π MIT 8.02x Electricity and Magnetism
Physics
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π Hi I'm Cherry! π
This deck contains everything taught in the free 3 part MITx - 8.02x Electricity and Magnetism course which can be found on openlearninglibrary.mit.edu (8.02.1x, 8.02.2x, 8.02.3x).
This course is made up of 14 weeks and 37 lessons (excluding optional week 15), but for AP Physics C: Electricity and Magnetism, you only need to up to week 11 and up to lesson 28 (the curriculum matches almost perfectly). I got a 5 on my AP exam self-studying using this course and this deck.
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βοΈ Features βοΈ:
- Cards in the deck contain plentiful derivations, proofs, images, and context on the back so that you know where formulas come from
- Every card is color-coded and math is written in MathJax
- Every card includes a link to and is tagged by their lesson # in the MITx - 8.02x Electricity and Magnetism course.
The cards in this deck work with the Clickable Tags addon. - All cards are ordered so that material that comes earlier in the course shows up as new cards before material that comes later
π Contents of the deck, organized by tags π:
βοΈ Prerequisites for the course and deck π:
- Physics Mechanics (force, energy, work, power)
- Calculus
- Multivariable calculus
- Vectors and vector products
- Double integrals are just integrals but for areas
- Line integrals are just normal integrals but for a line in 3d
- Surface integrals are just double integrals but for an area in 3d (which is called a surface)
- Gradient is just derivatives but in more dimensions
- Knowing the Divergence theorem and Stokes's theorem (only for lesson 35)
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Sample Data
| Text | The {{c2::electric potential difference between point \(A\) and point \(B\) within the electric field produced by a source charge \(q_s\)::what voltage}} is{{c2::\(\Delta V_{A,B}\)}} \(=\) {{c1::\(kq_s\left(\Large\frac{1}{r_B}-\frac{1}{r_A}\right)\)}} |
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| Back Extra | Where \(r_A\) and \(r_B\) are the distances that points \(A\) and \(B\) are from the charge \(q_s\)Since the electromagnetic force is a conservative force, the path taken doesn't matter. \[\Delta V_{A,B}=-\int_A^B \vec{E}\cdot d\vec{s}\] \[\Delta V_{A,B}=-\int_A^B k\frac{q_s}{r^2} \hat{r}_{s,t}\cdot d\vec{s}=-\int_A^B k\frac{q_s}{r^2} \hat{r}_{s,t}\cdot(dr\,\hat{r}+d\theta\,\hat\theta) =-\int_A^B k\frac{q_s}{r^2} dr\]\[\Delta V_{A,B}=k\frac{q_s}{r}\Big]^B_A=kq_s \Big(\frac{1}{r_B}-\frac{1}{r_A}\Big)\]
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| Text | In an AC circuit, the impedance \(Z\) \(=|Z|e^{i\delta}\) of a capacitor \(C\):\[Z\]\[|Z|\]\[\delta\]{{c1::\[\frac{-i\,\,}{\omega C}\]}}{{c1::\[\frac{1}{\omega C}\]}}{{c1::\[-\pi/2\]}} |
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| Summary 1 | Derivation |
| Back Extra | Let \(V=V_0\cos(\omega t)=V_0\,\text{Re}(e^{i\omega t} )\)and \(I=I_0\cos(\omega t-\delta)=I_0 \,\text{Re}(e^{i(\omega t-\phi)} )\)We know that:\(\begin{align}Q&=CV\\I_c &= C \frac{dV_c}{dt}\\I_c &= CV_0 \left(\frac{d}{dt}e^{i\omega t}\right)\\I_c &= i\omega C\,V_0 e^{i\omega t}\end{align}\)Then we find \(\begin{align}Z&=\frac{V_c}{I_c}\\&=\frac{V_0 e^{i\omega t} }{i\omega C\,V_0 e^{i\omega t}}\\&=\frac{1}{i\omega C}=\frac{-i}{\omega C}\end{align}\) |
| Summary 2 | Why current \(I_c=V_c/Z\) leads |
| Back Extra 2 | We see that when \(\Delta V\) across the capacitor is maximized, \(|Q|=C|\Delta V|\) is also maximized so the capacitor acts as an open circuit and \(I=0\).When voltage decreases after, \(|Q|=C|\Delta V|\) also decreases and the stored charge leaves the capacitor in the opposite direction, creating a negative \(I\).When voltage has decreased to \(0\), \(|I|\) is maximized as all the stored charge in the capacitor has been discharged as current \(\frac{dQ}{dt}\).However as voltage continues to become more negative, the capacitor stores up \(|Q|=C|\Delta V|\) on the other side and the magnitude of current \(|I|\) decreases.When \(\Delta V_C\) is minimized, \(|Q|=C|\Delta V|\) is maximized so the capacitor acts as an open circuit and \(I=0\).This repeats as the voltage alternates.
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| Summary 3 | Reference |
| Back Extra 3 | \[Z\]\[|Z|\]\[\delta\] resistor \(R\) \[R\]\[R\]\[0\] inductor \(L\) \[i\omega L\]\[\omega L\]\[\pi/2\] capacitor \(C\) \[\frac{-i\,\,}{\omega C}\]\[\frac{1}{\omega C}\]\[-\pi/2\]Where \(\omega\) is the driving frequency of the alternating voltage source. |
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| Back Extra 4 | Impedance \(Z=|Z|e^{i\delta}\) is the "effective resistance (\(\Omega\))" in AC circuits.Substituting the functions we get that \(V_0 e^{i \omega t}=|Z|e^{i\delta}\,I_0 e^{i (\omega t-\phi)}\quad\implies \quad V_0 e^{i \omega t}=|Z|I_0 e^{i (\omega t-\phi+\delta)}\)so \(V_0=|Z|I_0\) and \(\phi=\delta\)Where the phase shift \(\phi\) represents how many radians the current "lags" behind the voltage (\(2\pi\) rad is one entire cycle).
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| Text | In terms of the magnetic field, the {{c2::intensity of a planar EM wave::what}} is{{c2::\(I\) \(=\langle S \rangle\)::symbol}} \(=\) {{c1::\(\Large \frac{cB_0^2}{2\mu_0}\)::in terms of \(B\) only}} |
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| Back Extra | \(\vec{S}=\Large \frac{\vec{E} \,\times\, \vec{B} }{\mu_0}\)Using the identity \({\Large\frac EB}=c\), we find that these are all equal \(\begin{align}I=\langle S \rangle =\frac{E_0 B_0}{2\mu_0}=\frac{E_0^2}{2c\mu_0}=\frac{cB_0^2}{2\mu_0}\end{align}\)Then, we also have \(\begin{align}\left \langle \frac{dU}{dV} \right\rangle=\frac{I}{c}=\frac{\langle S \rangle}{c}\end{align}\) |
| Summary 2 | Poynting vector \(\vec S\) defined in terms of \(\text{Power}_{out}\) |
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 \[\begin{align}\text{Power}_{out}&=-\frac d{dt}\text{Energy stored in fields}-\text{Rate of the fields' work on charges (lost to other forms of energy)}\\&=-\frac d{dt}\iiint dU-(\vec{F}_{\text{E field},\, dq}\cdot \ \vec{v}_{dq}+\vec{F}_{\text{B field},\, dq}\cdot \vec{v}_{dq})\\&=-\frac d{dt}\iiint(u_E+u_B) \,dV-\left(\iiint (dq \vec{E}\cdot \vec{v}_{dq})+0\right)\\&=-\frac d{dt}\iiint(u_E+u_B) \,dV-\iiint ( \rho\, \vec{v}_{dq}\cdot\vec{E})\,dV\end{align}\] Remember that \(\vec{J}=nq\vec{v}_d=(\frac 1 A)q\vec v_d=\rho \vec v_d\)\[\text{Power}_{out}=-\frac d{dt}\iiint(u_E+u_B) \,dV-\iiint ( \vec J\cdot\vec{E})\,dV\]And the Poynting vector \(\vec{S}\) which represents the rate of energy flow per unit area (power per unit area) is defined so\(\text{Power}_{out}=\) \(-\frac d{dt}\iiint(u_E+u_B) \,dV-\iiint ( \vec J\cdot\vec{E})\,dV=\) \({\Large\unicode{x222f} }_{\!S}\,\,\vec S\cdot \hat n_{out} \,da \)
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| Summary 3 | Poynting vector \(\vec S\) defined in terms of \(\vec E\) and \(\vec B\) |
| Back Extra 3 | The magnitude of the Poynting vector \(S\) \(=\) \(\frac{dU}{A\,dt}=\frac{(u_E+u_B)dV}{A\,(dx/c)}=\frac{(\frac12 \varepsilon_0 E^2+\frac{B^2}{2\mu_0})\cdot(A\,dx)}{A\,(dx/c)}=\frac c2(\varepsilon_0E^2+\frac{B^2}{\mu_0})\)Substitute \(E=cB,\quad B=\frac Ec,\quad\)and \(c^2=\frac 1{\mu_0\varepsilon_0}\):\(S=\frac c2(\varepsilon_0E^2+\frac{B^2}{\mu_0})=\frac c2(c\varepsilon_0EB+\frac{EB}{c\mu_0})=\frac{EBc^2}2(\varepsilon_0+\frac{1}{c^2\mu_0})=\frac{EB}{2(\mu_0 \varepsilon_0)}(2\varepsilon_0)=\frac{EB}{\mu_0}\)Assigning the direction as the direction of propagation (like of EM waves), we get that the Poynting vector \(\vec{S}\) \(=\) \(\frac{\vec E \times \vec B}{\mu_0}\)
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| Summary 4 | Light intensity |
| Back Extra 4 | We define the intensity of EM-waves \(I\) as the time average of the magnitude of the poynting vector \(I=\langle S \rangle\) \(= \frac{\langle EB\rangle}{\mu_0}=\frac{\langle\,(E_0\sin(kx- \omega t))(B_0\sin(kx- \omega t))\,\rangle}{\mu_0}=\frac{E_0 B_0}{\mu_0}\langle \sin^2(kx- \omega t)\rangle\)\(\sin^2\) has a graph like this whose time average is just half of the amplitude, so intensity \(I=\langle S \rangle\) \(=\) \(\frac{E_0 B_0}{2\mu_0}\)
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| Summary 5 | Energy density |
| Back Extra 5 | To relate intensity to the energy density we find that \[\left\langle \frac{dU}{dV} \right\rangle=\langle u\rangle=\langle u_E+u_B\rangle=\left\langle \frac12 \varepsilon_0 E^2+\frac{B^2}{2\mu_0}\right\rangle=\left\langle \frac12 \varepsilon_0 EBc+\frac{EB}{2\mu_0c}\right\rangle=\left\langle \frac{EB}2 \right\rangle \left(\varepsilon_0 c+\frac{1}{\mu_0c}\right)=\left( \frac{E_0B_0}4 \right) \left(\frac{\mu_0\varepsilon_0 c^2+1}{\mu_0c}\right)=\left( \frac{E_0B_0}4 \right) \left(\frac{2}{\mu_0c}\right)=\frac{E_0 B_0}{2c\mu_0}=\frac{I}{c}\]So time averaged energy density \(\left\langle \frac{dU}{dV} \right\rangle\) \(=\) \(\frac Ic\quad\) and \(I=c\left\langle \frac{dU}{dV} \right\rangle=c\langle u \rangle\) |
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