Quizbank/University Physics Semester 2

Wright State University Lake Campus/2018-9/Phy2410 Template:EduV/announcement

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• The students can use the next section to see what "might" be on the test, and the equations shown can be displayed for the students to use:

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)

${\epsilon }_0=$ 8.85×10⁻¹² F/m = vacuum permittivity.

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

$k_e=\frac{1}{4\pi {\epsilon }_0}=$ = 8.99×10⁹ m/F

$\overrightarrow{F}=Q\overrightarrow{E}$ where $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\sum _{i=1}^N\frac{q_i}{r_{Pi}^2}{\hat{r}}_{Pi}$

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

$E=\frac{\sigma }{2{\epsilon }_0}$ = field above an infinite plane of charge.

$\overrightarrow{E}(\overrightarrow{r})=\frac{1}{4\pi {\epsilon }_0}\sum _{i=1}^N\frac{{\widehat{ℛ}}_i{Q}_i}{|{\overrightarrow{ℛ}}_i{|}^2}=\frac{1}{4\pi {\epsilon }_0}\sum _{i=1}^N\frac{{\overrightarrow{ℛ}}_i{Q}_i}{|{\overrightarrow{ℛ}}_i{|}^3}$ is the electric field at the field point, $\overrightarrow{r}$, due to point charges at the source points,${\overrightarrow{r}}_i$ , and ${\overrightarrow{ℛ}}_i=\overrightarrow{r}-{\overrightarrow{r}}_i,$ points from source points to the field point. $\overrightarrow{E}(\overrightarrow{r})=\frac{1}{4\pi {\epsilon }_0}\sum _{i=1}^N\frac{{\widehat{ℛ}}_i{Q}_i}{|{\overrightarrow{ℛ}}_i{|}^2}=\frac{1}{4\pi {\epsilon }_0}\sum _{i=1}^N\frac{{\overrightarrow{ℛ}}_i{Q}_i}{|{\overrightarrow{ℛ}}_i{|}^3}$ is the electric field at the field point, $\overrightarrow{r}$, due to point charges at the source points,${\overrightarrow{r}}_i$ , and ${\overrightarrow{ℛ}}_i=\overrightarrow{r}-{\overrightarrow{r}}_i,$ 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 10⁹ 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

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

$q_{enclosed}={\epsilon }_0\oint \overrightarrow{E}\cdot d\overrightarrow{A}$

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

$A_{\text{sphere}}=r^2{\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\varphi =4\pi r^2$

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

Calculating $\int fdA$ and $\int fdV$ with angular symmetry
Cyndrical: $dA=2\pi rdz;dV=dAdr$.  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

$\mathrm{\Delta }{V}_{AB}=V_A-V_B=-{\int }_A^B\overrightarrow{E}\cdot d\overrightarrow{\ell }$ = electric potential

$\overrightarrow{E}=-\frac{\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k}=-\overrightarrow{\mathrm{\nabla }}V$

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

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

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

$K=\frac{1}{2}m{v}^2$=kinetic energy. 1 eV = 1.602×10⁻¹⁹J

$V(r)=k\frac{q}{r}$ 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.

$C={\epsilon }_0\frac{A}{d}$ where A is area and d<<A^1/2 is gap length of parallel plate capacitor

$\text{Series}:\frac{1}{C_S}=\sum \frac{1}{C_i}.$   $\text{ Parallel:}{C}_P=\sum C_i.$

$u=\frac{1}{2}QV=\frac{1}{2}C{V}^2=\frac{1}{2C}{Q}^2$ = stored energy

$u_E=\frac{1}{2}{\epsilon }_0{E}^2$ = 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=$I=dQ/dt=nq{v}_{d}A$, where

$(n,q,v_d,A)$ = (density, charge, speed, Area)

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

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

$\rho ={\rho }_0[1+\alpha (T-T_0)]$, and $R=R_0[1+\alpha \mathrm{\Delta }T]$,

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

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

$V_{terminal}=\epsilon -I{r}_{eq}$ where $r_{eq}$=internal resistance and $\epsilon$=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: $q(t)=Q(1-e^{t/\tau })$ and $I=I_0{e}^{-t/\tau }$ where $\tau =RC$ is RC time, $Q=\epsilon C$ and $I_0=\epsilon /R$.

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

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

cross product

$|\overrightarrow{a}\times \overrightarrow{b}|$$=ab\sin \theta \iff$ $(\overrightarrow{a}\times \overrightarrow{b}{)}_x=(a_y{b}_z-a_z{b}_y)$, $(\overrightarrow{a}\times \overrightarrow{b}{)}_y=(a_z{b}_x-a_x{b}_z)$, $(\overrightarrow{a}\times \overrightarrow{b}{)}_z=(a_x{b}_y-a_y{b}_x)$
Magnetic force: $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B},$$d\overrightarrow{F}=I\overrightarrow{d\ell }\times \overrightarrow{B}$.
${\overrightarrow{v}}_d=\overrightarrow{E}\times \overrightarrow{B}/B^2$=EXB drift velocity
Circular motion (uniform B field): $r=\frac{mv}{qB}.$ Period=$T=\frac{2\pi m}{qB}.$

Hall effect

Dipole moment=$\overrightarrow{\mu }=NIA\hat{n}$. Torque=$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$. Stored energy=$U=\overrightarrow{\mu }\cdot \overrightarrow{B}$.
Hall field =$E=V/\ell =B{v}_d=\frac{IB}{neA}$
Lorentz force =$q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

Free space permeability ${\mu }_0=4\pi \times 10^{-7}$ T·m/A
Force between parallel wires $\frac{F}{\ell }=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}$
Biot–Savart law $\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\underset{wire}{\int }\frac{Id\overrightarrow{\ell }\times \hat{r}}{r^2}$
Ampère's Law:$\oint \overrightarrow{B}\cdot d\overrightarrow{\ell }=4\pi {\mu }_0{I}_{enc}$
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
${\epsilon }_0=$ 8.85×10⁻¹² F/m = vacuum permittivity.

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

$k_e=\frac{1}{4\pi {\epsilon }_0}=$ = 8.99×10⁹ m/F

$\overrightarrow{F}=Q\overrightarrow{E}$ where $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\sum _{i=1}^N\frac{q_i}{r_{Pi}^2}{\hat{r}}_{Pi}$

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

$E=\frac{\sigma }{2{\epsilon }_0}$ = field above an infinite plane of charge.

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

$q_{enclosed}={\epsilon }_0\oint \overrightarrow{E}\cdot d\overrightarrow{A}$

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

$A_{\text{sphere}}=r^2{\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\varphi =4\pi r^2$

$\mathrm{\Delta }{V}_{AB}=V_A-V_B=-{\int }_A^B\overrightarrow{E}\cdot d\overrightarrow{\ell }$ = electric potential

$\overrightarrow{E}=-\frac{\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k}=-\overrightarrow{\mathrm{\nabla }}V$

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

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

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

$K=\frac{1}{2}m{v}^2$=kinetic energy. 1 eV = 1.602×10⁻¹⁹J

$V(r)=k\frac{q}{r}$ 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.

$C={\epsilon }_0\frac{A}{d}$ where A is area and d<<A^1/2 is gap length of parallel plate capacitor

$\text{Series}:\frac{1}{C_S}=\sum \frac{1}{C_i}.$   $\text{ Parallel:}{C}_P=\sum C_i.$

$u=\frac{1}{2}QV=\frac{1}{2}C{V}^2=\frac{1}{2C}{Q}^2$ = stored energy

$u_E=\frac{1}{2}{\epsilon }_0{E}^2$ = energy density

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

Current=$I=dQ/dt=nq{v}_{d}A$, where

$(n,q,v_d,A)$ = (density, charge, speed, Area)

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

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

$\rho ={\rho }_0[1+\alpha (T-T_0)]$, and $R=R_0[1+\alpha \mathrm{\Delta }T]$,

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

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

$V_{terminal}=\epsilon -I{r}_{eq}$ where $r_{eq}$=internal resistance and $\epsilon$=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: $q(t)=Q(1-e^{t/\tau })$ and $I=I_0{e}^{-t/\tau }$ where $\tau =RC$ is RC time, $Q=\epsilon C$ and $I_0=\epsilon /R$.

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

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 ${\mathrm{\Phi }}_m={\int }_S\overrightarrow{B}\cdot \hat{n}dA$
Motional $\epsilon =B\ell v$ if $\overrightarrow{v}\perp \overrightarrow{B}$
Electromotive "force" (volts) $\epsilon =-N\frac{d{\mathrm{\Phi }}_m}{dt}=\oint \overrightarrow{E}\cdot d\overrightarrow{\ell }$
rotating coil $\epsilon =NBA\omega \sin \omega t$

Unit of inductance = Henry (H)=1V·s/A

Mutual inductance: $M\frac{d{I}_2}{dt}=N_1\frac{d{\mathrm{\Phi }}_{12}}{dt}=-{\epsilon }_1$ where ${\mathrm{\Phi }}_{12}$=flux through 1 due to current in 2. Reciprocity: $M\frac{d{I}_1}{dt}=-{\epsilon }_2$

Self-inductance: $N{\mathrm{\Phi }}_m=LI\to \epsilon =-L\frac{dI}{dt}$

$L_{\text{solenoid}}\approx {\mu }_0{N}^{2}A\ell$, $L_{\text{toroid}}\approx \frac{{\mu }_0{N}^{2}h}{2\pi }\ln \frac{R_2}{R_1}$, Stored energy=$\frac{1}{2}L{I}^2$

$I(t)=\frac{\epsilon }{R}(1-e^{-t/\tau })$ in LR circuit where $\tau =L/R$.

$q(t)=q_0\cos (\omega t+\varphi )$ in LC circuit where $\omega =\sqrt{\frac{1}{LC}}$

AC voltage and current $v=V_0\sin (\omega t-\varphi )$ if $i=I_0\sin \omega t.$
RMS values $I_{rms}=\frac{I_0}{\sqrt{2}}$ and $V_{rms}=\frac{V_0}{\sqrt{2}}$
Impedance $V_0=I_{0}X$
Resistor $V_0=I_0{X}_R,\varphi =0,$ where $X_R=R$
Capacitor $V_0=I_0{X}_C,\varphi =-\frac{\pi }{2},$ where $X_C=\frac{1}{\omega C}$
Inductor $V_0=I_0{X}_L,\varphi =+\frac{\pi }{2},$ where $X_L=\omega L$
RLC series circuit $V_0=I_{0}Z$ where $Z=\sqrt{R^2+{(X_L-X_C)}^2}$ and $\varphi ={\tan }^{-1}\frac{X_L-X_C}{R}$
Resonant angular frequency ${\omega }_0=\sqrt{\frac{1}{LC}}$
Quality factor $Q=\frac{{\omega }_0}{\mathrm{\Delta }\omega }=\frac{{\omega }_{0}L}{R}$
Average power $P_{ave}=\frac{1}{2}{I}_0{V}_0\cos \varphi =I_{rms}{V}_{rms}\cos \varphi$
Transformer voltages and currents $\frac{V_S}{V_P}=\frac{N_S}{N_P}=\frac{I_P}{I_S}$

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?)

cross product

$|\overrightarrow{a}\times \overrightarrow{b}|$$=ab\sin \theta \iff$ $(\overrightarrow{a}\times \overrightarrow{b}{)}_x=(a_y{b}_z-a_z{b}_y)$, $(\overrightarrow{a}\times \overrightarrow{b}{)}_y=(a_z{b}_x-a_x{b}_z)$, $(\overrightarrow{a}\times \overrightarrow{b}{)}_z=(a_x{b}_y-a_y{b}_x)$
Magnetic force: $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B},$$d\overrightarrow{F}=I\overrightarrow{d\ell }\times \overrightarrow{B}$.
${\overrightarrow{v}}_d=\overrightarrow{E}\times \overrightarrow{B}/B^2$=EXB drift velocity
Circular motion (uniform B field): $r=\frac{mv}{qB}.$ Period=$T=\frac{2\pi m}{qB}.$

Hall effect

Dipole moment=$\overrightarrow{\mu }=NIA\hat{n}$. Torque=$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$. Stored energy=$U=\overrightarrow{\mu }\cdot \overrightarrow{B}$.
Hall field =$E=V/\ell =B{v}_d=\frac{IB}{neA}$
Lorentz force =$q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

Free space permeability ${\mu }_0=4\pi \times 10^{-7}$ T·m/A
Force between parallel wires $\frac{F}{\ell }=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}$
Biot–Savart law $\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\underset{wire}{\int }\frac{Id\overrightarrow{\ell }\times \hat{r}}{r^2}$
Ampère's Law:$\oint \overrightarrow{B}\cdot d\overrightarrow{\ell }=4\pi {\mu }_0{I}_{enc}$
Magnetic field inside solenoid with paramagnetic material =$B=\mu nI$ where $\mu =(1+\chi ){\mu }_0$= permeability

Magnetic flux ${\mathrm{\Phi }}_m={\int }_S\overrightarrow{B}\cdot \hat{n}dA$
Motional $\epsilon =B\ell v$ if $\overrightarrow{v}\perp \overrightarrow{B}$
Electromotive "force" (volts) $\epsilon =-N\frac{d{\mathrm{\Phi }}_m}{dt}=\oint \overrightarrow{E}\cdot d\overrightarrow{\ell }$
rotating coil $\epsilon =NBA\omega \sin \omega t$

Unit of inductance = Henry (H)=1V·s/A

Mutual inductance: $M\frac{d{I}_2}{dt}=N_1\frac{d{\mathrm{\Phi }}_{12}}{dt}=-{\epsilon }_1$ where ${\mathrm{\Phi }}_{12}$=flux through 1 due to current in 2. Reciprocity: $M\frac{d{I}_1}{dt}=-{\epsilon }_2$

Self-inductance: $N{\mathrm{\Phi }}_m=LI\to \epsilon =-L\frac{dI}{dt}$

$L_{\text{solenoid}}\approx {\mu }_0{N}^{2}A\ell$, $L_{\text{toroid}}\approx \frac{{\mu }_0{N}^{2}h}{2\pi }\ln \frac{R_2}{R_1}$, Stored energy=$\frac{1}{2}L{I}^2$

$I(t)=\frac{\epsilon }{R}(1-e^{-t/\tau })$ in LR circuit where $\tau =L/R$.

$q(t)=q_0\cos (\omega t+\varphi )$ in LC circuit where $\omega =\sqrt{\frac{1}{LC}}$

AC voltage and current $v=V_0\sin (\omega t-\varphi )$ if $i=I_0\sin \omega t.$
RMS values $I_{rms}=\frac{I_0}{\sqrt{2}}$ and $V_{rms}=\frac{V_0}{\sqrt{2}}$
Impedance $V_0=I_{0}X$
Resistor $V_0=I_0{X}_R,\varphi =0,$ where $X_R=R$
Capacitor $V_0=I_0{X}_C,\varphi =-\frac{\pi }{2},$ where $X_C=\frac{1}{\omega C}$
Inductor $V_0=I_0{X}_L,\varphi =+\frac{\pi }{2},$ where $X_L=\omega L$
RLC series circuit $V_0=I_{0}Z$ where $Z=\sqrt{R^2+{(X_L-X_C)}^2}$ and $\varphi ={\tan }^{-1}\frac{X_L-X_C}{R}$
Resonant angular frequency ${\omega }_0=\sqrt{\frac{1}{LC}}$
Quality factor $Q=\frac{{\omega }_0}{\mathrm{\Delta }\omega }=\frac{{\omega }_{0}L}{R}$
Average power $P_{ave}=\frac{1}{2}{I}_0{V}_0\cos \varphi =I_{rms}{V}_{rms}\cos \varphi$
Transformer voltages and currents $\frac{V_S}{V_P}=\frac{N_S}{N_P}=\frac{I_P}{I_S}$

Displacement current $I_d={\epsilon }_0\frac{d{\mathrm{\Phi }}_E}{dt}$ where ${\mathrm{\Phi }}_E=\int \overrightarrow{E}\cdot d\overrightarrow{A}$ is the electric flux.

Maxwell's equations: ${ϵ}_0{\mu }_0=1/c^2$
${\oint }_S\overrightarrow{E}\cdot \mathrm{d}\overrightarrow{A}=\frac{1}{{ϵ}_0}{Q}_{in}$
${\oint }_S\overrightarrow{B}\cdot \mathrm{d}\overrightarrow{A}=0$
${\oint }_C\overrightarrow{E}\cdot \mathrm{d}\overrightarrow{\ell }=-{\int }_S\frac{\partial \overrightarrow{B}}{\partial t}\cdot \mathrm{d}\overrightarrow{A}$
${\oint }_C\overrightarrow{B}\cdot \mathrm{d}\overrightarrow{\ell }={\mu }_{0}I+{ϵ}_0{\mu }_0\frac{\mathrm{d}{\mathrm{\Phi }}_E}{\mathrm{d}t}$

$\frac{{\partial }^2{E}_y}{\partial x^2}={\epsilon }_0{\mu }_0\frac{{\partial }^2{E}_y}{\partial t^2}$ and $\frac{E_0}{B_0}=c$

Poynting vector $\overrightarrow{S}=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}$=energy flux

Average intensity $I=S_{ave}=\frac{c{\epsilon }_0}{2}{E}_0^2=\frac{c}{2{\mu }_0}{B}_0^2=\frac{1}{2{\mu }_0}{E}_0{B}_0$

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
  

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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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${\epsilon }_0=$ 8.85×10⁻¹² F/m = vacuum permittivity.

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

$k_e=\frac{1}{4\pi {\epsilon }_0}=$ = 8.99×10⁹ m/F

$\overrightarrow{F}=Q\overrightarrow{E}$ where $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\sum _{i=1}^N\frac{q_i}{r_{Pi}^2}{\hat{r}}_{Pi}$

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

$E=\frac{\sigma }{2{\epsilon }_0}$ = field above an infinite plane of charge.

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

$q_{enclosed}={\epsilon }_0\oint \overrightarrow{E}\cdot d\overrightarrow{A}$

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

$A_{\text{sphere}}=r^2{\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\varphi =4\pi r^2$

$\mathrm{\Delta }{V}_{AB}=V_A-V_B=-{\int }_A^B\overrightarrow{E}\cdot d\overrightarrow{\ell }$ = electric potential

$\overrightarrow{E}=-\frac{\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k}=-\overrightarrow{\mathrm{\nabla }}V$

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

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

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

$K=\frac{1}{2}m{v}^2$=kinetic energy. 1 eV = 1.602×10⁻¹⁹J

$V(r)=k\frac{q}{r}$ 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.

$C={\epsilon }_0\frac{A}{d}$ where A is area and d<<A^1/2 is gap length of parallel plate capacitor

$\text{Series}:\frac{1}{C_S}=\sum \frac{1}{C_i}.$   $\text{ Parallel:}{C}_P=\sum C_i.$

$u=\frac{1}{2}QV=\frac{1}{2}C{V}^2=\frac{1}{2C}{Q}^2$ = stored energy

$u_E=\frac{1}{2}{\epsilon }_0{E}^2$ = energy density

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

Current=$I=dQ/dt=nq{v}_{d}A$, where

$(n,q,v_d,A)$ = (density, charge, speed, Area)

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

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

$\rho ={\rho }_0[1+\alpha (T-T_0)]$, and $R=R_0[1+\alpha \mathrm{\Delta }T]$,

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

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

$V_{terminal}=\epsilon -I{r}_{eq}$ where $r_{eq}$=internal resistance and $\epsilon$=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: $q(t)=Q(1-e^{t/\tau })$ and $I=I_0{e}^{-t/\tau }$ where $\tau =RC$ is RC time, $Q=\epsilon C$ and $I_0=\epsilon /R$.

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

cross product

$|\overrightarrow{a}\times \overrightarrow{b}|$$=ab\sin \theta \iff$ $(\overrightarrow{a}\times \overrightarrow{b}{)}_x=(a_y{b}_z-a_z{b}_y)$, $(\overrightarrow{a}\times \overrightarrow{b}{)}_y=(a_z{b}_x-a_x{b}_z)$, $(\overrightarrow{a}\times \overrightarrow{b}{)}_z=(a_x{b}_y-a_y{b}_x)$
Magnetic force: $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B},$$d\overrightarrow{F}=I\overrightarrow{d\ell }\times \overrightarrow{B}$.
${\overrightarrow{v}}_d=\overrightarrow{E}\times \overrightarrow{B}/B^2$=EXB drift velocity
Circular motion (uniform B field): $r=\frac{mv}{qB}.$ Period=$T=\frac{2\pi m}{qB}.$

Hall effect

Dipole moment=$\overrightarrow{\mu }=NIA\hat{n}$. Torque=$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$. Stored energy=$U=\overrightarrow{\mu }\cdot \overrightarrow{B}$.
Hall field =$E=V/\ell =B{v}_d=\frac{IB}{neA}$
Lorentz force =$q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

Free space permeability ${\mu }_0=4\pi \times 10^{-7}$ T·m/A
Force between parallel wires $\frac{F}{\ell }=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}$
Biot–Savart law $\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\underset{wire}{\int }\frac{Id\overrightarrow{\ell }\times \hat{r}}{r^2}$
Ampère's Law:$\oint \overrightarrow{B}\cdot d\overrightarrow{\ell }=4\pi {\mu }_0{I}_{enc}$
Magnetic field inside solenoid with paramagnetic material =$B=\mu nI$ where $\mu =(1+\chi ){\mu }_0$= permeability

Magnetic flux ${\mathrm{\Phi }}_m={\int }_S\overrightarrow{B}\cdot \hat{n}dA$
Motional $\epsilon =B\ell v$ if $\overrightarrow{v}\perp \overrightarrow{B}$
Electromotive "force" (volts) $\epsilon =-N\frac{d{\mathrm{\Phi }}_m}{dt}=\oint \overrightarrow{E}\cdot d\overrightarrow{\ell }$
rotating coil $\epsilon =NBA\omega \sin \omega t$

Unit of inductance = Henry (H)=1V·s/A

Mutual inductance: $M\frac{d{I}_2}{dt}=N_1\frac{d{\mathrm{\Phi }}_{12}}{dt}=-{\epsilon }_1$ where ${\mathrm{\Phi }}_{12}$=flux through 1 due to current in 2. Reciprocity: $M\frac{d{I}_1}{dt}=-{\epsilon }_2$

Self-inductance: $N{\mathrm{\Phi }}_m=LI\to \epsilon =-L\frac{dI}{dt}$

$L_{\text{solenoid}}\approx {\mu }_0{N}^{2}A\ell$, $L_{\text{toroid}}\approx \frac{{\mu }_0{N}^{2}h}{2\pi }\ln \frac{R_2}{R_1}$, Stored energy=$\frac{1}{2}L{I}^2$

$I(t)=\frac{\epsilon }{R}(1-e^{-t/\tau })$ in LR circuit where $\tau =L/R$.

$q(t)=q_0\cos (\omega t+\varphi )$ in LC circuit where $\omega =\sqrt{\frac{1}{LC}}$

AC voltage and current $v=V_0\sin (\omega t-\varphi )$ if $i=I_0\sin \omega t.$
RMS values $I_{rms}=\frac{I_0}{\sqrt{2}}$ and $V_{rms}=\frac{V_0}{\sqrt{2}}$
Impedance $V_0=I_{0}X$
Resistor $V_0=I_0{X}_R,\varphi =0,$ where $X_R=R$
Capacitor $V_0=I_0{X}_C,\varphi =-\frac{\pi }{2},$ where $X_C=\frac{1}{\omega C}$
Inductor $V_0=I_0{X}_L,\varphi =+\frac{\pi }{2},$ where $X_L=\omega L$
RLC series circuit $V_0=I_{0}Z$ where $Z=\sqrt{R^2+{(X_L-X_C)}^2}$ and $\varphi ={\tan }^{-1}\frac{X_L-X_C}{R}$
Resonant angular frequency ${\omega }_0=\sqrt{\frac{1}{LC}}$
Quality factor $Q=\frac{{\omega }_0}{\mathrm{\Delta }\omega }=\frac{{\omega }_{0}L}{R}$
Average power $P_{ave}=\frac{1}{2}{I}_0{V}_0\cos \varphi =I_{rms}{V}_{rms}\cos \varphi$
Transformer voltages and currents $\frac{V_S}{V_P}=\frac{N_S}{N_P}=\frac{I_P}{I_S}$

Displacement current $I_d={\epsilon }_0\frac{d{\mathrm{\Phi }}_E}{dt}$ where ${\mathrm{\Phi }}_E=\int \overrightarrow{E}\cdot d\overrightarrow{A}$ is the electric flux.

Maxwell's equations: ${ϵ}_0{\mu }_0=1/c^2$
${\oint }_S\overrightarrow{E}\cdot \mathrm{d}\overrightarrow{A}=\frac{1}{{ϵ}_0}{Q}_{in}$
${\oint }_S\overrightarrow{B}\cdot \mathrm{d}\overrightarrow{A}=0$
${\oint }_C\overrightarrow{E}\cdot \mathrm{d}\overrightarrow{\ell }=-{\int }_S\frac{\partial \overrightarrow{B}}{\partial t}\cdot \mathrm{d}\overrightarrow{A}$
${\oint }_C\overrightarrow{B}\cdot \mathrm{d}\overrightarrow{\ell }={\mu }_{0}I+{ϵ}_0{\mu }_0\frac{\mathrm{d}{\mathrm{\Phi }}_E}{\mathrm{d}t}$

$\frac{{\partial }^2{E}_y}{\partial x^2}={\epsilon }_0{\mu }_0\frac{{\partial }^2{E}_y}{\partial t^2}$ and $\frac{E_0}{B_0}=c$

Poynting vector $\overrightarrow{S}=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}$=energy flux

Average intensity $I=S_{ave}=\frac{c{\epsilon }_0}{2}{E}_0^2=\frac{c}{2{\mu }_0}{B}_0^2=\frac{1}{2{\mu }_0}{E}_0{B}_0$

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

transclustion problem with cp2.5 and 2.6

{{#lsth:OpenStax University Physics/E&M/Electric Charges and Fields|For quiz at [[QB/d_cp2.5]]}}<ref>Not transcluding for some reason</ref> $\mathrm{\Phi }=\overrightarrow{E}\cdot \overrightarrow{A}$ $\to \int \overrightarrow{E}\cdot d\overrightarrow{A}=\int \overrightarrow{E}\cdot \hat{n}dA$ = electric flux $q_{enclosed}={\epsilon }_0\oint \overrightarrow{E}\cdot d\overrightarrow{A}$ $d\text{Vol}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\varphi d\theta$ $A_{\text{sphere}}=r^2{\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\varphi =4\pi r^2$