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Sections 3.7, 13.2 - 13.3 , and 13.6 of Matter and Interactions (4th edition)
Electric Force
You have already read about the kind of interaction you expect when you place two charges next to each other: either they are attracted to each other (charges have different signs) or repelled from each other (charges have the same sign). As you learned in your mechanics course, we can think about these kinds of pulls or pushes as a force acting on the charge(s). This force results from the interaction of a charge with the electric field produced by the other charge(s). We will call this new force the electric force. In general, there are two ways to think about the electric force: either one charge interacts with another charge or one charge interacts with the electric field that is produced by another charge. We will usually favor thinking about how charges interact with the field (rather than how charges interact with other charges) because it is through that field that the interaction occurs and we have already developed a straight-forward mathematical description (superposition) when there are multiple charges or complicated shapes of charge. These notes will introduce the general relationship between electric force and electric field, discuss the example for two point charges, and describe what happens if you have multiple point charges.
General Relationship
In general, the electric force on any charge (q) is equal to the charge times the net electric field ($\vec{E}_{net}$) at the location of q. This is the most general way to write the electric force on a charge, which means it is always true - if you know what the electric field is at a location due to all the other charges (no matter what shape or type of charge produced that electric field), you can easily determine the electric force on a charge would be at that location.
$$\vec{F}_{net \rightarrow q}=q*\vec{E}_{net}$$
Where we are using the notation of $net \rightarrow q$ to show that this is the force from the net electric field on the charge q. Because electric field is a vector and charge is a scalar, this means that when multiplied together, you get a vector (electric force). This vector points in the same direction as the electric field for positive charges and in the opposite direction of the field in the case of negative charges. Note that this is the electric force from the external electric field on the charge. A charge's own electric field cannot affect the charge itself (for the same reason that you can't lift a board while standing on it and fly). Also note that electric force is not the same thing as electric field, but they are related. The electric field is the electric force per unit charge at a given location; in a sense, it tells you how large (magnitude) and which way (direction) the force could be. Note there's some ambiguity about the direction of the force as it would matter which kind of charge is placed in the field.
The electric force works in same way as any other force that you learned about in Mechanics:
- The electric force has units of newtons (N).
- The electric force can be combined with any other forces acting on an object to find the net force on that object
- The electric force can contribute to a change in the object's momentum
- You can still use free-body diagrams (or force diagrams) to determine the net force on an object. The electric force would simply be one of the arrows in the diagram.
- Newton's third law still applies - The electric force from one charge on a second charge is equal and opposite to the electric force from the second charge on the first.
The electric force is a conservative force
In addition to the general results above, the electric force is also a **conservative force**. A conservative force is a force: - for which we can define a potential energy and - for which changes in that potential energy are path-independent (only the initial and final states matter - not how you went from the initial to final state). In fact, one of these implies the other, but that's for later physics courses.
Two examples of conservative forces from mechanics include: the gravitational force and the spring force. It doesn't matter how you go from $y=3m$ to $y=5m$, the change in gravitational energy (which is easy to define) is the same. This is in contrast to non-conservative force, which result changes in changes to the system energy that depend on the path the system takes. You cannot define a potential energy for non-conservative forces. Examples of non-conservative forces that you may know from mechanics are: the friction force and air drag. In these cases we cannot define what the “energy of friction” would be, in part because it depends on what path you take (e.g., the thermal energy change due to friction on a squiggly path between two points would be much higher than on a straight path between those same points).
Two Point Charges
Let's return to our example of a fixed electric dipole - two charges one positive and one negative located a small distance apart. In this case, we are interested in what the electric force from the negative charge is on the positive charge and vice versa. Starting with the electric force on the positive charge, we can write the general relationship:
$$\vec{F}_{net \rightarrow q}=q\vec{E}_{net}$$
where $q$ here is the positive charge and $\vec{F}_{q}$ is then the force on the positive charge. In this case, $\vec{E}_{net}$ is net electric field at the location of $q_{+}$. Since the negative charge $q_{-}$ is the only other charge around $q_+$, the net electric field at $q_+$ is equal to the electric field from the negative charge $\vec{E}_{net}=\vec{E}_{q-}$.
$$\vec{F}_{q- \rightarrow +q}=q_{+}\vec{E}_{q-}$$
When we make the point charge assumption, we know from last week that electric field for a negative point is given by:
$$E_{q-}=\frac{1}{4\pi\epsilon_0}\frac{q_{-}}{r^3}\vec{r}_{- \rightarrow +}$$
where $\vec{r}_{- \rightarrow +}$ points from the charge producing the field ($q_{-}$ in this case) to the location of interest (where $q_{+}$ is). If we plug this electric field into our equation for the force on the positive charge, we get:
$$\vec{F}_{q- \rightarrow +q}=q_{+}\vec{E}_{q-}=q_{+}\frac{1}{4\pi\epsilon_0}\frac{q_{-}}{r^3}\vec{r}_{- \rightarrow +}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}q_{-}}{r^3}\vec{r}_{- \rightarrow +}$$
By following the same process, we can find the electric force on the negative charge to be: $$\vec{F}_{q+ \rightarrow q-}=q_{-}\vec{E}_{q+}=q_{-}\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{r^3}\vec{r}_{+ \rightarrow -}=\frac{1}{4\pi\epsilon_0}\frac{q_{-}q_{+}}{r^3}\vec{r}_{+ \rightarrow -}$$
The only thing that is different between the $\vec{F}_{+q}$ and $\vec{F}_{-q}$ is the direction that the $\vec{r}$ points. (Here $\vec{F}_{+q}$=-$\vec{F}_{-q}$ and they are describing the same electric interaction so these forces are a Newton's third law pair.)
The general equation for the force between two point charges is then: $$\vec{F}_{1 \rightarrow 2}=\frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{(r_{1 \rightarrow 2})^3}\vec{r}_{1 \rightarrow 2}=\frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{(r_{1 \rightarrow 2})^2}\hat{r}_{1 \rightarrow 2}$$
Where here we have used the definition of the unit vector ($\hat{r}=\frac{\vec{r}}{r}$) to get the two different versions of the equation. There are a few things to notice about this equation. First, this equation is only true for the electric force between two point charges. Second, this force is not a constant force - it depends on the separation distance between the two charges. The closer the two charges are, the stronger the push/pull will be. Finally, this equation may look familiar from mechanics - if you change the charges into masses and change the constant, you will get the equation for Newtonian gravity that describes the gravitational interaction between two large masses. It turns out that there are many parallels between the gravitational force and the electric force.
More than two charges
If you have more than two point charges, you can still use the same process to find the electric force. Suppose you have a third positive charge in between the positive and negative charges of the dipole as shown. What would be the force on this third charge?
We start by finding the net electric field at the location of $q_3$ using superposition, which is given by the electric field from the positive charge and the electric field from the negative charge. $$\vec{E}_{net}=\vec{E}_{+}+\vec{E}_{-}$$
The electric field from the positive charge is given by: $$ E_{+}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(r_{+ \rightarrow 3})^3}\vec{r}_{+ \rightarrow 3}$$ where $\vec{r}_{+ \rightarrow 3}= \langle d/2, h,0 \rangle $ because it points from the positive charge to the location of $q_3$. $r_{+ \rightarrow 3}$ is the magnitude of $\vec{r}_{+ \rightarrow 3}$ so $$r_{+ \rightarrow 3}=\sqrt{(d/2)^2+h^2}$$
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: $$ E_{-}=\frac{1}{4\pi\epsilon_0}\frac{q_{-}}{(r_{- \rightarrow 3})^3}\vec{r}_{- \rightarrow 3}$$ where $\vec{r}_{- \rightarrow 3}= \langle -d/2, h,0 \rangle $ because it points from the negative charge to the location of $q_3$. $r_{- \rightarrow 3}$ is the magnitude of $\vec{r}_{- \rightarrow 3}$ so $$r_{- \rightarrow 3}=\sqrt{(-d/2)^2+h^2}=\sqrt{(d/2)^2+h^2}$$
So this means that $\vec{E}_{-}$ is given by: $$ E_{-}=\frac{1}{4\pi\epsilon_0}\frac{q_{-}}{(\sqrt{(d/2)^2+h^2})^3}\langle -d/2, h,0 \rangle $$
So the net electric field is given by: $$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(\sqrt{(d/2)^2+h^2})^3}\langle d/2, h,0 \rangle + \frac{1}{4\pi\epsilon_0}\frac{q_{-}}{(\sqrt{(d/2)^2+h^2})^3}\langle -d/2, h,0 \rangle $$
If we assume the dipole charges are equal in magnitude (but still opposite in sign) then this becomes: $$\vec{E}_{net}=+\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle d/2, h,0 \rangle - \frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle -d/2, h,0 \rangle $$
$$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}(\langle d/2, h,0 \rangle - \langle -d/2, h,0 \rangle) $$
$$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle (d/2+d/2), (h-h),0 \rangle $$
$$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/2)^2+h^2})^3}\langle d, 0,0 \rangle $$
So this is the net electric field at $q_3$ from both the positive and negative dipole charges. Finally to find the force on $q_3$, we use: $$\vec{F}_{q_3}=q_3\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q_3*q}{(\sqrt{(d/2)^2+h^2})^3}\langle d, 0,0 \rangle$$
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.
General Electric Field from 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. If we 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 $h^2$ 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: $$\vec{E}_{net} \approx \frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(h^2})^3}\langle d, 0,0 \rangle $$ $$\vec{E}_{net} \approx \frac{1}{4\pi\epsilon_0}\frac{q}{h^3}\langle d, 0,0 \rangle $$ Note, this electric field equation is only true for points far away from the dipole and perpendicular to the dipole axis.
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.)