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--- 184_notes:examples:week3_particle_in_field [2018/01/24 23:16] – [Solution] tallpaul
+++ 184_notes:examples:week3_particle_in_field [2021/05/19 15:01] (current) – schram45
@@ Line 1: / Line 1: @@
-=====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?
@@ Line 7: / Line 9: @@
   * 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.
-  * 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*}
 \Delta U &= -\int_i^f\vec{F}\bullet d\vec{r} &&&&&& (1) \\
 \Delta U &= q\Delta V                        &&&&&& (2)
 \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*}
 \Delta V=-\int_i^f \vec{E}\bullet d\vec{r} &&&&&& (3)
@@ Line 22: / Line 24: @@
 ===Representations===
-{{ 184_notes:3_particle_acceleration_field.png?200 |Particle in the Electric Field}}
+[{{ 184_notes:3_particle_acceleration_field.png?200 |Particle Q in the Electric Field}}]
+<WRAP TIP>
+===Assumption===
+No gravitational effects are being considered in this problem. Typically point charges are really small and have negligible masses. This means that the gravitational force would be very small compared to the electric force acting on the particle in the accelerator and can be excluded from the calculations and representation.
+</WRAP>
 ===Goal===
@@ Line 31: / Line 38: @@
 <WRAP TIP>
 === Approximation ===
-We will approximate the particle as a point charge. We already know it is a "particle" which is a pretty small thing, so our approximation seems reasonable. We want to make this approximation because it allows us to use certain tools, like the equation for electric force in the Facts, tools that can only be applied to point charges.
+We will approximate the particle as a //__point charge__//. We already know it is a "particle" which is a pretty small thing, so our approximation seems reasonable. We want to make this approximation because it allows us to use certain tools, like the equation for electric force in the Facts, tools that can only be applied to point charges.
 </WRAP>
@@ Line 52: / Line 59: @@
 </WRAP>
-We have listed two representations (Equations (1) and (2)) for the electric potential energy. You can check for yourself that both lead us to the following equation (using Equations (3) and (4)):  $\Delta U=-q\int_i^f \vec{E}\bullet d\vec{r}$
+We have listed two expressions (Equations (1) and (2)) for the electric potential energy of the particle. You can check for yourself that both lead us to the following equation (using Equations (3) and (4)):  $\Delta U=-q\int_i^f \vec{E}\bullet d\vec{r}$
-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, 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*}
-\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} \\
          &= -QE_0(x_0+L-x_0) \\
          &= -QE_0L
 \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. $$0=\Delta E_{\text{sys}}=\Delta U+\Delta K=-QE_0L+\frac{1}{2}M(v_f^2-v_i^2)$$
+<WRAP TIP>
+===Assumption===
+Assuming the electric field is constant within the accelerator allows the  $E_0$  to be taken out of the integral in this problem.
+</WRAP>
+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. $$0=\Delta E_{\text{sys}}=\Delta U+\Delta K=-QE_0L+\frac{1}{2}m(v_f^2-v_i^2)$$
+<WRAP TIP>
+===Assumption===
+Assuming there is a conservation of energy allows the total change in energy of the system to be zero.
+</WRAP>
 Remember that  $\vec{v}_i=0$ , so we can solve for the unknown  $\vec{v}_f$ :
 \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*}
 We reasoned earlier that the particle will be traveling in the  $+x$ -direction, so the final velocity will be
-$$\vec{v}_f=\sqrt{\frac{2QE_0L}{M}}\hat{x}$$
+$$\vec{v}_f=\sqrt{\frac{2QE_0L}{m}}\hat{x}$$