Projectile Motion in an Electric Field

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion in a uniform electric field. The scenario describes an electron shot between two charged plates, with specific parameters including the electric field strength, plate dimensions, and initial velocity of the electron. Participants are tasked with determining whether the electron strikes one of the plates and, if not, the vertical position at which it exits the space between the plates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of time required for the electron to travel horizontal and vertical distances, questioning the correctness of their attempts and the implications of their findings. There is exploration of the effects of acceleration due to the electric field and the signs of initial velocities and accelerations. Some participants express confusion about the physical setup and the resulting motion of the electron.

Discussion Status

The discussion is active, with participants providing various attempts at solving the problem and seeking clarification on their reasoning. Some have offered calculations and interpretations, while others are questioning the assumptions made regarding directionality and the effects of the electric field on the electron's motion. There is no explicit consensus on the correct approach, but several productive lines of inquiry are being explored.

Contextual Notes

Participants are navigating the complexities of kinematics in an electric field, with specific attention to the signs of acceleration and velocity. There are indications of confusion regarding the implications of the electric field on the electron's trajectory, as well as the need to consider the effects of both horizontal and vertical motion simultaneously.

  • #31
y=(vt(sinθ))−0.5gt^2
y=((5.15E6 m/s)(7.77e-9 s)) - 0.5(8.78E14 m/s^2)(7.77e-9 s)^2
y=0.0135 m

I just checked this, and this is the correct answer (! yay!). Why is the 0.04 not involved in this last step like I thought? Is it because it is involved in the calculation for Tx, which is the time we used for this part?
 
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  • #32
rlc said:
Why is the 0.04 not involved in this last step like I thought? Is it because it is involved in the calculation for Tx, which is the time we used for this part?
Yes.
 
  • #33
Thank you very much for helping me through this problem. I probably wouldn't have found those mistakes if you hadn't pointed them out to me.
 

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