Finding the electric field magnitude in a parallel plate compacitor.

nemisisnik

Problem:
The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 9.42 x 10^6 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

What i did:

x=.5(Vo + V)t
.02m=.5(9.42 x 10^6 m/s + 9.42 x 10^6 m/s)t
solve for t=2 x 10^-9
then:
x=Vot + .5at^2
.0015m=0 + .5(a)(2 x 10^-9)^2
solve for a=8.87 x 10^15
then:
F=ma
F=(9.11 x 10^-31)(8.87 x 10^15)
F=8.08 x 10^-15
then:
E=F/q
E=8.08 x 10^-15/(1.6 x 10^-19)
E=50524.3 N/C

But that is the wrong answer, i was just wondering what i did wrong and how to correct myself.


In Addition i was wondering how to do this one?:
A long, thin rod (length = 3.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a +7.0 C point charge is fixed to one end of the rod, and a -7.0 C point charge is fixed to the other end. Everywhere in the x, y plane there is a constant external electric field (magnitude 2.0 x 103 N/C) that is perpendicular to the rod. With respect to the z axis, find the magnitude of the net torque applied to the rod.

Thank you in advanced!!

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Doc Al

Redo this arithmetic.

Also: Don't round off intermediate values (such as your value for t); wait until the last step.

Doc Al

Start by finding the force on each end of the rod. Once you have the forces, you should be able to compute the torque about the z-axis. (What direction does the field point?)

nemisisnik

How do you find the force on each end of the rod?
Should i just us the equation: E=kq/r^2 and then use E=F/q ?

Doc Al

No, that formula gives you the field from a point charge. Not relevant here, since you are given the field.
That's the one you want. (F = Eq.)

nemisisnik

ahh ok, i got it right, thank you so much for your help :)