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--- 184_notes:examples:week3_particle_in_field [2021/05/19 14:52] – schram45
+++ 184_notes:examples:week3_particle_in_field [2021/05/19 15:01] (current) – schram45
@@ Line 22: / Line 22: @@
 \vec{F}=q\vec{E} &&&&&&&& (4)
 \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: Makes electric field constant along the distance L that the charge will travel through the 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, it will gain kinetic energy.
 ===Representations===
@@ Line 77: / Line 70: @@
          &= -QE_0L
 \end{align*}
+<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*}