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--- 184_notes:dipole_sup [2018/05/29 14:54] – curdemma
+++ 184_notes:dipole_sup [2020/08/17 17:29] (current) – dmcpadden
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-[[184_notes:comp_super|Next Page: Superposition and the Computer]]
+/*[[184_notes:comp_super|Next Page: Superposition and the Computer]]
-[[184_notes:superposition|Previous Page: Superposition]]
+[[184_notes:superposition|Previous Page: Superposition]]*/
 ===== Dipole Superposition Example =====
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 === 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 ($\vec{E}_{net}$) should be equal to the electric field from the positive charge at Point P ($\vec{E}_{+}$) plus the electric field from the negative charge at Point P ($\vec{E}_{-}$):
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 First, we will find the electric field from the positive charge, which is given by:
-$$ E_{+}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(r_{+ \rightarrow P})^3}\vec{r}_{+ \rightarrow P}$$
+$$ E_{+}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(r_{+ \rightarrow P})^3}\vec{r}_{+ \rightarrow P}$$ where $\vec{r}_{+ \rightarrow P}= \langle d/2, h,0 \rangle $ 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 $\vec{r}_{+ \rightarrow P}= \langle d/2, h,0 \rangle $ because it points from the positive charge to the location of Point P. In this equation, $r_{+ \rightarrow P}$ is the magnitude of $\vec{r}_{+ \rightarrow P}$ so
+In the electric field equation, $r_{+ \rightarrow P}$ is the magnitude of $\vec{r}_{+ \rightarrow P}$ so
 $$r_{+ \rightarrow P}=\sqrt{(d/2)^2+h^2}$$
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 So this final result is the net electric field at Point P from both the positive and negative dipole charges. There is only an x-component to the electric field, so we know that the electric field will point directly to the right (from the positive charge to the negative charge).
-=== Electric Field far away from a Dipole ===
-[{{  184_notes:dipolefar.png?100|Point P is very far away from the dipole}}]
-Since dipoles occur frequently in nature (we can model any atom as a dipole), it is often useful to have simplified equation for the electric field of a dipole. We can start with the equation that we found for the electric field of the dipole above:
-$$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle d, 0,0 \rangle $$
-Now if we //__assume that we are really far away from the dipole__//, then this would mean that $h$ is much much larger than the separation of $d$. This allows us to simplify the denominator. If $h>>d$, then the $h^2$ portion of $(d/2)^2+h^2$ will be significantly larger than the $(d/2)^2$ term. So we can approximate:
-$$(d/2)^2+h^2 \approx h^2$$
-Using this with our electric field equation gives a simplified answer of:
-$$\vec{E}_{net} \approx \frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(h^2})^3}\langle d, 0,0 \rangle $$
-$$\vec{E}_{net} \approx \frac{1}{4\pi\epsilon_0}\frac{q}{h^3}\langle d, 0,0 \rangle $$
-Note, //this electric field equation is only true for points far away from the dipole and perpendicular to the dipole axis//.
-=== Electric Field on Axis of the Dipole ===
-[{{  184_notes:dipolefaraxis.png?300|Point P is far away from the dipole but along the same axis}}]
-We could also follow a similar process to find the electric field for points far away but on the same axis as the dipole. This would consist of finding the electric field from both the positive and negative charge, adding those fields together through superposition, and making an approximation that the separation between the charges is much smaller than the distance to the point of interest. This will give you an //approximate// electric field on the axis of:
-$$|\vec{E}_{axis}|=\frac{1}{4\pi\epsilon_0}\frac{2qd}{r^3}$$
-where $r$ is the distance from the middle of the dipole to the point of interest, $d$ is separation between the positive and negative charges, and $q$ is the magnitude of one of the charges.
-Note this is only the magnitude of the electric field. The direction depends on which side of the dipole you are considering. (If you are closer to the negative charge, the electric field will point toward the negative charge. If you are closer to the positive charge, the electric field will point away from the positive charge.)
 ==== Finding the force from the dipole ====