1)
$ F = q (\bar{E} + \cancelto{0}{\bar{\mu}\times\bar{B}} ) = qE $
$ \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)
$ 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}| $
