Line 6: Line 6:
  
 
<math>\bar{p} = q\bar{d}</math>
 
<math>\bar{p} = q\bar{d}</math>
 +
 
<math>V_1 - V_2 = -\int_1^2\bar{E}\cdot d\bar{l} = 10,000\hspace{2cm} D= \epsilon\bar{E} = \epsilon_0\bar{E} + \bar{p}</math>
 
<math>V_1 - V_2 = -\int_1^2\bar{E}\cdot d\bar{l} = 10,000\hspace{2cm} D= \epsilon\bar{E} = \epsilon_0\bar{E} + \bar{p}</math>
 +
 
<math>-E_zl = 10,000</math>
 
<math>-E_zl = 10,000</math>
 +
 
<math>\bar{E} = - \frac{10,000}{l}\hat{z}</math>
 
<math>\bar{E} = - \frac{10,000}{l}\hat{z}</math>
 +
 
<math>\bar{F} = - \frac{10,000q}{l}\hat{z} </math>
 
<math>\bar{F} = - \frac{10,000q}{l}\hat{z} </math>
 
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2)
 
2)
  
[[Image:A1FO32013.png|Alt text|550x232px]]
+
[[Image:A2FO32013.png|Alt text|550x232px]]
  
 
<math>P = VI = I^2R = \frac{V^2}{R}</math>
 
<math>P = VI = I^2R = \frac{V^2}{R}</math>
 +
 
<math>F = q(\bar{E} + \bar{\mu} \times\bar{B} ) = 0</math>
 
<math>F = q(\bar{E} + \bar{\mu} \times\bar{B} ) = 0</math>
 +
 
<math>\bar{E} = \frac{-\bar{\mu}\times\bar{B}}{q} = \frac{-\bar{v}\times\bar{B}}{q}\leftarrow\hat{z}</math>
 
<math>\bar{E} = \frac{-\bar{\mu}\times\bar{B}}{q} = \frac{-\bar{v}\times\bar{B}}{q}\leftarrow\hat{z}</math>
 +
 
<math>\nabla\times\bar{H} = j\omega\epsilon\bar{E}+\bar{J}</math>
 
<math>\nabla\times\bar{H} = j\omega\epsilon\bar{E}+\bar{J}</math>
 +
 
<math>V_1 - V_2 = -\int_1^2\bar{E}\cdot d\bar{l} = \frac{-d}{q}|\bar{v}\times\bar{B}|^2</math>
 
<math>V_1 - V_2 = -\int_1^2\bar{E}\cdot d\bar{l} = \frac{-d}{q}|\bar{v}\times\bar{B}|^2</math>
 +
 
<math>p = \frac{d^2}{q^2}|\bar{v}\times\bar{B}|</math>
 
<math>p = \frac{d^2}{q^2}|\bar{v}\times\bar{B}|</math>

Revision as of 16:49, 17 June 2017

1)

$ F = q (\bar{E} + \cancelto{0}{\bar{\mu}\times\bar{B}} ) = qE $

Alt text

$ \bar{p} = q\bar{d} $

$ V_1 - V_2 = -\int_1^2\bar{E}\cdot d\bar{l} = 10,000\hspace{2cm} D= \epsilon\bar{E} = \epsilon_0\bar{E} + \bar{p} $

$ -E_zl = 10,000 $

$ \bar{E} = - \frac{10,000}{l}\hat{z} $

$ \bar{F} = - \frac{10,000q}{l}\hat{z} $


\\ 2)

Alt text

$ P = VI = I^2R = \frac{V^2}{R} $

$ F = q(\bar{E} + \bar{\mu} \times\bar{B} ) = 0 $

$ \bar{E} = \frac{-\bar{\mu}\times\bar{B}}{q} = \frac{-\bar{v}\times\bar{B}}{q}\leftarrow\hat{z} $

$ \nabla\times\bar{H} = j\omega\epsilon\bar{E}+\bar{J} $

$ V_1 - V_2 = -\int_1^2\bar{E}\cdot d\bar{l} = \frac{-d}{q}|\bar{v}\times\bar{B}|^2 $

$ p = \frac{d^2}{q^2}|\bar{v}\times\bar{B}| $

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett