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184_notes:dipole_sup [2018/05/29 14:48] – [Dipole Superposition Example] curdemma | 184_notes:dipole_sup [2018/06/27 20:01] – dmcpadden | ||
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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). | 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). | ||
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- | === Electric Field far away from a Dipole === | ||
- | [{{ 184_notes: | ||
- | Since dipoles occur frequently in nature (we can model any atom as 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: | ||
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- | $$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/ | ||
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- | Now if we //__assume that we are really far away from the dipole__//, then this would mean that $h$ is much much larger than the separation of $d$. This allows us to simplify the denominator. If $h>> | ||
- | $$(d/ | ||
- | 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// | ||
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- | === Electric Field on Axis of the Dipole === | ||
- | [{{ 184_notes: | ||
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- | 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, | ||
- | $$|\vec{E}_{axis}|=\frac{1}{4\pi\epsilon_0}\frac{2qd}{r^3}$$ | ||
- | 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. | ||
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- | 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 ==== | ==== Finding the force from the dipole ==== | ||
Once we have the electric field from the dipole, it becomes relatively simple to find the force from that dipole on any other charge that we would have near by simply by using the electric force relationship: | Once we have the electric field from the dipole, it becomes relatively simple to find the force from that dipole on any other charge that we would have near by simply by using the electric force relationship: | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
For example, if we had a positive charge $q_3$ that was centered a distance h above our dipole (as in the first example), we could find the force on $q_3$ by using the electric field that we calculated before: | For example, if we had a positive charge $q_3$ that was centered a distance h above our dipole (as in the first example), we could find the force on $q_3$ by using the electric field that we calculated before: | ||
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If $q_3$ happened to be negative, then the force would instead point to left (but we would not have to rewrite $\vec{E}_{net}$ because we didn't change anything about the dipole): | If $q_3$ happened to be negative, then the force would instead point to left (but we would not have to rewrite $\vec{E}_{net}$ because we didn't change anything about the dipole): | ||
$$\vec{F}_{q_3}=-q_3\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{-q_3*q}{(\sqrt{(d/ | $$\vec{F}_{q_3}=-q_3\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{-q_3*q}{(\sqrt{(d/ | ||
- | This is often why we care about the electric field - **it doesn' | + | This is often why we care about the electric field - **it doesn' |
==== Superposition of Potential and Energy ==== | ==== Superposition of Potential and Energy ==== | ||
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Once you have the total electric potential around the dipole, it is then very easy to find the electric potential energy between a third charge ($q_3$) and the dipole: | Once you have the total electric potential around the dipole, it is then very easy to find the electric potential energy between a third charge ($q_3$) and the dipole: | ||
$$U=q_3 V_{tot}$$ | $$U=q_3 V_{tot}$$ | ||
- | Again, this is often why we care about the electric potential - **it doesn' | + | Again, this is often why we care about the electric potential - **it doesn' |