# Quizbank/University Physics Semester 2

Wright State University Lake Campus/2018-9/Phy2410 Template:EduV/announcement
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** Physics equations/Equations
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## University Physics Semester 2

This unit has 11 exams (tests) that can viewed by clicking the links (e.g., T1). They are classroom-ready exams based on the collection of quizzes shown below each exam. The fraction indicates the ratio of the number of questions randomly selected to the number of questions on each quiz. Students can access these quizzes using either the (uneditable) permalink or directly via links to subpages of QB. Students and instructors can also view all the questions on this unit's /Questions list. Ideas for use by instructors can be found at Quizbank/Instructions.

### Quizbank/University Physics Semester 2/T1 Thu 9/6

* Note change of date for this quiz
3/6 from Special:Permalink/1894334 to QB/d_cp2.5 | Equations 2/4 from Special:Permalink/1863337 to QB/a18ElectricChargeField_findE | Equations (solutions) 5/13 from Special:Permalink/1863397 to QB/c18ElectricChargeField_lineCharges | Equations (solution)
$\varepsilon _{0}=$ 8.85×10$^{−12}$ F/m = vacuum permittivity.
e = 1.602×10$^{−19}$C: negative (positive) charge for electrons (protons)
$k_{e}={\tfrac {1}{4\pi \varepsilon _{0}}}=$ = 8.99×10$^9$ 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.
${\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{2}}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\vec {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{3}}}$ 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. ${\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{2}}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\vec {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{3}}}$ 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.

#### Errata
##### QB/d_cp2.5

   * Question 2: The angle is above the x-axis not the −x axis.
   * Question 4: There is an error of 10$^9$ which can be removed by changing nC to C (nano-Coulombs to Coulombs) in the question.
   * Question 5: I forgot to give you a value for z. Take z=1 until this gets fixed.
   * Question 6: Perhaps the student should be informed that the approximation of the field near an infinite plane should be used here.
* Equations for this test

##### QB/d_cp2.6

    Question 1: Replace C by nC for both charges in the question.S

### Quizbank/University Physics Semester 2/T2 Tue 9/18

4/6 from Special:Permalink/1894335 to QB/d_cp2.6 | Equations 2/3 from Special:Permalink/1863399 to QB/c19ElectricPotentialField_SurfaceIntegral |Equations 4/6 from Special:Permalink/1863398 to QB/c19ElectricPotentialField_GaussLaw | Equations
$\Phi ={\vec {E}}\cdot {\vec {A}}$ $\to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA$ = electric flux
$q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$
$d\,{\text{Vol}}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\phi d\theta$
$A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}$
Calculating $\int fdA$ and $\int fdV$ with angular symmetry Cyndrical: $dA=2\pi r\,dz;\,dV=dA\,dr$. Spherical: $\int dA=4\pi r^{2},\;dV=4\pi r^{2}\,dr$
Calculating $\int fdA$ and $\int fdV$ with angular symmetry Cyndrical: $dA=2\pi r\,dz;\,dV=dA\,dr$. Spherical: $\int dA=4\pi r^{2},\;dV=4\pi r^{2}\,dr$

### Quizbank/University Physics Semester 2/T3

3/11 from Special:Permalink/1893815 to QB/d_cp2.7 | Equations 2/4 from Special:Permalink/1893633 to QB/d_cp2.8 | Equations 3/5 from Special:Permalink/1863338 to QB/a19ElectricPotentialField_Capacitance | Equations 2/4 from Special:Permalink/1863339 to QB/a19ElectricPotentialField_KE_PE | Equations
$\Delta V_{AB}=V_{A}-V_{B}=-\int _{A}^{B}{\vec {E}}\cdot d{\vec {\ell }}$ = 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$^{−31}$kg (1.67× 10$^{−27}$kg). Elementary charge = e = 1.602×10$^{−19}$C.
$K={\tfrac {1}{2}}mv^{2}$=kinetic energy. 1 eV = 1.602×10$^{−19}$J
$V(r)=k{\tfrac {q}{r}}$ near isolated point charge
Many charges: $V_{P}=k\sum _{1}^{N}{\frac {q_{i}}{r_{i}}}\to k\int {\frac {dq}{r}}$.
The alpha-particle is made up of two protons and two neutrons.
$Q=CV$ 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
Available at special:permalink/1925995

### Quizbank/University Physics Semester 2/T4

2/10 from Special:Permalink/1893634 to QB/d_cp2.9|Equations 2/9 from Special:Permalink/1895273 to QB/d_cp2.10|Equations 2/4 from Special:Permalink/1863340 to QB/a20ElectricCurrentResistivityOhm_PowerDriftVel | Equations 2/21 from Special:Permalink/1863341 to QB/a21CircuitsBioInstDC_circAnalQuiz1 | Equations 2/5 from Special:Permalink/1863342 to QB/a21CircuitsBioInstDC_circuits | Equations
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=$P=IV=I^{2}R=V^{2}/R$
$V_{terminal}=\varepsilon -Ir_{eq}$ 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 $I(t)=-{\tfrac {Q}{RC}}e^{-t/\tau }$

### Quizbank/University Physics Semester 2/T5

2/9 from Special:Permalink/1902372 to QB/d_cp2.11 | Equations | Textbook links for d_cp2.11
2/11 from Special:Permalink/1892310 to QB/d_cp2.12 | Equations | Textbook links for d_cp2.12 2/4 from Special:Permalink/1863344 to QB/a22Magnetism_forces | Equations 2/6 from Special:Permalink/1828922 to QB/c22Magnetism_ampereLaw | Equations 2/4 from Special:Permalink/1863400 to QB/c22Magnetism_ampereLawSymmetry | Equations
$|{\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 =$q\left({\vec {E}}+{\vec {v}}\times {\vec {B}}\right)$
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

### Quizbank/University Physics Semester 2/T6 Wright State Lake skips T6

1/6 from Special:Permalink/1894334 to QB/d_cp2.5 | Equations 2/6 from Special:Permalink/1894335 to QB/d_cp2.6 | Equations 2/11 from Special:Permalink/1893815 to QB/d_cp2.7 | Equations 1/4 from Special:Permalink/1893633 to QB/d_cp2.8 | Equations 2/10 from Special:Permalink/1893634 to QB/d_cp2.9 | Equations 2/9 from Special:Permalink/1895273 to QB/d_cp2.10 | Equations $\varepsilon _{0}=$ 8.85×10$^{−12}$ F/m = vacuum permittivity.
e = 1.602×10$^{−19}$C: negative (positive) charge for electrons (protons)
$k_{e}={\tfrac {1}{4\pi \varepsilon _{0}}}=$ = 8.99×10$^9$ 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.
$\Phi ={\vec {E}}\cdot {\vec {A}}$ $\to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA$ = electric flux
$q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$
$d\,{\text{Vol}}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\phi d\theta$
$A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}$
$\Delta V_{AB}=V_{A}-V_{B}=-\int _{A}^{B}{\vec {E}}\cdot d{\vec {\ell }}$ = 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$^{−31}$kg (1.67× 10$^{−27}$kg). Elementary charge = e = 1.602×10$^{−19}$C.
$K={\tfrac {1}{2}}mv^{2}$=kinetic energy. 1 eV = 1.602×10$^{−19}$J
$V(r)=k{\tfrac {q}{r}}$ near isolated point charge
Many charges: $V_{P}=k\sum _{1}^{N}{\frac {q_{i}}{r_{i}}}\to k\int {\frac {dq}{r}}$.
$Q=CV$ 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
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=$P=IV=I^{2}R=V^{2}/R$
$V_{terminal}=\varepsilon -Ir_{eq}$ 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 $I(t)=-{\tfrac {Q}{RC}}e^{-t/\tau }$

### Quizbank/University Physics Semester 2/T7

3/9 from Special:Permalink/1893631 to QB/d_cp2.13 | Equations 2/6 from Special:Permalink/1892308 to QB/d_cp2.14 | Equations 3/8 from Special:Permalink/1894891 to QB/d_cp2.15 | Equations 1/2 from Special:Permalink/1863345 to QB/a23InductionACcircuits_Q1 | Equations 1/4 from Special:Permalink/1863343 to QB/a21CircuitsBioInstDC_RCdecaySimple | Equations
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 $\varepsilon =NBA\omega \sin \omega t$
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 $\omega ={\sqrt {\tfrac {1}{LC}}}$
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 ${\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}$

### Next test: Quizbank/University Physics Semester 2/T8 Tues 4 Dec

5/8 from Special:Permalink/1863346 to QB/a25GeometricOptics_image
2/4 from Special:Permalink/1863347 to QB/a25GeometricOptics_thinLenses | Equations 3/4 from Special:Permalink/1863348 to QB/a25GeometricOptics_vision

### Final exam: Quizbank/University Physics Semester 2/T9

2/9 from Special:Permalink/1902372 to QB/d_cp2.11 | Equations 2/11 from Special:Permalink/1892310 to QB/d_cp2.12 | Equations 1/9 from Special:Permalink/1893631 to QB/d_cp2.13 | Equations 1/6 from Special:Permalink/1892308 to QB/d_cp2.14 | Equations 1/8 from Special:Permalink/1894891 to QB/d_cp2.15 | Equations 2/6 from Special:Permalink/1895295 to QB/d_cp2.16 | Equations 2/4 from Special:Permalink/1863401 to QB/c24ElectromagneticWaves_displacementCurrent | Equations (there might be a bit more?)
$|{\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 =$q\left({\vec {E}}+{\vec {v}}\times {\vec {B}}\right)$
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
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 $\varepsilon =NBA\omega \sin \omega t$
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 $\omega ={\sqrt {\tfrac {1}{LC}}}$
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 ${\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}$
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).
To be continued

### Quizbank/University Physics Semester 2/T10

* See also WikiJournal of Science/A card game for Bell's theorem and its loopholes
1/8 from Special:Permalink/1878340 to QB/d_Bell.solitaire | Discussion 5/18 from Special:Permalink/1878339 to QB/d_Bell.polarization | Discussion 5/25 from Special:Permalink/1885266 to QB/d_Bell.photon | Discussion 1/10 from Special:Permalink/1878495 to QB/d_Bell.Venn | Discussion
We might not need an equation sheet for this quiz. These are "conceptual" quizzes so instead of "Equations" we have "Discussion".

### Quizbank/University Physics Semester 2/T11

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$\varepsilon _{0}=$ 8.85×10$^{−12}$ F/m = vacuum permittivity.
e = 1.602×10$^{−19}$C: negative (positive) charge for electrons (protons)
$k_{e}={\tfrac {1}{4\pi \varepsilon _{0}}}=$ = 8.99×10$^9$ 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.
$\Phi ={\vec {E}}\cdot {\vec {A}}$ $\to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA$ = electric flux
$q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$
$d\,{\text{Vol}}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\phi d\theta$
$A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}$
$\Delta V_{AB}=V_{A}-V_{B}=-\int _{A}^{B}{\vec {E}}\cdot d{\vec {\ell }}$ = 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$^{−31}$kg (1.67× 10$^{−27}$kg). Elementary charge = e = 1.602×10$^{−19}$C.
$K={\tfrac {1}{2}}mv^{2}$=kinetic energy. 1 eV = 1.602×10$^{−19}$J
$V(r)=k{\tfrac {q}{r}}$ near isolated point charge
Many charges: $V_{P}=k\sum _{1}^{N}{\frac {q_{i}}{r_{i}}}\to k\int {\frac {dq}{r}}$.
$Q=CV$ 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
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=$P=IV=I^{2}R=V^{2}/R$
$V_{terminal}=\varepsilon -Ir_{eq}$ 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 $I(t)=-{\tfrac {Q}{RC}}e^{-t/\tau }$
$|{\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 =$q\left({\vec {E}}+{\vec {v}}\times {\vec {B}}\right)$
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
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 $\varepsilon =NBA\omega \sin \omega t$
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 $\omega ={\sqrt {\tfrac {1}{LC}}}$
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 ${\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}$
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).
Check transclusion

## See also:

* Questions list (useful for students as they prepare for the tests)
* Quizbank/Instructions (classroom use suggestions)
* Subpages

### Transclusion problem:Copied from Talk

* ↑ But perhaps not. How difficult do we what these questions to be?
* ↑ Not transcluding for some reason