at the field point, ${\vec {r}}$ , due to point charges at the source points, ${\vec {r}}_{i}$ , and ${\vec {\mathcal {R}}}_{i}={\vec {r}}-{\vec {r}}_{i},$ points from source points to the field point.

T2

2.6 Gauss's law:

$\Phi ={\vec {E}}\cdot {\vec {A}}$ $\to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA$ = electric flux

$q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$

$d\,{\text{Vol}}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\phi d\theta$

$A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}$

Surface Integrals

Calculating $\int fdA$ and $\int fdV$ with angular symmetry
Cyndrical: $dA=2\pi r\,dz;\,dV=dA\,dr$ . Spherical: $\int dA=4\pi r^{2},\;dV=4\pi r^{2}\,dr$

More Gauss Law

Calculating $\int fdA$ and $\int fdV$ with angular symmetry
Cyndrical: $dA=2\pi r\,dz;\,dV=dA\,dr$ . Spherical: $\int dA=4\pi r^{2},\;dV=4\pi r^{2}\,dr$

T3

2.7 Electric potential The alpha-particle is made up of two protons and two neutrons.

$\Delta V_{AB}=V_{A}-V_{B}=-\int _{A}^{B}{\vec {E}}\cdot d{\vec {\ell }}$ = electric potential

${\vec {E}}=-{\tfrac {\partial V}{\partial x}}{\hat {i}}-{\tfrac {\partial V}{\partial y}}{\hat {j}}-{\tfrac {\partial V}{\partial z}}{\hat {k}}=-{\vec {\nabla }}V$

$q\Delta V$ = change in potential energy (or simply $U=qV$ )

$Power={\tfrac {\Delta U}{\Delta t}}={\tfrac {\Delta q}{\Delta t}}V=IV=e{\tfrac {\Delta N}{\Delta t}}$

Electron (proton) mass = 9.11×10⁻³¹kg (1.67× 10⁻²⁷kg). Elementary charge = e = 1.602×10⁻¹⁹C.

$K={\tfrac {1}{2}}mv^{2}$ =kinetic energy. 1 eV = 1.602×10⁻¹⁹J

$V(r)=k{\tfrac {q}{r}}$ near isolated point charge

Many charges: $V_{P}=k\sum _{1}^{N}{\frac {q_{i}}{r_{i}}}\to k\int {\frac {dq}{r}}$ .

2.8 Capacitance

$Q=CV$ defines capacitance.

$C=\varepsilon _{0}{\tfrac {A}{d}}$ where A is area and d<<A^1/2 is gap length of parallel plate capacitor

${\text{Series}}:\;{\tfrac {1}{C_{S}}}=\sum {\tfrac {1}{C_{i}}}.$ ${\text{ Parallel:}}\;C_{P}=\sum C_{i}.$

$u={\tfrac {1}{2}}QV={\tfrac {1}{2}}CV^{2}={\tfrac {1}{2C}}Q^{2}$ = stored energy

$u_{E}={\tfrac {1}{2}}\varepsilon _{0}E^{2}$ = energy density

T4

2.9 Current and resistors

Electric current: 1 Amp (A) = 1 Coulomb (C) per second (s)

Current= $I=dQ/dt=nqv_{d}A$ , where

$(n,q,v_{d},A)$ = (density, charge, speed, Area)

$I=\int {\vec {J}}\cdot d{\vec {A}}$ where ${\vec {J}}=nq{\vec {v}}_{d}$ =current density.

${\vec {E}}=\rho {\vec {J}}$ = electric field where $\rho$ = resistivity

$\rho =\rho _{0}\left[1+\alpha (T-T_{0})\right]$ , and $R=R_{0}\left[1+\alpha \Delta T\right]$ ,

where $R=\rho {\tfrac {L}{A}}$ is resistance

$V=IR$ and Power= $P=IV=I^{2}R=V^{2}/R$

2.10 Direct current circuits

$V_{terminal}=\varepsilon -Ir_{eq}$ where $r_{eq}$ =internal resistance and $\varepsilon$ =emf.

$R_{series}=\sum _{i=1}^{N}R_{i}$ and $R_{parallel}^{-1}=\sum _{i=1}^{N}R_{i}^{-1}$

Kirchhoff Junction: $\sum I_{in}=\sum I_{out}$ and Loop: $\sum V=0$

Charging an RC (resistor-capacitor) circuit: $q(t)=Q\left(1-e^{t/\tau }\right)$ and $I=I_{0}e^{-t/\tau }$ where $\tau =RC$ is RC time, $Q=\varepsilon C$ and $I_{0}=\varepsilon /R$ .

Discharging an RC circuit: $q(t)=Qe^{-t/\tau }$ and $I(t)=-{\tfrac {Q}{RC}}e^{-t/\tau }$

T5

2.11 Magnetic forces and fields

$|{\vec {a}}\times {\vec {b}}|$ $=ab\sin \theta \Leftrightarrow$ $({\vec {a}}\times {\vec {b}})_{x}=(a_{y}b_{z}-a_{z}b_{y})$ , $({\vec {a}}\times {\vec {b}})_{y}=(a_{z}b_{x}-a_{x}b_{z})$ , $({\vec {a}}\times {\vec {b}})_{z}=(a_{x}b_{y}-a_{y}b_{x})$
Magnetic force: ${\vec {F}}=q{\vec {v}}\times {\vec {B}},\;$ $d{\vec {F}}=I{\overrightarrow {d\ell }}\times {\vec {B}}$ .
${\vec {v}}_{d}={\vec {E}}\times {\vec {B}}/B^{2}$ =EXB drift velocity
Circular motion (uniform B field): $r={\tfrac {mv}{qB}}.\;$ Period= $T={\tfrac {2\pi m}{qB}}.\;$

Dipole moment= ${\vec {\mu }}=NIA{\hat {n}}$ . Torque= ${\vec {\tau }}={\vec {\mu }}\times {\vec {B}}$ . Stored energy= $U={\vec {\mu }}\cdot {\vec {B}}$ .
Hall field = $E=V/\ell =Bv_{d}={\tfrac {IB}{neA}}$
Lorentz force = $q\left({\vec {E}}+{\vec {v}}\times {\vec {B}}\right)$

2.12 Sources of magnetic fields

Free space permeability $\mu _{0}=4\pi \times 10^{-7}$ T·m/A
Force between parallel wires ${\tfrac {F}{\ell }}={\tfrac {\mu _{0}I_{1}I_{2}}{2\pi r}}$
Biot–Savart law ${\vec {B}}={\tfrac {\mu _{0}}{4\pi }}\int \limits _{wire}{\frac {Id{\vec {\ell }}\times {\hat {r}}}{r^{2}}}$
Ampère's Law: $\oint {\vec {B}}\cdot d{\vec {\ell }}=4\pi \mu _{0}I_{enc}$
Magnetic field inside solenoid with paramagnetic material = $B=\mu nI$ where $\mu =(1+\chi )\mu _{0}$ = permeability

(we skip T6 because it was a review of previous chapters)

T7

2.13 Electromagnetic induction

Magnetic flux $\Phi _{m}=\int _{S}{\vec {B}}\cdot {\hat {n}}dA$
Motional $\varepsilon =B\ell v$ if ${\vec {v}}\perp {\vec {B}}$
Electromotive "force" (volts) $\varepsilon =-N{\tfrac {d\Phi _{m}}{dt}}=\oint {\vec {E}}\cdot d{\vec {\ell }}$
rotating coil $\varepsilon =NBA\omega \sin \omega t$

2.14 Inductance

Unit of inductance = Henry (H)=1V·s/A

Mutual inductance: $M{\tfrac {dI_{2}}{dt}}=N_{1}{\tfrac {d\Phi _{12}}{dt}}=-\varepsilon _{1}$ where $\Phi _{12}$ =flux through 1 due to current in 2. Reciprocity: $M{\tfrac {dI_{1}}{dt}}=-\varepsilon _{2}$

Self-inductance: $N\Phi _{m}=LI\rightarrow \varepsilon =-L{\tfrac {dI}{dt}}$

$L_{\text{solenoid}}\approx \mu _{0}N^{2}A\ell$ , $L_{\text{toroid}}\approx {\tfrac {\mu _{0}N^{2}h}{2\pi }}\ln {\tfrac {R_{2}}{R_{1}}}$ , Stored energy= ${\tfrac {1}{2}}LI^{2}$

$I(t)={\tfrac {\varepsilon }{R}}\left(1-e^{-t/\tau }\right)$ in LR circuit where $\tau =L/R$ .

$q(t)=q_{0}\cos(\omega t+\phi )$ in LC circuit where $\omega ={\sqrt {\tfrac {1}{LC}}}$

2.15 Alternating current circuits

AC voltage and current $v=V_{0}\sin(\omega t-\phi )$ if $i=I_{0}\sin \omega t.$
RMS values $I_{rms}={\tfrac {I_{0}}{\sqrt {2}}}$ and $V_{rms}={\tfrac {V_{0}}{\sqrt {2}}}$
Impedance $V_{0}=I_{0}X$
Resistor $V_{0}=I_{0}X_{R},\;\phi =0,$ where $X_{R}=R$
Capacitor $V_{0}=I_{0}X_{C},\;\phi =-{\tfrac {\pi }{2}},$ where $X_{C}={\tfrac {1}{\omega C}}$
Inductor $V_{0}=I_{0}X_{L},\;\phi =+{\tfrac {\pi }{2}},$ where $X_{L}=\omega L$
RLC series circuit $V_{0}=I_{0}Z$ where $Z={\sqrt {R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ and $\phi =\tan ^{-1}{\frac {X_{L}-X_{C}}{R}}$
Resonant angular frequency $\omega _{0}={\sqrt {\tfrac {1}{LC}}}$
Quality factor $Q={\tfrac {\omega _{0}}{\Delta \omega }}={\tfrac {\omega _{0}L}{R}}$
Average power $P_{ave}={\frac {1}{2}}I_{0}V_{0}\cos \phi =I_{rms}V_{rms}\cos \phi$
Transformer voltages and currents ${\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}$

T8

3.2 Geometric optics and image formation

${\frac {1}{S_{1}}}+{\frac {1}{S_{2}}}={\frac {1}{f}}$ relates the focal length f of the lens, the image distance S₁, and the object distance S₂. The figure depicts the situation for which (S₁, S₂, f) are all positive: (1)The lens is converging (convex); (2) The real image is to the right of the lens; and (3) the object is to the left of the lens. If the lens is diverging (concave), then f < 0. If the image is to the left of the lens (virtual image), then S₂ < 0 .

T9

2.16 Electromagnetic waves

Displacement current $I_{d}=\varepsilon _{0}{\tfrac {d\Phi _{E}}{dt}}$ where $\Phi _{E}=\int {\vec {E}}\cdot d{\vec {A}}$ is the electric flux.

Maxwell's equations: $\epsilon _{0}\mu _{0}=1/c^{2}$
$\oint _{S}{\vec {E}}\cdot \mathrm {d} {\vec {A}}={\frac {1}{\epsilon _{0}}}Q_{in}\qquad$
$\oint _{S}{\vec {B}}\cdot \mathrm {d} {\vec {A}}=0$
$\oint _{C}{\vec {E}}\cdot \mathrm {d} {\vec {\ell }}=-\int _{S}{\frac {\partial {\vec {B}}}{\partial t}}\cdot \mathrm {d} {\vec {A}}$
$\oint _{C}{\vec {B}}\cdot \mathrm {d} {\vec {\ell }}=\mu _{0}I+\epsilon _{0}\mu _{0}{\frac {\mathrm {d} \Phi _{E}}{\mathrm {d} t}}$

${\frac {\partial ^{2}E_{y}}{\partial x^{2}}}=\varepsilon _{0}\mu _{0}{\frac {\partial ^{2}E_{y}}{\partial t^{2}}}$ and ${\tfrac {E_{0}}{B_{0}}}=c$

Poynting vector ${\vec {S}}={\tfrac {1}{\mu _{0}}}{\vec {E}}\times {\vec {B}}$ =energy flux

Average intensity $I=S_{ave}={\tfrac {c\varepsilon _{0}}{2}}E_{0}^{2}={\tfrac {c}{2\mu _{0}}}B_{0}^{2}={\tfrac {1}{2\mu _{0}}}E_{0}B_{0}$

Radiation pressure $p=I/c$ (perfect absorber) and $p=2I/c$ (perfect reflector).