# Finding the electric field magnitude in a parallel plate compacitor.

1. Sep 24, 2008

### 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.

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2. Sep 24, 2008

### Staff: Mentor

Redo this arithmetic.

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

3. Sep 24, 2008

### Staff: Mentor

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?)

4. Sep 24, 2008

### 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 ?

5. Sep 24, 2008

### Staff: Mentor

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.)

6. Sep 24, 2008

### nemisisnik

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