184_notes:examples:week3_particle_in_field

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184_notes:examples:week3_particle_in_field [2018/05/24 14:58] – curdemma184_notes:examples:week3_particle_in_field [2021/05/19 15:01] (current) – schram45
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-[184_notes:pc_energy|Return to Electric Potential Energy]+[[184_notes:pc_energy|Return to Electric Potential Energy]]
  
 =====Example: Particle Acceleration through an Electric Field===== =====Example: Particle Acceleration through an 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 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=== ===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 "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> </WRAP>
  
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          &= -QE_0L          &= -QE_0L
 \end{align*} \end{align*}
 +
 +<WRAP TIP>
 +===Assumption===
 +Assuming the electric field is constant within the accelerator allows the E0 to be taken out of the integral in this problem.
 +</WRAP>
 +
 The physical significance of this result is that the particle "loses" QE0L 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=ΔEsys=ΔU+ΔK=−QE0L+12m(v2f−v2i) The physical significance of this result is that the particle "loses" QE0L 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=ΔEsys=ΔU+ΔK=−QE0L+12m(v2f−v2i)
 +
 +<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 →vi=0, so we can solve for the unknown →vf: Remember that →vi=0, so we can solve for the unknown →vf:
 \begin{align*} \begin{align*}
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