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

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


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

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

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

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