184_notes:examples:week3_particle_in_field [Projects & Practices in Physics]

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--- 184_notes:examples:week3_particle_in_field [2018/05/24 15:02] – [Solution] curdemma
+++ 184_notes:examples:week3_particle_in_field [2021/05/19 15:01] (current) – schram45
@@ Line 24: / 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 65: / 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*}