184_notes:extra_sup

In the last page of notes, you read about how superposition applies in general for electric field and electric potential. These notes will go into more detail about how to calculate the electric field using superposition for the specific example of an electric dipole. (We are showing this example of electric field, rather than electric potential, because the vector math is more complicated and worth describing in more detail.)

Electric Field between a Dipole

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 ($\vec{E}_{-}$):

$$\vec{E}_{net}=\vec{E}_{+}+\vec{E}_{-}$$

First, we will find the electric field from the positive charge, which is given by: $$ E_{+}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(r_{+ \rightarrow P})^3}\vec{r}_{+ \rightarrow P}$$ 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 $$r_{+ \rightarrow P}=\sqrt{(d/2)^2+h^2}$$

So this means that $\vec{E}_{+}$ is given by: $$ E_{+}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(\sqrt{(d/2)^2+h^2})^3}\langle d/2, h,0 \rangle $$

Similarly, we can find the electric field from the negative charge: $$ E_{-}=\frac{1}{4\pi\epsilon_0}\frac{q_{-}}{(r_{- \rightarrow P})^3}\vec{r}_{- \rightarrow P}$$ where $\vec{r}_{- \rightarrow P}= \langle -d/2, h,0 \rangle $ because it points from the negative charge to the location of Point P. In this equation, $r_{- \rightarrow P}$ is the magnitude of $\vec{r}_{- \rightarrow P}$ so $$r_{- \rightarrow P}=\sqrt{(-d/2)^2+h^2}=\sqrt{(d/2)^2+h^2}$$

So this means that $\vec{E}_{-}$ is given by: $$ E_{-}=\frac{1}{4\pi\epsilon_0}\frac{q_{-}}{(\sqrt{(d/2)^2+h^2})^3}\langle -d/2, h,0 \rangle $$

The net electric field is then found by adding the two electric fields together: $$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(\sqrt{(d/2)^2+h^2})^3}\langle d/2, h,0 \rangle + \frac{1}{4\pi\epsilon_0}\frac{q_{-}}{(\sqrt{(d/2)^2+h^2})^3}\langle -d/2, h,0 \rangle $$

If we assume the dipole charges are equal in magnitude (but still opposite in sign) then this becomes: $$\vec{E}_{net}=+\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle d/2, h,0 \rangle - \frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle -d/2, h,0 \rangle $$

We can simplify the math some to find the general equation for the electric field between the two charges:

$$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}(\langle d/2, h,0 \rangle - \langle -d/2, h,0 \rangle) $$

$$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle (d/2+d/2), (h-h),0 \rangle $$

$$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle d, 0,0 \rangle $$

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

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:

$$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle d, 0,0 \rangle $$

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

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.)

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: $\vec{F}=q\vec{E}$.

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: $$\vec{F}_{q_3}=q_3\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q_3*q}{(\sqrt{(d/2)^2+h^2})^3}\langle d, 0,0 \rangle$$

This force only points in the $\hat{x}$ direction, which makes sense. Because $q_3$ is positive, we know that it should be repelled from the positive charge and attracted toward the negative charge. As much as the positive charge wants to push it away, the negative charge wants to pull it back so there is no change in the $\hat{y}$ direction, but both charges want to send $q_3$ to the right.

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): $$\vec{F}_{q_3}=-q_3\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{-q_3*q}{(\sqrt{(d/2)^2+h^2})^3}\langle d, 0,0 \rangle$$ This is often why we care about the electric field - it doesn't matter what change about $q_3$, the electric field from the dipole is going to stay the same. This idea will save you time when doing your calculations, especially for calculations involving more than three charges. (Since we already have the electric field for when Point P is far away from the dipole and when it is on the axis of the dipole, we could also easily find the force on $q_3$ at those locations too.)

We could have also done the same procedure for finding the total electric potential at Point P rather than finding the electric field - find the electric potential at Point P from the positive charge, find the electric potential at Point P from the negative charge and add them together to get $V_{tot}$.

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: $$U=q_3 V_{tot}$$ Again, this is often why we care about the electric potential - it doesn't matter what change about $q_3$, the electric potential from the dipole is going to stay the same.

  • 184_notes/extra_sup.txt
  • Last modified: 2018/01/18 19:50
  • by dmcpadden