# User talk:Guy vandegrift/1

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${\displaystyle \varepsilon _{0}=}$ 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.

${\displaystyle {\mathcal {ANNOUNCEMENT}}}$   This is only one of three closely related resources relevant to OpenStax ${\displaystyle {\mathcal {E}}\&{\mathcal {M}}}$
To view the other two resources visit Category:Quizbank/Units

• To use this resource you need to download 11 large pdf files. Here are links to them:   T1    T2    T3   T4    T5   T6   T7    T8    T9    T10    T11
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• Equations sheets are under construction and are based on
• 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

${\displaystyle {\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{2}}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\vec {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{3}}}}$  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.

### Quizbank/University Physics Semester 2/T2

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

### Quizbank/University Physics Semester 2/T3

${\displaystyle \Delta V_{AB}=V_{A}-V_{B}=-\int _{A}^{B}{\vec {E}}\cdot d{\vec {\ell }}}$  = 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: ${\displaystyle V_{P}=k\sum _{1}^{N}{\frac {q_{i}}{r_{i}}}\to k\int {\frac {dq}{r}}}$ .

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

${\displaystyle Q=CV}$  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

### Quizbank/University Physics Semester 2/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 ${\displaystyle I(t)=-{\tfrac {Q}{RC}}e^{-t/\tau }}$

### Quizbank/University Physics Semester 2/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

### Quizbank/University Physics Semester 2/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

transclustion problem with cp2.5 and 2.6

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

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

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

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

[2]

${\displaystyle \varepsilon _{0}=}$  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.

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

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

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

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

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

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 ${\displaystyle I(t)=-{\tfrac {Q}{RC}}e^{-t/\tau }}$

### Quizbank/University Physics Semester 2/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 ${\displaystyle {\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}}$

To be continued

To be continued

### Quizbank/University Physics Semester 2/T9

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

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 ${\displaystyle {\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}}$

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).

To be continued

### Quizbank/University Physics Semester 2/T11

1/6 from Special:Permalink/1894334 to QB/d_cp2.5 | Equations
1/6 from Special:Permalink/1894335 to QB/d_cp2.6 | Equations
1/11 from Special:Permalink/1893815 to QB/d_cp2.7 | Equations
1/4 from Special:Permalink/1893633 to QB/d_cp2.8 | Equations
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1/6 from Special:Permalink/1895295 to QB/d_cp2.16 | Equations

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

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

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

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

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

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 ${\displaystyle I(t)=-{\tfrac {Q}{RC}}e^{-t/\tau }}$

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

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 ${\displaystyle {\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}}$

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).

Check transclusion

### Transclusion problem:Copied from Talk

transclustion problem with cp2.5 and 2.6

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

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

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

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

[4]

## Try it here

transclustion problem with cp2.5 and 2.6

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

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

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

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

[6]

1. Not transcluding for some reason
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4. Not transcluding for some reason
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