Percent correction for Magnetic Field vs. Electric Field

In summary, the conversation discusses how to calculate the percent correction in the presence of gravity in a voltage equation involving an electron. The participants also discuss the definition of percent correction and the importance of considering gravity in the calculation. The conversation ends with a suggestion to check private messages for further clarification.
  • #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)
 

Attachments

  • photo.jpg
    photo.jpg
    43.2 KB · Views: 484
Physics news on Phys.org
  • #2
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.
 
  • #3
i did a little work, but idk if it is right?
 

Attachments

  • photo (1).jpg
    photo (1).jpg
    38.6 KB · Views: 510
  • #4
You missed out defining what "percentage correction" means.
What are you correcting and why?

The fbd and calculation seems OK.

qvlB = qV + mgl

So V= vlB-gl(m/q) with gravity

Without gravity: V0=vlB

If you used B=V0/vl with V0=V then you have introduced a systematic error into your equation for determining B.
 
  • #5
michaelle1991 said:
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)

Please check your PMs...
 
Back
Top