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bbbbbev

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An electron accelerated from rest through potential difference 1.6 kV enters the gap between two parallel plates having separation 20.0 mm and potential difference 120 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. What uniform magnetic field allows the electron to travel in a straight line in the gap?

I was thinking that all I really needed to do was figure out the velocity of the electron and plug all the values into the equation

F = qvBsin(phi)

phi = 90 degrees in this problem, so the equation can just be written as

F = qvB

To find v, I used the fact that change in kinetic energy = change in potential energy.

KE = PE

(1/2)mv^2 = q(deltaV)

v = sqrt((2q(deltaV))/m)

After plugging in all the numbers I got:

v = sqrt(2(1.6e-19C)(1600V)/(9.11e-31kg))

v = 2.37e7 m/s

When I plug that into the magnetic field equation I get:

B = F/qv, so B = Eq/qv. Since E = V/d and the q's cancel, I then used

B = V/(distance)(v)

B = (120 V)/(0.02 m)(2.37e7m/s) = 2.53e-4 T

But that was not the correct answer. I know this is a lot of work to look through, but I'd appreciate any help anyone could give me because I'm really not sure I'm looking at this problem correctly. What am I missing.

Thanks,

Beverly