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Introductory Physics Homework Help
Field transformations in the z-direction
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[QUOTE="milkism, post: 6871244, member: 734134"] Okay I have tried to do it with different configurations. [B]New solution:[/B] [ATTACH type="full" alt="1680354446217.jpeg"]324324[/ATTACH] At configuration 1 (top configuration), I concluded the following: $$E_y = \frac{\sigma}{\epsilon _0}$$ $$\mathbf{K_{\pm}} = \pm \sigma v_0 \mathbf{\hat{z}}$$ $$B_x = -\mu _0 \sigma v_0$$ $$\overline{E_y} = \gamma \left( E_y - vB_x \right)$$ $$\overline{B_x} = \gamma \left( B_x - \frac{v}{c^2} E_y \right)$$ At configuration 2 (middle left), I concluded the following: $$\overline{E_z}=E_z$$ At configuration 3 (bottom left), I concluded the following: $$\overline{B_z} = B_z$$ At configuration 4 (bottom right), I concluded the following: $$E_x = \frac{\sigma}{\epsilon_0}$$ $$B_y = \mu _0 \sigma v_0$$ $$\overline{E_x} = \gamma \left( E_x + vB_y \right)$$ $$\overline{B_y} = \gamma \left( B_y + \frac{v}{c^2} E_x \right)$$ [B]Personal questions:[/B] Since in griffiths they have boosted at a negative x directio, should I also have boosted in a negative z direction? I think my second solution is more correct than my first one. [/QUOTE]
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Field transformations in the z-direction
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