Mass-to-charge ratio in a mass spectrometer missing factor of 2

[BecauseICan]

Edit: Sorry I just realized this should be in Introductory Physics. Can a moderator could please move it?

Homework Statement

Show that in a mass spectrometer, the mass to charge ratio, m/q is equal to:

m/q = (B²r²)/(2$$\Delta$$V)

Homework Equations

F = qvBsin$$\theta$$

The Attempt at a Solution

I started with F = qvBsin$$\theta$$

An ion moves in a circular path in a spectrometer, so F = mv²/r

mv²/r = qvBsin$$\theta$$

Rearranging and simplifying gets:

m/q = Br/v, and since v = E/B I get m/q = B₂r/E

E = V/r and my final result is:

m/q = (B²r²)/$$\Delta$$V

So I'm wondering where I lost the 2 in the Potential Difference term. Any help is appreciated.

 

[Delphi51]

More information should have been given!
v = E/B implies a speed selector, usually a separate section of the apparatus whose B is not the same as the one causing the circular motion.
E = V/r is only true if the speed selector has a pair of parallel plates with separation r = radius of curvature. Very strange! Maybe the plate separation is r/2 or something.