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184_notes:examples:week3_particle_in_field [2018/01/24 23:21] – [Solution] tallpaul | 184_notes:examples:week3_particle_in_field [2021/01/26 21:22] – [Example: Particle Acceleration through an Electric Field] bartonmo | ||
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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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===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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& | & | ||
\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, | ||
- | $$\vec{v}_f=\sqrt{\frac{2QE_0L}{M}}\hat{x}$$ | + | $$\vec{v}_f=\sqrt{\frac{2QE_0L}{m}}\hat{x}$$ |