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184_notes:examples:week5_flux_tilted_surface [2021/05/29 21:05] – schram45 | 184_notes:examples:week5_flux_tilted_surface [2021/06/01 15:23] – schram45 | ||
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* $\Phi_e$ | * $\Phi_e$ | ||
* $\vec{A}$ | * $\vec{A}$ | ||
- | |||
- | ===Approximations & Assumptions=== | ||
- | * The electric field is constant. | ||
- | * The surface is flat. | ||
- | * The electric flux through the surface is due only to $\vec{E}$. | ||
===Representations=== | ===Representations=== | ||
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* We represent the situation with the following diagram. Note that the top of the rectangle aligns along the $z$-direction, | * We represent the situation with the following diagram. Note that the top of the rectangle aligns along the $z$-direction, | ||
[{{ 184_notes: | [{{ 184_notes: | ||
+ | |||
+ | <WRAP TIP> | ||
+ | ===Approximations & Assumptions=== | ||
+ | * The electric field is constant. | ||
+ | * The surface is flat. | ||
+ | * The electric flux through the surface is due only to $\vec{E}$. | ||
+ | </ | ||
====Solution==== | ====Solution==== | ||
In order to find electric flux, we must first find $\vec{A}$. Remember in the [[184_notes: | In order to find electric flux, we must first find $\vec{A}$. Remember in the [[184_notes: | ||
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&= 60\text{ Vm} | &= 60\text{ Vm} | ||
\end{align*} | \end{align*} | ||
- | This makes sense cause the max flux through the surface would occur when the the surface was perpendicular to the electric field and the area vector was in the same direction as the electric field. This would result in a flux of $120Vm$. Since the surface in this case is $\theta=30^\circ$ | + | This makes sense cause the max flux through the surface would occur when the the surface was perpendicular to the electric field and the area vector was in the same direction as the electric field. This would result in a flux of $120Vm$. Since the surface in this case is $30^\circ$ away from the electric field we would expect there to be a much smaller flux due to the large angle between the area vector and the electric field. |