Magnetic Force to balance out Electric Force

[mbrmbrg]

Homework Statement

In the figure (see attatched), an electron accelerated from rest through potential difference 2.0 kV enters the gap between two parallel plates having separation 20.0 mm and potential difference 130 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?
(answer is in Telsa, and the field is in the positive-k direction.)

Homework Equations

When a system is in equilibrium,
$$\sum F=0$$

Magnetic (Lorentz?) Force is given by:
$$F_B=qv\times B$$

Electric Force is given by:
$$F=qE$$

The electric field produced by parallel plates is:
$$E=\frac{V}{d}$$

The work-energy theorem states that W=K.
For electric potential, W=qV.
$$K=\frac{mv^2}{2}$$

The Attempt at a Solution

Finding the Velocity of the Electron as it enters the gap:
K=W
$$\frac{1}{2}mv^2=qV_1$$
where V_1 is the potential through which the electron is accelerated.
Isolate v to get
$$v=\sqrt{\frac{2qV_1}{m}}$$

Onward:
We want a situation of equilibrium where $$F_B=F_E$$.
So $$qv\times B=qE$$
the q's cancel, and the cross product is maximum, so
$$vB=E$$
plug in the velocity which we found above and the electric field of parallel plates to get
$$\sqrt{\frac{2qV_1}{m}}B=\frac{V_2}{d}$$
where V_2 is the potential between the two plates.
Isolate B to get
$$B = \frac{V_2}{d}\sqrt{\frac{m}{2qV_1}}$$

Now plug in my numbers
$$B = \frac{130}{20\times10^{-3}}\sqrt{\frac{9.109\times10^{-31}}{(2)(1.602\times10^{-19})(2000)}}=2.45\times10^{-4}T$$
Wrong

Doubly odd because when I used the values:
particle=electron
V_1=1.0kV
d=20.0mm
V_2=100V

I got the correct answer of 2.67e-4 T.
Weird...

 

[cepheid]

I can't find your mistake, if there even is one.

 

[mbrmbrg]

Thanks, nice to know it's not another one of my "redface" errors.
I guess I'll take it up with my professor, ask him to check that WebAssign is expecting a reasonable answer.