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184_notes:pc_force [2017/08/30 22:15] – [Electric Force] pwirving | 184_notes:pc_force [2021/01/26 21:15] – bartonmo | ||
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Sections 3.7, 13.2 - 13.3 , and 13.6 of Matter and Interactions (4th edition) | Sections 3.7, 13.2 - 13.3 , and 13.6 of Matter and Interactions (4th edition) | ||
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+ | [[184_notes: | ||
===== Electric Force ===== | ===== Electric Force ===== | ||
+ | Last week, you have read about the [[184_notes: | ||
- | 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. These notes will introduce the general relationship between electric force and electric field and discuss the example |
{{youtube> | {{youtube> | ||
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$$\vec{F}_{net \rightarrow q}=q*\vec{E}_{net}$$ | $$\vec{F}_{net \rightarrow q}=q*\vec{E}_{net}$$ | ||
- | 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 | + | 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. Since electric field is a vector and charge is a scalar, when they are multiplied together, |
- | The electric force works in same way as any other force that you learned about in Mechanics: | + | The electric force works in the same way as any other force that you learned about in Mechanics: |
* The electric force has units of newtons (N). | * 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 [[183_notes: | + | * The electric force can be combined with any other forces acting on an object to find the [[183_notes: |
- | * The electric force can contribute to a [[183_notes: | + | * The electric force can contribute to a [[183_notes: |
* You can still use [[183_notes: | * You can still use [[183_notes: | ||
* [[https:// | * [[https:// | ||
- | === The electric force is a conservative force === | + | ==== The electric force is a conservative force ==== |
In addition to the general results above, the electric force is also a [[https:// | In addition to the general results above, the electric force is also a [[https:// | ||
- | - 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). | + | - 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' | + | Two examples of conservative forces from mechanics include: the gravitational force and the spring force. It doesn' |
- | === Two Point Charges === | + | ==== Two Point Charges |
- | {{ 184_notes: | + | [{{ 184_notes: |
- | Let's return to our example of a fixed electric dipole - two charges one positive and one negative located a small distance apart. | + | As an example of electric force, we will talk about an electric dipole - two charges |
$$\vec{F}_{net \rightarrow q}=q\vec{E}_{net}$$ | $$\vec{F}_{net \rightarrow q}=q\vec{E}_{net}$$ | ||
- | 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 $\vec{E}_{net}=\vec{E}_{q-}$. | + | 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 the 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 $\vec{E}_{net}=\vec{E}_{q-}$. |
$$\vec{F}_{q- \rightarrow +q}=q_{+}\vec{E}_{q-}$$ | $$\vec{F}_{q- \rightarrow +q}=q_{+}\vec{E}_{q-}$$ | ||
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$$\vec{F}_{1 \rightarrow 2}=\frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{(r_{1 \rightarrow 2})^3}\vec{r}_{1 \rightarrow 2}=\frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{(r_{1 \rightarrow 2})^2}\hat{r}_{1 \rightarrow 2}$$ | $$\vec{F}_{1 \rightarrow 2}=\frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{(r_{1 \rightarrow 2})^3}\vec{r}_{1 \rightarrow 2}=\frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{(r_{1 \rightarrow 2})^2}\hat{r}_{1 \rightarrow 2}$$ | ||
- | Where here we have used the [[184_notes: | + | Where here we have used the [[184_notes: |
- | + | ||
- | === 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? | + | |
- | + | ||
- | {{ 184_notes: | + | |
- | + | ||
- | We start by finding the net electric field at the location of $q_3$ using superposition, | + | |
- | $$\vec{E}_{net}=\vec{E}_{+}+\vec{E}_{-}$$ | + | |
- | + | ||
- | 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 | + | |
- | $$r_{+ \rightarrow 3}=\sqrt{(d/ | + | |
- | + | ||
- | So this means that $\vec{E}_{+}$ is given by: | + | |
- | $$ E_{+}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(\sqrt{(d/ | + | |
- | + | ||
- | 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 | + | |
- | $$r_{- \rightarrow 3}=\sqrt{(-d/ | + | |
- | + | ||
- | So this means that $\vec{E}_{-}$ is given by: | + | |
- | $$ E_{-}=\frac{1}{4\pi\epsilon_0}\frac{q_{-}}{(\sqrt{(d/ | + | |
- | + | ||
- | So the net electric field is given by: | + | |
- | $$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(\sqrt{(d/ | + | |
- | + | ||
- | If we //__assume the dipole charges are equal in magnitude__// | + | |
- | $$\vec{E}_{net}=+\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/ | + | |
- | + | ||
- | $$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/ | + | |
- | + | ||
- | $$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/ | + | |
- | + | ||
- | $$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/ | + | |
- | + | ||
- | So this is the net electric field at $q_3$ from both the positive and negative dipole charges. Finally to find the force on $q_3$, we use: | + | |
- | $$\vec{F}_{q_3}=q_3\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q_3*q}{(\sqrt{(d/ | + | |
- | + | ||
- | 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 === | + | |
- | {{ 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. If we start with the equation that we found for the electric field of the dipole above: | + | |
- | + | ||
- | $$\vec{E}_{net}=\frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(d/ | + | |
- | + | ||
- | Now if we //__assume that we are really far away from the dipole__//, then this would mean that $h$ is much much larger than the separation of $d$. This allows us to simplify the denominator. If $h>> | + | |
- | $$(d/ | + | |
- | Using this with our electric field equation gives: | + | |
- | $$\vec{E}_{net} \approx \frac{1}{4\pi\epsilon_0}\frac{q}{(\sqrt{(h^2})^3}\langle d, 0,0 \rangle $$ | + | |
- | $$\vec{E}_{net} \approx \frac{1}{4\pi\epsilon_0}\frac{q}{h^3}\langle d, 0,0 \rangle $$ | + | |
- | Note, //this electric field equation is only true for points far away from the dipole and perpendicular to the dipole axis//. | + | |
- | + | ||
- | {{184_notes: | + | |
- | + | ||
- | 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. | + | |
- | + | ||
- | 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.) | + | |
==== Examples ==== | ==== Examples ==== | ||
[[: | [[: |