Differences

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--- 184_notes:examples:week5_flux_tilted_surface [2021/05/29 21:06] – schram45
+++ 184_notes:examples:week5_flux_tilted_surface [2021/06/04 00:41] (current) – schram45
@@ Line 12: / Line 12: @@
   * $\Phi_e$
   * $\vec{A}$
-===Approximations & Assumptions===
-  * The electric field is constant.
-  * The surface is flat.
-  * The electric flux through the surface is due only to $\vec{E}$.
 ===Representations===
@@ Line 23: / Line 18: @@
   * We represent the situation with the following diagram. Note that the top of the rectangle aligns along the $z$-direction, but we represent using a perspective that also shows area. The little coordinate axes drawn may be unfamiliar. The circle with the dot simply means that the $+z$-axis comes straight out of the page. A circle with an X would mean into the page. Just think of an arrow either seen head-on (all you see is the point), or seen from behind (you see the feathers).
 [{{ 184_notes:5_flux_flat_angle.png?400 |Tilted Surface}}]
+<WRAP TIP>
+===Approximations & Assumptions===
+There are a few simplifying approximations and assumptions we should make before solving this problem.
+  * The electric field is constant: Allows the electric field to be constant through our area which simplifies down the flux equation.
+  * The surface is flat: Allows all the area vectors associated with the surface to point in the same direction.
+  * The electric flux through the surface is due only to $\vec{E}$.
+</WRAP>
 ====Solution====
 In order to find electric flux, we must first find $\vec{A}$. Remember in the [[184_notes:e_flux#Area_as_a_Vector|notes on flux]] that area can be a vector when we define it as a [[184_notes:math_review#Vector_Multiplication|cross product]] of width and length vectors. Here, we can use the following for width and length, with width being the top of the rectangle, and pointing out of the page, and length being the longer side, and pointing at an upwards angle:
@@ Line 40: / Line 43: @@
         &= 60\text{ Vm}
 \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 $\30^\circ$ away from the electric field we would expect there to be a much smaller flux die to the large angle between the area vector and the electric field.
+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.