The discussion focuses on calculating the percent correction for the electric field when considering the effects of gravity on an electron's motion. Participants emphasize the need to define "percent correction" to understand what is being corrected and why. A free-body diagram is suggested to visualize the forces acting on the electron, including gravity. The equations presented show how to incorporate gravity into the voltage calculation, leading to a modified voltage equation. The conversation highlights the importance of recognizing systematic errors when determining magnetic field strength.
#1
michaelle1991
2
0
Hi guys,
In the photo is the problem. We set Fb =Fe to show for E then out E into the Voltage equation to get V=vlB. However, if we won;t ignore the gravity force, what is the percent correction? (Note: q =Electron)
Welcome to PF;
What is the definition of "percent correction" - that will tell you how to work it out.
In the presence of gravity - draw a free-body diagram for the electron.
Hi guys,
In the photo is the problem. We set Fb =Fe to show for E then out E into the Voltage equation to get V=vlB. However, if we won;t ignore the gravity force, what is the percent correction? (Note: q =Electron)
Here is what I tried
This question was actually asked in one of our engineering entrances.
The answer was 1D.
My teachers say that we have to use μₘ/fₘ to get to this answer. I cannot understand why. I'll be really glad if you could tell me the exact definition of power (numerically) that works in all scenarios.
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