# on 17-Nov-2023 (Fri)

#### Annotation 7597276663052

 Divergence of gradient Divergence of gradient: \begin{align}\nabla\cdot(\nabla T)=\frac{\partial^{2} T}{\partial x^{2}} +\frac{\partial^{2} T}{\partial y^{2}} +\frac{\partial^{2} T}{\partial z^{2}}\end{align}

#### Flashcard 7597278498060

Question

\begin{align}\nabla\cdot(\nabla T)=\frac{\partial^{2} T}{\partial x^{2}} +\frac{\partial^{2} T}{\partial y^{2}} +\frac{\partial^{2} T}{\partial z^{2}}\end{align}

status measured difficulty not learned 37% [default] 0

Divergence of gradient: \begin{align}\nabla\cdot(\nabla T)=\frac{\partial^{2} T}{\partial x^{2}} +\frac{\partial^{2} T}{\partial y^{2}} +\frac{\partial^{2} T}{\partial z^{2}}\end{align}

#### Annotation 7597280070924

 Curl of gradient Curl of gradient: The curl of a gradient is always zero:$$\nabla\times(\nabla T)=\mathbf{0}$$

#### Flashcard 7597282954508

Question

The curl of a gradient is always zero:$$\nabla\times(\nabla T)=\mathbf{0}$$

status measured difficulty not learned 37% [default] 0

Curl of gradient: The curl of a gradient is always zero:$$\nabla\times(\nabla T)=\mathbf{0}$$

#### Annotation 7597284527372

 Divergence of curl Divergence of curl: The divergence of a curl, like the curl of a gradient, is always zero: $$\nabla\cdot(\nabla\times T)=\mathbf{0}$$

#### Flashcard 7597287148812

Question
Divergence of curl: [...]
The divergence of a curl, like the curl of a gradient, is always zero: $$\nabla\cdot(\nabla\times T)=\mathbf{0}$$

status measured difficulty not learned 37% [default] 0

Divergence of curl
Divergence of curl: The divergence of a curl, like the curl of a gradient, is always zero: $$\nabla\cdot(\nabla\times T)=\mathbf{0}$$

#### Annotation 7597288721676

 Curl of curl Curl of curl: Curl-of-curl gives nothing new: $$\nabla\times(\nabla\times T)=\nabla(\nabla\cdot\mathbf{v})-\nabla^{2}\mathbf{v}$$

#### Flashcard 7597290032396

Question

Curl of curl: Curl-of-curl gives nothing new: [...]

$$\nabla\times(\nabla\times T)=\nabla(\nabla\cdot\mathbf{v})-\nabla^{2}\mathbf{v}$$

status measured difficulty not learned 37% [default] 0

Curl of curl
Curl of curl: Curl-of-curl gives nothing new: $$\nabla\times(\nabla\times T)=\nabla(\nabla\cdot\mathbf{v})-\nabla^{2}\mathbf{v}$$

#### Annotation 7597295799564

 Differential Form of Maxwell's Equations Differential Form of Maxwell's Equations: $$\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}$$

#### Flashcard 7597297634572

Question
Differential Form of Maxwell's Equations: [...]
$$\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}$$

status measured difficulty not learned 37% [default] 0

Differential Form of Maxwell's Equations
Differential Form of Maxwell's Equations: $$\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}$$

#### Annotation 7597299207436

 Integrational Form of Maxwell's Equations Integrational Form of Maxwell's Equations: $$\begin{array}{rrl}\\ (\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&\displaystyle\frac{1}{\varepsilon_{0}}\int_{V}\rho\ \mathrm{d}\tau\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_{0}\int_{S}\mathbf{J}\cdot\mathrm{d}\mathbf{a}+\mu_0 \epsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\ \end{array}$$

#### Flashcard 7597300518156

Question
Integrational Form of Maxwell's Equations: [...]
$$\begin{array}{rrl}\\ (\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&\displaystyle\frac{1}{\varepsilon_{0}}\int_{V}\rho\ \mathrm{d}\tau\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_{0}\int_{S}\mathbf{J}\cdot\mathrm{d}\mathbf{a}+\mu_0 \epsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\ \end{array}$$

status measured difficulty not learned 37% [default] 0

Integrational Form of Maxwell's Equations
Integrational Form of Maxwell's Equations: $$\begin{array}{rrl}\\ (\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&\displaystyle\frac{1}{\varepsilon_{0}}\int_{V}\rho\ \mathrm{d}\tau\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_{0}\int_{S}\mathbf{J}\cdot\mathrm{d}\mathbf{a}+\mu_0 \epsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\ \end{array}$$

#### Annotation 7597302615308

 Differential Form of Maxwell's Equations in vacuum Differential Form of Maxwell's Equations in Vacuum: $$\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}$$

#### Flashcard 7597304450316

Question
Differential Form of Maxwell's Equations in Vacuum: [...]
$$\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}$$

status measured difficulty not learned 37% [default] 0

Differential Form of Maxwell's Equations in vacuum
Differential Form of Maxwell's Equations in Vacuum: $$\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}$$

#### Annotation 7597306023180

 Integrational Form of Maxwell's Equations in Vacuum Integrational Form of Maxwell's Equations in Vacuum: $$\begin{array}{rrl}\\(\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\\end{array}$$

#### Flashcard 7597309955340

Question
Integrational Form of Maxwell's Equations in Vacuum: [...]

$$\begin{array}{rrl}\\(\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\\end{array}$$

status measured difficulty not learned 37% [default] 0

Integrational Form of Maxwell's Equations in Vacuum
Integrational Form of Maxwell's Equations in Vacuum: $$\begin{array}{rrl}\\(\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\\end{array}$$

#### Annotation 7597317819660

 Plane wave form of Electromagnatic wave The Plane wave form of Electromagnetic wave is $$\begin{array}{rl}\vec{E}&=&E_{0}\exp[i(kz-\omega t)]\cdot\hat{x}\\\vec{B}&=&B_{0}\exp[i(kz-\omega t)]\cdot\hat{y}\end{array}$$ where - $$E_{0}$$ is the amplitude of the electric field. - $$B_{0}$$ is the amplitude of the magnetic field. - We also have $$B_{0}=\frac{k}{\omega}E_{0}=\frac{1}{c}E_{0}=\sqrt{\mu_{0}\varepsilon_{0}}E_{0}$$ - $$k$$ is the wave number, or wave vector, representing the propagation direction. - $$\omega$$ is the angular frequency. - $$\hat{x},\hat{y}$$ is the polarization vector, representing the polarization direction.

#### Flashcard 7597319130380

Question
The plane wave form of Electromagnetic wave is [...]

$$\begin{array}{rl}\vec{E}&=&E_{0}\exp[i(kz-\omega t)]\cdot\hat{x}\\\vec{B}&=&B_{0}\exp[i(kz-\omega t)]\cdot\hat{y}\end{array}$$

where
- $$E_{0}$$ is the amplitude of the electric field.
- $$B_{0}$$ is the amplitude of the magnetic field.

- We also have $$B_{0}=\frac{k}{\omega}E_{0}=\frac{1}{c}E_{0}=\sqrt{\mu_{0}\varepsilon_{0}}E_{0}$$
- $$k$$ is the wave number, or wave vector, representing the propagation direction.
- $$\omega$$ is the angular frequency.
- $$\hat{x},\hat{y}$$ is the polarization vector, representing the polarization direction.

status measured difficulty not learned 37% [default] 0

Plane wave form of Electromagnatic wave
Plane wave form of Electromagnatic wave is $$\begin{array}{rl}\vec{E}&=&E_{0}\exp[i(kz-\omega t)]\cdot\hat{x}\\\vec{B}&=&B_{0}\exp[i(kz-\omega t)]\cdot\hat{y}\end{array}$$ where - $$E_{0}$$ is the amplitude of the electric field. - $$B_{0}$$ is the amplitude of the magnatic field. - $$k$$ is the wave number, or wave vector, represent the propagation direction. - $$\omega$$ is the angular frequency. - $$\hat{x},\hat{y}$$ is the polarization vector, represent the polarization direction.