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184_notes:examples:week3_particle_in_field [2018/01/24 23:10] – [Solution] tallpaul | 184_notes:examples:week3_particle_in_field [2021/05/19 14:28] – schram45 | ||
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- | =====Particle Acceleration through an Electric Field===== | + | [[184_notes: |
+ | |||
+ | =====Example: | ||
Suppose you have a particle with a mass $m$, charge $Q$, initially at rest in an electric field $\vec{E}=E_0\hat{x}$. The electric field extends for a distance $L$ in the $+\hat{x}$-direction before dropping off abruptly to 0. So, the magnitude of the electric field is exactly $E_0$, and then exactly $0$. There is no in-between. What happens to the particle if $Q>0$? What if $Q<0$? What if $Q=0$? In one of these cases, the particle exits the electric field. What is its velocity when it reaches the region of zero electric field? | Suppose you have a particle with a mass $m$, charge $Q$, initially at rest in an electric field $\vec{E}=E_0\hat{x}$. The electric field extends for a distance $L$ in the $+\hat{x}$-direction before dropping off abruptly to 0. So, the magnitude of the electric field is exactly $E_0$, and then exactly $0$. There is no in-between. What happens to the particle if $Q>0$? What if $Q<0$? What if $Q=0$? In one of these cases, the particle exits the electric field. What is its velocity when it reaches the region of zero electric field? | ||
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* It has charge $Q$, which can be positive or negative or zero. | * It has charge $Q$, which can be positive or negative or zero. | ||
* The particle is a distance $L$ from the boundary of the electric field. | * The particle is a distance $L$ from the boundary of the electric field. | ||
- | * We can write the change in electric potential energy (from an initial location " | + | * We can write the change in electric potential energy (from an initial location "$i$" to a final location "$f$") for a point charge two ways here: |
\begin{align*} | \begin{align*} | ||
\Delta U &= -\int_i^f\vec{F}\bullet d\vec{r} &&&&&& | \Delta U &= -\int_i^f\vec{F}\bullet d\vec{r} &&&&&& | ||
\Delta U &= q\Delta V &&&&&& | \Delta U &= q\Delta V &&&&&& | ||
\end{align*} | \end{align*} | ||
- | * We can write the change in electric potential (from an initial location " | + | * We can write the change in electric potential (from an initial location "$i$" to a final location "$f$") as |
\begin{align*} | \begin{align*} | ||
\Delta V=-\int_i^f \vec{E}\bullet d\vec{r} &&&&&& | \Delta V=-\int_i^f \vec{E}\bullet d\vec{r} &&&&&& | ||
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\vec{F}=q\vec{E} &&&&&&&& | \vec{F}=q\vec{E} &&&&&&&& | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | ===Assumptions=== | ||
+ | * Point Charge: Allows us to use the electric potential equation and the problem does not specify anything otherwise. | ||
+ | * Constant charge: Simplifies the value of charge meaning it is not charging or discharging over time. | ||
+ | * Electric field is constant in accelerator: | ||
+ | * No gravitational effects: Gravity would be another force acting on our charge in this situation, however for simplicity we are not told any mass and neglect gravity for this problem. | ||
+ | * Conservation of energy: no energy is being added or taken out of the system. This means as the charge loses electric potential energy as it leaves the accelerator, | ||
===Representations=== | ===Representations=== | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
===Goal=== | ===Goal=== | ||
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<WRAP TIP> | <WRAP TIP> | ||
=== Approximation === | === Approximation === | ||
- | We will approximate the particle as a point charge. We already know it is a " | + | We will approximate the particle as a //__point charge__//. We already know it is a " |
</ | </ | ||
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Okay, so now the only thing we need to find is the velocity of the particle ($Q>0$) when it reaches the boundary of the electric field. Where can we start? Let's break down what is happening to the particle. It starts from rest. Since it has charge, and the electric field at its location is nonzero, it feels an electric force. The force acts on the particle, and the particle accelerates. When it leaves the electric field, its acceleration drops to zero, so it is moving at constant velocity. | Okay, so now the only thing we need to find is the velocity of the particle ($Q>0$) when it reaches the boundary of the electric field. Where can we start? Let's break down what is happening to the particle. It starts from rest. Since it has charge, and the electric field at its location is nonzero, it feels an electric force. The force acts on the particle, and the particle accelerates. When it leaves the electric field, its acceleration drops to zero, so it is moving at constant velocity. | ||
- | It would be easy to find the electric force (like we did above), but it's not obvious how we'd find out the amount of time the particle spends in the electric field. We would need to know the timing in order to use force to find the change in momentum ($\vec{F}=\Delta \vec{p}/ | + | It would be easy to find the electric force (like we did above), but it's not obvious how we'd find out the amount of time the particle spends in the electric field. We would need to know the timing in order to use force to find the change in momentum ($\vec{F}=\Delta \vec{p}/ |
+ | |||
+ | <WRAP TIP> | ||
+ | === Plan === | ||
+ | We will use conservation of energy to find the final velocity of the particle. We'll go through the following steps. | ||
+ | * Our system will be the single particle. | ||
+ | * The initial state is when the particle is at rest in the electric field. | ||
+ | * The final state is when the particle exits the electric field. | ||
+ | * We expect the system to experience a decrease in electric potential energy, and an equivalent increase in kinetic energy. | ||
+ | * We can use the new kinetic energy to find the corresponding velocity. | ||
+ | </ | ||
- | We have listed two representations | + | We have listed two expressions |
- | We choose the path of integration to start at the current location of the particle, and end at place where the particle exits the electric field. Based on our assumptions, the particle will follow a straight line in the $+x$-direction (when $Q>0$, as explained earlier). So we can set $d\vec{r}=\hat{x}dx$. We also need to define the path endpoints, so we'll say the particle starts at $x_i=x_0$, which means the integration will end at $x_f=x_0+L$. You'll see that it won't matter in the end what $x_0$ is, since it drops out. In fact, it is common to set $x_0=0$, for simplicity of calculation. In this example, we'll leave it as $x_0$. | + | We choose the path of integration to start at the current location of the particle, and end at place where the particle exits the electric field, |
\begin{align*} | \begin{align*} | ||
- | \Delta U &= -q\int_i^f \vec{E}\bullet d\vec{r} \\ | + | \Delta U &= -q\int_i^f \vec{E}\bullet |
- | & | + | & |
- | & | + | & |
& | & | ||
& | & | ||
& | & | ||
\end{align*} | \end{align*} | ||
- | The physical significance of this result is that the particle " | + | The physical significance of this result is that the particle " |
Remember that $\vec{v}_i=0$, | Remember that $\vec{v}_i=0$, | ||
\begin{align*} | \begin{align*} | ||
- | & 0=-QE_0L+\frac{1}{2}M(v_f^2-0) \\ | + | & 0=-QE_0L+\frac{1}{2}m(v_f^2-0) \\ |
- | \Rightarrow \text{ } & QE_0L=\frac{1}{2}Mv_f^2 \\ | + | \Rightarrow \text{ } & QE_0L=\frac{1}{2}mv_f^2 \\ |
- | \Rightarrow \text{ } & \frac{2QE_0L}{M}=v_f^2 | + | \Rightarrow \text{ } & \frac{2QE_0L}{m}=v_f^2 |
\end{align*} | \end{align*} | ||
We reasoned earlier that the particle will be traveling in the $+x$-direction, | We reasoned earlier that the particle will be traveling in the $+x$-direction, | ||
- | $$\vec{v}_f=\sqrt{\frac{2QE_0L}{M}}\hat{x}$$ | + | $$\vec{v}_f=\sqrt{\frac{2QE_0L}{m}}\hat{x}$$ |