Talk:MyOpenMath/Solutions/Maxwell's integral equations

https://en.wikiversity.org/w/index.php?title=Talk:MyOpenMath/Solutions/Maxwell%27s_integral_equations&oldid=2174778

Spherical

$\text{If }r>R:{\epsilon }_0\underset{4\pi r^{2}E}{\underbrace{\oint \overrightarrow{E}\cdot d\overrightarrow{A}}}=\underset{\frac{4\pi R^3}{3}{\rho }_0}{\underbrace{\int \rho d\tau }}$

$\text{If }r<R:{\epsilon }_0\underset{4\pi r^{2}E}{\underbrace{\oint \overrightarrow{E}\cdot d\overrightarrow{A}}}=\underset{\frac{4}{3}\pi r^3{\rho }_0}{\underbrace{\int \rho d\tau }}$

Cylindrical

$\text{If }r>R:{\epsilon }_0\underset{2\pi rLE}{\underbrace{\oint \overrightarrow{E}\cdot d\overrightarrow{A}}}=\underset{\pi R^{2}L{\rho }_0}{\underbrace{{\rho }_0\int d\tau }}$

$\text{If }r<R:{\epsilon }_0\underset{2\pi rLE}{\underbrace{\oint \overrightarrow{E}\cdot d\overrightarrow{A}}}=\underset{\pi r^{2}L{\rho }_0}{\underbrace{{\rho }_0\int d\tau }}$

slab

${\epsilon }_0\underset{A(E_{+}-E_{-})}{\underbrace{\oint \overrightarrow{E}\cdot d\overrightarrow{A}}}=\underset{Ab{\rho }_0=A\sigma }{\underbrace{{\rho }_0\int d\tau }},$

  where: $\{\begin{array}{cc}{\overrightarrow{E}}_{+}=& E(x_0+L/2)\hat{x}\\ {\overrightarrow{E}}_{-}=& E(x_0-L/2)\hat{x}\end{array}$

$\to {\epsilon }_0\frac{\partial E_x}{\partial x}=\rho (x)$

cylindrical wire

$\text{If }r>R:\underset{2\pi r{B}_{\theta }}{\underbrace{\oint \overrightarrow{B}\cdot d\overrightarrow{l}}}={\mu }_{0}I$

$\text{If }r<R:\underset{2\pi r{B}_{\theta }}{\underbrace{\oint \overrightarrow{B}\cdot d\overrightarrow{l}}}={\mu }_0\underset{J_0\pi r^2}{\underbrace{\int J_{0}dA}}$

$\text{To find }{J}_0:I=\underset{\pi R^2{J}_0}{\underbrace{\int {\overrightarrow{J}}_0\cdot d\overrightarrow{A}}}$

Solenoid

$\underset{B_{z}L}{\underbrace{\oint \overrightarrow{B}\cdot d\overrightarrow{l}}}={\mu }_{0}NI$

Faraday

$\underset{E_{\theta }2\pi r}{\underbrace{\oint \overrightarrow{E}\cdot d\overrightarrow{l}}}=-\frac{d}{dt}\underset{BA}{\underbrace{\int \overrightarrow{B}\cdot d\overrightarrow{A}}}$

Maxwell

${\epsilon }_0\oint \overrightarrow{E}\cdot d\overrightarrow{A}=Q$

$I_{displacement}={\epsilon }_0\frac{d}{dt}\oint \overrightarrow{E}\cdot d\overrightarrow{A}$

$I_{conventional}={\int }_S{\overrightarrow{J}}_0\cdot d\overrightarrow{A}$

${\oint }_{\partial S}\overrightarrow{B}\cdot d\overrightarrow{l}=$

${\mu }_0({\int }_S\overrightarrow{J}\cdot d\overrightarrow{A}+{\epsilon }_0\frac{d}{dt}{\int }_S\overrightarrow{E}\cdot d\overrightarrow{A})$

toroid=

$\underset{B_{\theta }2\pi r}{\underbrace{\oint \overrightarrow{B}\cdot d\overrightarrow{l}}}={\mu }_{0}NI$