184_notes:examples:week3_particle_in_field

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184_notes:examples:week3_particle_in_field [2018/01/24 23:19] – [Solution] tallpaul184_notes:examples:week3_particle_in_field [2018/05/24 15:00] – [Example: Particle Acceleration through an Electric Field] curdemma
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-=====Particle Acceleration through an Electric Field=====+[[184_notes:pc_energy|Return to Electric Potential Energy]] 
 + 
 +=====Example: Particle Acceleration through an 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? 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 "i" to a final location "f") for a point charge two ways here:+  * 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} &&&&&& (1) \\ \Delta U &= -\int_i^f\vec{F}\bullet d\vec{r} &&&&&& (1) \\
 \Delta U &= q\Delta V                        &&&&&& (2) \Delta U &= q\Delta V                        &&&&&& (2)
 \end{align*} \end{align*}
-  * We can write the change in electric potential (from an initial location "i" to a final location "f") as+  * 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} &&&&&& (3) \Delta V=-\int_i^f \vec{E}\bullet d\vec{r} &&&&&& (3)
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 ===Representations=== ===Representations===
-{{ 184_notes:3_particle_acceleration_field.png?200 |Particle in the Electric Field}}+[{{ 184_notes:3_particle_acceleration_field.png?200 |Particle in the Electric Field}}]
  
 ===Goal=== ===Goal===
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 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, as outlined in the Plan. The particle will follow a straight line in the $+x$-direction (when $Q>0$, as explained earlier). So we can set $\text{d}\vec{r}=\text{d}x \cdot \hat{x}$. 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, as outlined in the Plan. The particle will follow a straight line in the $+x$-direction (when $Q>0$, as explained earlier). So we can set $\text{d}\vec{r}=\text{d}x \cdot \hat{x}$. 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 $x_0$.
 \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 \text{d}\vec{r} \\ 
-         &= -Q\int_{x_0}^{x_0+L} E_0\hat{x}\bullet \hat{x}dx \\ +         &= -Q\int_{x_0}^{x_0+L} E_0\text{d}x (\hat{x}\bullet \hat{x}) \\ 
-         &= -QE_0\int_{x_0}^{x_0+L} dx \\+         &= -QE_0\int_{x_0}^{x_0+L} \text{d}x \\
          &= \left.-QE_0x\right|_{x_0}^{x_0+L} \\          &= \left.-QE_0x\right|_{x_0}^{x_0+L} \\
          &= -QE_0(x_0+L-x_0) \\          &= -QE_0(x_0+L-x_0) \\
          &= -QE_0L          &= -QE_0L
 \end{align*} \end{align*}
-The physical significance of this result is that the particle "loses" $QE_0L$ of electric potential energy as it travels through the electric field. We do not transfer any energy to the surroundings (through heat, friction, etc.), so this "lost" potential energy must have just been converted to kinetic energy. $$0=\Delta E_{\text{sys}}=\Delta U+\Delta K=-QE_0L+\frac{1}{2}M(v_f^2-v_i^2)$$+The physical significance of this result is that the particle "loses" $QE_0L$ of electric potential energy as it travels through the electric field. We do not transfer any energy to the surroundings (through heat, friction, etc.), so this "lost" potential energy must have just been converted to kinetic energy. $$0=\Delta E_{\text{sys}}=\Delta U+\Delta K=-QE_0L+\frac{1}{2}m(v_f^2-v_i^2)$$
 Remember that $\vec{v}_i=0$, so we can solve for the unknown $\vec{v}_f$: Remember that $\vec{v}_i=0$, so we can solve for the unknown $\vec{v}_f$:
 \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, so the final velocity will be We reasoned earlier that the particle will be traveling in the $+x$-direction, so the final velocity will be
-$$\vec{v}_f=\sqrt{\frac{2QE_0L}{M}}\hat{x}$$+$$\vec{v}_f=\sqrt{\frac{2QE_0L}{m}}\hat{x}$$
  
  • 184_notes/examples/week3_particle_in_field.txt
  • Last modified: 2021/05/19 15:01
  • by schram45