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--- 184_notes:examples:week5_flux_dipole [2017/09/25 14:06] – [Solution] tallpaul
+++ 184_notes:examples:week5_flux_dipole [2017/09/25 14:08] – [Solution] tallpaul
@@ Line 21: / Line 21: @@
 ====Solution====
 First, notice that we probably do not want to do any calculations here, since the it will not be fun to take a dot-product of the dipole's electric field and the area-vector, and it will get very messy very quickly when we start integrating over the surface of the cylinder. Instead, we evaluate the situation more qualitatively. Consider the electric field vectors of the dipole near the surface of the cylinder:
-{{ 184_notes:5_dipole_field_lines.png?400 |Dipole Electric Field Lines}}
+{{ 184_notes:5_dipole_field_lines.png?300 |Dipole Electric Field Lines}}
 Notice that the vectors near the positive charge are leaving the cylinder, and the vectors near the negative charge are entering. Not only this, but they are mirror images of each other. Wherever an electric field vector points out of the cylinder on the right side, there is another electric field vector on the left that is pointing into the cylinder at the same angle. These mirror image vectors also have the same magnitude, though it is a little tougher to visualize.
-We could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored for left vs. right, and the $\vec{E}$-vectors are mirrored, but opposite ($\vec{E}_{left}=-\vec{E}_{right}$). We tentatively write the equality,
+We could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored for left vs. right, and the $\vec{E}$-vectors are mirrored, but opposite $\left(\vec{E}_{left}=-\vec{E}_{right}\right)$. We tentatively write the equality,
 $$\Phi_{left}=-\Phi_{right}$$