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