Electron in uniform field caused by two plates

[Fluxthroughme]

Homework Statement

An electron is projected with an initial speed [itex]v_0 = 1.60 * 10^6 m/s[/itex] into a uniform field between two parallel plates, separated by a distance of 1cm, and each plate is 2cm across. Assume the field outside the plates is zero. The electron enters the field at a point midway between the plates. If the electron just missing the upper plate as it emerges from the field (So it travels 2cm in the field before leaving it), find the magnitude of the electric field


2. The attempt at a solution

For the velocity component perpendicular to the field lines, we get the time it spends in the field: [itex]s = \frac{u + v}{2} t. s = 2 * 10^{-2} = (1.6*10^6)t \Rightarrow t = \frac{2*10^{-2}}{1.6*10^6}[/itex]
So I take the component parallel to the field lines [itex]v = u + at. u = 0 \Rightarrow v = at. v^2 = u^2 + 2as \Rightarrow a^2t^2 = 2as \Rightarrow a = \frac{2s}{t^2}[/itex]
Now since I want the value of the electric field, I use the value of a and t I just worked out to get [itex]F = ma = Eq \Rightarrow E = \frac{ma}{q}[/itex].
Using m, mass of the electron, a from the values above, and q, the charge on the electron. Saving time typing out all the fractions, I get [itex]a = 2.56 * 10^14. m = 9.1*10^{-31}. q = 1.6 * 10^{-19}. E = 1456 N/C[/itex]

The textbook gives an answer of 364N/C - exactly 1/4 of that, so I've clearly missed a factor of four somewhere. But I don't know where?

 

[Doc Al]

What did you get for the time? What distance did you use in calculating the acceleration?

 

[Fluxthroughme]

What did you get for the time? What distance did you use in calculating the acceleration?

Ahh! There's my factor of four; I had forgotten to change the distance from [itex]s = 2*10^{-2}[/itex] to [itex]0.5*10^{-2}[/itex]. Thank you!