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)