diff --git a/docs/physics/electrodynamics/9-conservation-laws.md b/docs/physics/electrodynamics/9-conservation-laws.md index 81741ad..f171433 100644 --- a/docs/physics/electrodynamics/9-conservation-laws.md +++ b/docs/physics/electrodynamics/9-conservation-laws.md @@ -53,9 +53,9 @@ If the dyadics are symmetric, the energy functions uniquely specify the energy i Consider a long coaxial cable, bridged by a constant voltage $V$ on one side and a resistor $R$ on the other. Then, we know that between the conductors, $\vb{H} = \frac{R}{2\pi s R}\vb{\varphi}$. Additionally, we know that -$$V=\int_a^b \vb{E} \vdot d\vb{l} = \frac{Q}{2\pi \ell \epsilon_0} \ln(\frac{b}{a}) \Rightarrow \frac{Q}{\ell} = \frac{2\pi\epsilon_0}{\ln(\frac{b}{a})} V$$ +$$V=\int_a^b \vb{E} \vdot d\vb{l} = \frac{Q}{2\pi \ell \varepsilon_0} \ln(\frac{b}{a}) \Rightarrow \frac{Q}{\ell} = \frac{2\pi\varepsilon_0}{\ln(\frac{b}{a})} V$$ -This then implies that $\vb{E} = \frac{V}{\ln(\frac{b}{a})s} \vu(s)$. We can then solve for both the energy density and Poynting vector, as well as $\vb{v} = \vb{S} / u$, the speed at which energy moves through the cable. With the impedance for a coaxial cable $Z_{C0} = \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{\ln(\frac{b}{a})}{2\pi}$, we see that +This then implies that $\vb{E} = \frac{V}{\ln(\frac{b}{a})s} \vu(s)$. We can then solve for both the energy density and Poynting vector, as well as $\vb{v} = \vb{S} / u$, the speed at which energy moves through the cable. With the impedance for a coaxial cable $Z_{C0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} \frac{\ln(\frac{b}{a})}{2\pi}$, we see that $$\vb{v} = \frac{2c \vu{z}}{\frac{R}{Z_{C0}} + \frac{Z_{C0}}{R}}$$ @@ -87,7 +87,7 @@ $$\vb{f}(\vb{r}) = \rho_e(\vb{r})\vb{E}(\vb{r}) + \vb{J}_e(\vb{r}) \times \vb{B} Substituting in Maxwell's Equations, we see that -$$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \epsilon_0 (\div \vb{E})\vb{E} + (\curl \vb{E})\times\vb{D} + \mu_0(\div \vb{H})\vb{H} + (\curl \vb{H})\times\vb{B}$$ +$$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \varepsilon_0 (\div \vb{E})\vb{E} + (\curl \vb{E})\times\vb{D} + \mu_0(\div \vb{H})\vb{H} + (\curl \vb{H})\times\vb{B}$$ Now, we claim that the right-hand side is the divergence of some tensor $\overleftrightarrow{\vb{T}}$, so that @@ -96,20 +96,20 @@ $$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \div \ove This tensor is the Maxwell Stress Tensor. We claim that the divergence of this tensor is composed of both an electric and magnetic part, so that $\div \overleftrightarrow{\vb{T}} = \div \overleftrightarrow{\vb{T}}_e + \div \overleftrightarrow{\vb{T}}_m$. Then, we can state $$\begin{align} -\div \overleftrightarrow{\vb{T}}_e &= \epsilon_0 [(\div \vb{E})\vb{E} + (\curl \vb{E})\times \vb{E}] \\ -\div \overleftrightarrow{\vb{T}}_m &= \epsilon_0 [(\div \vb{H})\vb{H} + (\curl \vb{H})\times \vb{H}] +\div \overleftrightarrow{\vb{T}}_e &= \varepsilon_0 [(\div \vb{E})\vb{E} + (\curl \vb{E})\times \vb{E}] \\ +\div \overleftrightarrow{\vb{T}}_m &= \varepsilon_0 [(\div \vb{H})\vb{H} + (\curl \vb{H})\times \vb{H}] \end{align}$$ We know that $\div(\vb{EE}) = (\div \vb{E})\vb{E} + (\vb{E} \vdot \nabla)\vb{E}$ and $\div(\overleftrightarrow{\vb{I}}f) = \grad f$. If we let $f = \frac{1}{2}\vb{E} \vdot \vb{E}$, we see that $\grad(\frac{1}{2}\vb{E} \vdot \vb{E}) = (\vb{E} \vdot \nabla)\vb{E} + (\curl \vb{E})\vb{E}$. Then, we see that $$\begin{align} -\overleftrightarrow{\vb{T}}_e &= \epsilon_0 \vb{EE} - \frac{\epsilon_0}{2} \overleftrightarrow{\vb{I}}(\vb{E} \vdot \vb{E}) \\ +\overleftrightarrow{\vb{T}}_e &= \varepsilon_0 \vb{EE} - \frac{\varepsilon_0}{2} \overleftrightarrow{\vb{I}}(\vb{E} \vdot \vb{E}) \\ \overleftrightarrow{\vb{T}}_m &= \mu_0 \vb{HH} - \frac{\mu_0}{2} \overleftrightarrow{\vb{I}}(\vb{H} \vdot \vb{H}) \end{align}$$ -Knowing that $\overleftrightarrow{\vb{T}} = \overleftrightarrow{\vb{T}}_e + \overleftrightarrow{\vb{T}}_m$, and that $u = \frac{1}{2}(\epsilon_0 E^2 + \mu_0 H^2)$ is the energy density of the electromagnetic fields in a vacuum, +Knowing that $\overleftrightarrow{\vb{T}} = \overleftrightarrow{\vb{T}}_e + \overleftrightarrow{\vb{T}}_m$, and that $u = \frac{1}{2}(\varepsilon_0 E^2 + \mu_0 H^2)$ is the energy density of the electromagnetic fields in a vacuum, -$$\overleftrightarrow{\vb{T}} = \epsilon_0 \vb{EE} + \mu_0 \vb{HH} - \overleftrightarrow{\vb{I}}u$$ +$$\overleftrightarrow{\vb{T}} = \varepsilon_0 \vb{EE} + \mu_0 \vb{HH} - \overleftrightarrow{\vb{I}}u$$ Additionally, we denote the time rate of change of the momentum density of the electromagnetic fields as $\vb{g}(\vb{r}) = \vb{D}(\vb{r}) \times \vb{B}(\vb{r})$. Thus, diff --git a/docs/physics/electrostatics/2-coulomb.md b/docs/physics/electrostatics/2-coulomb.md index 6d936ba..8f3edde 100644 --- a/docs/physics/electrostatics/2-coulomb.md +++ b/docs/physics/electrostatics/2-coulomb.md @@ -8,7 +8,7 @@ Thus, according to Coulomb's Law, $$ \begin{align} - F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\ + F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\ F^m_{qQ}(\vb{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \end{align} $$ @@ -17,7 +17,7 @@ Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\v $$ \begin{align} - E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\ + E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\ H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \end{align} $$ @@ -26,7 +26,7 @@ Now, let $Q$ be at point $\vb{r'}$. Then, the unit vector becomes $\frac{\vb{r}- $$ \begin{align} - E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\ + E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\ H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \end{align} $$ @@ -35,7 +35,7 @@ With multiple charges, we can apply the *superposition principal* to see the fol $$ \begin{align} - E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\ + E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\ H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \end{align} $$ @@ -44,7 +44,7 @@ We can convert this to an integral as $N$ goes to infinity. $$ \begin{align} - E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\ + E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\ H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \end{align} $$ @@ -61,7 +61,7 @@ Now, apply the divergence operator over $\vb{r}$ to the electrostatic and magnet $$ \begin{align} - \div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\ + \div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\ \div{H(\vb{r})} &= \div{(\frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \end{align} $$ @@ -70,9 +70,9 @@ As the divergence operator does not operate on $\vb{r'}$, we see that $$ \begin{align} - \div{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\ - &= \frac{1}{4 \pi \epsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\ - &= \frac{\rho_e(\vb{r})}{\epsilon_0} \\ + \div{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\ + &= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\ + &= \frac{\rho_e(\vb{r})}{\varepsilon_0} \\ \div{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\ &= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\ &= \frac{\rho_m(\vb{r})}{\mu_0} @@ -83,7 +83,7 @@ The curl of an electrostatic or magnetostatic is relatively simple. $$ \begin{align} - \curl{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\ + \curl{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\ \curl{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\ \end{align} $$ @@ -100,30 +100,30 @@ We can verify that $\curl{(\vb{r}-\vb{r'})} = 0$, cancelling the first term. Add ## Section 2.4 - Electric and Magnetic Flux Densities -The electric and magnetic flux density vectors are given by $\epsilon_0 \vb{E}$ and $\mu_0 \vb{H}$. +The electric and magnetic flux density vectors are given by $\varepsilon_0 \vb{E}$ and $\mu_0 \vb{H}$. Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following: $$ -\Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m +\Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m $$ Thus, applying divergence theorem, $$ -Q_e = \Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{E}} \dd V +Q_e = \Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{E}} \dd V $$ $$ -Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{H}} \dd V +Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{H}} \dd V $$ Since $Q_e = \int_V \rho_e \dd V$ and $Q_m = \int_V \rho_m \dd V$, we see that $$ \begin{align} - \int_V \rho_e \dd V &= \epsilon_0 \int_V \div{\vb{E}} \dd V \\ - \rho_e &= \epsilon_0 \int_V \div{\vb{E}} \dd V \\ + \int_V \rho_e \dd V &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\ + \rho_e &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\ \int_V \rho_m \dd V &= \mu_0 \int_V \div{\vb{H}} \dd V \\ \rho_m &= \mu_0 \int_V \div{\vb{H}} \dd V \\ \end{align} diff --git a/docs/physics/electrostatics/3-electro-magnetic-potentials.md b/docs/physics/electrostatics/3-electro-magnetic-potentials.md index 1818561..bfa9dea 100644 --- a/docs/physics/electrostatics/3-electro-magnetic-potentials.md +++ b/docs/physics/electrostatics/3-electro-magnetic-potentials.md @@ -31,7 +31,7 @@ $$ Strictly speaking, this is a potential difference. To find the absolute potential, assume a point charge $Q$ at the origin, and a charge $q$. We take the work as $q$ moves from $\vb{r'} = \vb{\infty}$ to $\vb{r'} = \vb{r}$. Thus, $$ -W = -q_e \frac{Q_e}{4 \pi \epsilon_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_e \frac{Q_e}{4 \pi \epsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \epsilon_0} \frac{1}{r} +W = -q_e \frac{Q_e}{4 \pi \varepsilon_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_e \frac{Q_e}{4 \pi \varepsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \varepsilon_0} \frac{1}{r} $$ $$ @@ -41,25 +41,25 @@ $$ Letting the potential as $\vb{r} \rightarrow \infty$ equal $0$ be our reference and dividing out $q$, we find that the voltage for arrangement is the following: $$ -V_e(\vb{r}) = \frac{Q_e}{4 \pi \epsilon_0 r} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 r} +V_e(\vb{r}) = \frac{Q_e}{4 \pi \varepsilon_0 r} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 r} $$ Now, if we let the stationary charge $Q$ be located at $\vb{r'}$, we see that $$ -V_e(\vb{r}) = \frac{Q_e}{4 \pi \epsilon_0 \abs{\vb{r} - \vb{r'}}} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 \abs{\vb{r} - \vb{r'}}} +V_e(\vb{r}) = \frac{Q_e}{4 \pi \varepsilon_0 \abs{\vb{r} - \vb{r'}}} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 \abs{\vb{r} - \vb{r'}}} $$ If we allow multiple charges, this becomes $$ -V_e(\vb{r}) = \frac{1}{4\pi \epsilon_0} \sum_{i=1}^N \frac{Q_ei}{\abs{\vb{r}-\vb{r_i}}} +V_e(\vb{r}) = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^N \frac{Q_ei}{\abs{\vb{r}-\vb{r_i}}} $$ Taking this to its natural limit, $$ -V_e(\vb{r}) = \frac{1}{4 \pi \epsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'} +V_e(\vb{r}) = \frac{1}{4 \pi \varepsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'} $$ $$ @@ -71,25 +71,25 @@ $$ Given two point charges $Q_{e1}, Q_{e2}$ we know the work to bring them together is $$ -W_2 = W_{21} = \frac{1}{4 \pi \epsilon_0} \frac{Q_{e1} Q_{e2}}{\abs{\vb{r_2} - \vb{r_1}}} +W_2 = W_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_{e1} Q_{e2}}{\abs{\vb{r_2} - \vb{r_1}}} $$ Superposition applies here. The energy to create $N$ charges is $$ -W_n = \frac{1}{2} \frac{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}} +W_n = \frac{1}{2} \frac{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}} $$ For the sake of symmetry, sum overall charges and divide by 2. $$ -W_n = \frac{1}{2} \frac{1}{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}} +W_n = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}} $$ Rearranging, we see the following: $$ -W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \epsilon_0} \frac{Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}} +W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \varepsilon_0} \frac{Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}} = \frac{1}{2}\sum_{i = 1}^{N} Q_{ei} V(\vb{r_i}) $$ @@ -103,7 +103,7 @@ $$ We can also express this as $$ -W_e = \frac{1}{2} \frac{1}{4 \pi \epsilon_0} \int_V \int_{V'} \frac{\rho_e(\vb{r})\rho_e(\vb{r'})}{\abs{\vb{r} - \vb{r'}}} \dd{V'} \dd{V} +W_e = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \int_V \int_{V'} \frac{\rho_e(\vb{r})\rho_e(\vb{r'})}{\abs{\vb{r} - \vb{r'}}} \dd{V'} \dd{V} $$ $$ @@ -115,13 +115,13 @@ Note the $\frac{1}{2}$ is the same anti-double-counting factor introduced previo We can now write an expression for energy of a charge density in terms of the field that it produces. $$ -W = \frac{\epsilon_0}{2} \int_V (\div{\vb{E}(\vb{r})}) V(\vb{r}) \dd V +W = \frac{\varepsilon_0}{2} \int_V (\div{\vb{E}(\vb{r})}) V(\vb{r}) \dd V $$ Simplifying, we see that $$ -W_e = \frac{\epsilon_0}{2} \int_{V} E^2(\vb{r}) \dd V ; \quad +W_e = \frac{\varepsilon_0}{2} \int_{V} E^2(\vb{r}) \dd V ; \quad W_m = \frac{\mu_0}{2} \int_{V} H^2(\vb{r}) \dd V $$ @@ -132,7 +132,7 @@ We know that $\vb{E}(\vb{r}) = -\div{V_e(\vb{r})}$ and $\vb{H}(\vb{r}) = -\div{V Combined this, as well as the first of the Maxwell equations, we see that $$ -\div{\vb{E}} = -\div{\grad{V_e}} = - \laplacian{V_e} = \frac{\rho_e}{\epsilon_0} +\div{\vb{E}} = -\div{\grad{V_e}} = - \laplacian{V_e} = \frac{\rho_e}{\varepsilon_0} $$ $$ @@ -150,13 +150,13 @@ $$ Now, we can construct a potential function in terms of said green function that satisfies the Laplace equation. $$ -V_e(\vb{r}) = - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\epsilon_0} \dd{V'} +V_e(\vb{r}) = - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\varepsilon_0} \dd{V'} $$ This is the specific solution. Let $\psi(\vb{r})$ be a solution to the homogenous equation. We can state the following: $$ -V_e(\vb{r}) = \psi(\vb{r}) - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\epsilon_0} \dd{V'} +V_e(\vb{r}) = \psi(\vb{r}) - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\varepsilon_0} \dd{V'} $$ We will consider the potential of a point charge. THat is, the limit of potential is zero as distance approaches infinity. @@ -164,20 +164,20 @@ We will consider the potential of a point charge. THat is, the limit of potentia Recall the potential of a point charge: $$ -V_e(\vb{r}) = \frac{Q_e}{\epsilon_0} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} +V_e(\vb{r}) = \frac{Q_e}{\varepsilon_0} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} $$ We know that $- \laplacian{V(\vb{r})} = \div{\vb{E}(\vb{r})}$. Thus, recall the electric field of a point charge. $$ -\vb{E}(\vb{r}) = -\grad{V(\vb{r})} = \frac{Q_e}{\epsilon_0} \grad({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) = \frac{Q_e}{\epsilon_0} \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} +\vb{E}(\vb{r}) = -\grad{V(\vb{r})} = \frac{Q_e}{\varepsilon_0} \grad({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) = \frac{Q_e}{\varepsilon_0} \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} $$ Taking the divergence, we find that $$ -- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\epsilon_0} = \frac{Q_e}{\epsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) -= \frac{Q_e}{\epsilon_0} \div \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} = \frac{Q_e}{\epsilon_0} \delta(\vb{r} - \vb{r'}) +- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\varepsilon_0} = \frac{Q_e}{\varepsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) += \frac{Q_e}{\varepsilon_0} \div \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} = \frac{Q_e}{\varepsilon_0} \delta(\vb{r} - \vb{r'}) $$ Thus, we see that @@ -189,7 +189,7 @@ $$ Finally, $$ -V_e(\vb{r}) = \int_{V'} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} \frac{\rho_e}{\epsilon_0} \dd{V'} +V_e(\vb{r}) = \int_{V'} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} \frac{\rho_e}{\varepsilon_0} \dd{V'} $$ $$ @@ -230,9 +230,9 @@ $$ Applying this to the equation for voltage, we see that $$ -V_e(r) = \frac{1}{\epsilon_0} \int_{V'} G(\vb{r}, \vb{r'}) p_e(\vb{r'}) \dd{V'} \approx -\frac{1}{4 \pi \epsilon_0 r} \int_{V'} (1 + \frac{\vu{r} \vdot \vb{r'}}{r}) p_e(\vb{r'}) \dd{V'} -= \frac{Q_e}{4 \pi \epsilon_0 r} + \frac{\vu{r} \vdot \vb{p}}{4 \pi \epsilon_0 r^2} +V_e(r) = \frac{1}{\varepsilon_0} \int_{V'} G(\vb{r}, \vb{r'}) p_e(\vb{r'}) \dd{V'} \approx +\frac{1}{4 \pi \varepsilon_0 r} \int_{V'} (1 + \frac{\vu{r} \vdot \vb{r'}}{r}) p_e(\vb{r'}) \dd{V'} += \frac{Q_e}{4 \pi \varepsilon_0 r} + \frac{\vu{r} \vdot \vb{p}}{4 \pi \varepsilon_0 r^2} $$ **Definition**. The first and second terms of this equation are the *monopole* and *dipole* terms respectively. diff --git a/docs/physics/electrostatics/4-conductors.md b/docs/physics/electrostatics/4-conductors.md index 70cdba9..3425b9a 100644 --- a/docs/physics/electrostatics/4-conductors.md +++ b/docs/physics/electrostatics/4-conductors.md @@ -16,7 +16,7 @@ Consider any two points internal to the conductor. The voltage between said poin The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $\dd{\vb{l}}$. Now, $\vb{E} \vdot \dd{\vb{l}} = \vb{E}_s \vdot \dd{\vb{l}}_s + \vb{E}_p \vdot \dd{\vb{l}}_p = \dd{V_s} + \dd{V_p}$, in terms of parallel and perpendicular components. The parallel voltage difference is zero, so the electric field must be zero. -Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\epsilon_0} = \frac{A\sigma_e}{\epsilon_0}$. Thus, $\sigma_e = \epsilon_0 E$. +Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\varepsilon_0} = \frac{A\sigma_e}{\varepsilon_0}$. Thus, $\sigma_e = \varepsilon_0 E$. ## Section 4.3 - Exercises involving conductors at fixed potentials @@ -48,19 +48,19 @@ This field is zero outside of a material, and if non-zero inside a material, wil **Definition**. Charges not bound are called *free*, with density denoted as $\rho_{ef}$. Combined with $\rho_{eb}$, they form the basis of the electric field. THat is, $$ -\epsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb} +\varepsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb} $$ **Definition**. The electric flux density field $\vb{D}$ is defined as $$ -\vb{D} = \epsilon_0 \vb{E} + \vb{P} +\vb{D} = \varepsilon_0 \vb{E} + \vb{P} $$ Both $\vb{D}$ and $\vb{P}$ have units of Coulombs/m^2. Additionally, we see that $$ -\div \vb{D} = \div (\epsilon_0 \vb{E} + \vb{P}) = \div \epsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef} +\div \vb{D} = \div (\varepsilon_0 \vb{E} + \vb{P}) = \div \varepsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef} $$ ## Section 4.5 - Induced Electric Charges, their Potentials and Fields @@ -75,7 +75,7 @@ $$ C := \frac{Q}{V} $$ -This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \epsilon_0 R$. For a parallel plate capacitor, this reduces to $C = \frac{epsilon_0 A}{d}$. +This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \varepsilon_0 R$. For a parallel plate capacitor, this reduces to $C = \frac{epsilon_0 A}{d}$. ## Section 4.7 - Forces on Charged Conductors in Electric Fields diff --git a/docs/physics/mechanics/2-projectiles-charged-particles.md b/docs/physics/mechanics/2-projectiles-charged-particles.md new file mode 100644 index 0000000..676bfcd --- /dev/null +++ b/docs/physics/mechanics/2-projectiles-charged-particles.md @@ -0,0 +1,20 @@ +# Chapter 2 - Projectiles and Charged Particles + +## Section 2.1 - Air Resistance + +**Definition**. The *drag*, or resistive force on an object due to the atmosphere, is denoted as $\vb{f}$. Note that this is **not** the force density, but the overall force. In most cases, this force directly opposes the direction of motion. If not, the other component is known as *lift*, however this is mostly negligible. + +We define air resistance as $\vb{f} = -f(v) \vu{v}$. We consider two types in this text: linear, where $f(v) = f_{lin} = bv$, and quadratic, where $f(v) = f_{quad} = cv^2$. Note that often times we consider both, and state that $f(v) = f_{lin} + f_{quad} = bv + cv^2$. + +The linear term comes from viscous drag and is generally proportional to the viscosity of the medium, while quadratic drag tends to arise from the particle needing to accelerate the mass of air which it is constantly colliding against. + +In some cases, we can calculate these coefficients. With $D$ as the diameter of a spherical object, and $\beta$ and $\gamma$ as properties of the medium, we can state that $b = \beta D$ and $c = \gamma D^2$. In air at STP, $\beta = 1.6 \times 10^{-4} \text{N} \vdot \text{s}/\text{m}^2$, and $\gamma = 0.25 \text{N} \vdot \text{s}^2/\text{m}^4$. + +Oftentimes, one factor is far more impactful than the other, meaning that the smaller of the two may be neglected. To do so, compute the following ratio: + +$$\frac{f_{quad}}{f_{lin}} = \frac{cv^2}{bv} = \frac{\gamma D}{\beta}v$$ + +Note that the result is expected to be of the same order of magnitude as the *Reynolds number* $R = Dv \mathcal{Q}/\eta$, where $\mathcal{Q}$ is the density of the medium and $\eta$ the viscosity. + +## Section 2.2 - Linear Air Resistance +