Cathode Ray Tube Lab: Proving D=Vd/Va{(L1^2+2L1L2)/4d}

[darkchoco]

I'm not even sure if it belongs in here...
but my class is assigned to do this lab
the objective is to determine that D=Vd/Va{(L1^2+2L1L2)/4d} is correct...
so my teacher also want us to do 2 graphical analysis based on this formula
I think i found one way of doing graphical analysis (if i put (L^2 +2LL)/4d*1/Va = m
and m is from y=mx+b which means that (L^2 +2LL)/4d*1/Va is slope... so that means D=mVd would this work?? - If I confused you, just ignore this part... DX)

so my question is how should I approach to prove the formula...
how can I rearrange the formula into two linear equation to do graphical analysis?

 

[jbsteele]

To prove the formula, you will need to use algebraic methods. Start by multiplying both sides of the equation by 4d. This will give you D*4d = Vd*(L1^2 + 2L1L2). From here, you can factor out the Vd from the right side of the equation, giving you D*4d = Vd*(L1 + L2)*(L1 - L2). Then, rearrange the equation to isolate D on one side and the two linear equations on the other side. This will give you two equations of the form y = mx + b which you can then use to graph.