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

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Also available as File:Wsul file pdf 04.pdf

## Contents

##### T1

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

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

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

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

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

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

Find E

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

##### T2

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

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

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

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

Surface Integrals

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

More Gauss Law

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

##### T3

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

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

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

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

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

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

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

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

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

${\displaystyle Q=CV}$ defines capacitance.

${\displaystyle C=\varepsilon _{0}{\tfrac {A}{d}}}$ where A is area and d<<A1/2 is gap length of parallel plate capacitor

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

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

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

##### T4

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

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

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

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

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

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

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

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

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

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

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

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

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

##### T5
cross product

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

Hall effect

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

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

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

##### T7

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

Unit of inductance = Henry (H)=1V·s/A

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

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

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

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

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

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

##### T8

${\displaystyle {\frac {1}{S_{1}}}+{\frac {1}{S_{2}}}={\frac {1}{f}}}$ 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

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

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

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

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

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

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