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| 184_notes:dipole_sup [2018/06/27 20:01] – dmcpadden | 184_notes:dipole_sup [2020/08/17 17:29] (current) – dmcpadden | ||
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| - | [[184_notes: | + | /*[[184_notes: |
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| ===== Dipole Superposition Example ===== | ===== Dipole Superposition Example ===== | ||
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| === Electric Field between a Dipole === | === Electric Field between a Dipole === | ||
| - | [{{ 184_notes:dipolepointp.png|Problem setup: r vectors between the charges in the dipole and point P}}] | + | [{{ 184_notes:dipolepointp2.png|Problem setup: r vectors between the charges in the dipole and point P}}] |
| We will start by finding the net electric field at the location of Point P (shown in the figure to the right) using superposition. Here we have P positioned a height h above the two charges in the dipole and centered between the positive and negative charge horizontally. From the superposition principle, we know that the total electric field at Point P (→Enet) should be equal to the electric field from the positive charge at Point P (→E+) plus the electric field from the negative charge at Point P (→E−): | We will start by finding the net electric field at the location of Point P (shown in the figure to the right) using superposition. Here we have P positioned a height h above the two charges in the dipole and centered between the positive and negative charge horizontally. From the superposition principle, we know that the total electric field at Point P (→Enet) should be equal to the electric field from the positive charge at Point P (→E+) plus the electric field from the negative charge at Point P (→E−): | ||
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| First, we will find the electric field from the positive charge, which is given by: | First, we will find the electric field from the positive charge, which is given by: | ||
| - | E+=14πϵ0q+(r+→P)3→r+→P | + | E+=14πϵ0q+(r+→P)3→r+→P where →r+→P=⟨d/2,h,0⟩ because it points from the positive charge to the location of Point P. Note that this equation for the r-vector is highly dependent on your choice of origin. In this case, we have placed the origin in between the two point charges and a distance h below Point P. |
| - | FIXME where →r+→P=⟨d/2,h,0⟩ because it points from the positive charge to the location of Point P. In this equation, r+→P is the magnitude of →r+→P so | + | |
| + | In the electric field equation, r+→P is the magnitude of →r+→P so | ||
| r+→P=√(d/2)2+h2 | r+→P=√(d/2)2+h2 | ||