Recent content by Fleur_weasley

Okay, then I guess that's what we'll be going with. Thank you so much for all your help, we appreciate it very much! have a good one.

Knowing our teacher, it's entirely possible that this is beyond curriculum standards. We're going to try using the equations F/m = a = -w^2x and F = -(nkq1q2)/r^2 to solve for magnitude of w. Can we do that?

The whole unit is called energy and momentum. We were doing simple harmonic motion, pendulums, springs, and static electricity as well as energy transformations.

Oh! that's right, oops. In that case, our Ee2 is (9.22*10^-24)*2 = 1.84*10^-23 (since it's 0.25m in relation to each of the two plates). Our velocity is then 2.8*10^3. Thank you so much. There's still the issue of solving for the angular frequency (w) though.

for Ee1 (sum of Ee when r =0.15 and r= 0.35, initial displacement ) we got 2.2*10^-23 and for Ee2 (using r=0.25, the center) we got 9.22*10^-24

using the following: Ee1 +EK1 = Ee2 + EK2 Ee = (n*k*q1*q2)/r EK = (mv^2)/2 however, at EK1, the velocity is 0 because the electron is at the turning point and therefore has no velocity so the equation ends up looking like this: Ee1 + 0 = Ee2 + EK2 We were given, n (10000 electron equivalent on...

using the following: Ee1 +EK1 = Ee2 + EK2 Ee = (n*k*q1*q2)/r EK = (mv^2)/2 however, at EK1, the velocity is 0 because the electron is at the turning point and therefore has no velocity so the equation ends up looking like this: Ee1 + 0 = Ee2 + EK2 We were given, n (10000 electron equivalent...

There is no further information about the plates. This is all the information we were given.

1. There are 2 negatively charged plates opposite each other. In between them, there is a vacuum tube (50 cm long), containing only 1 electron. Assume it is completely isolated. The charge value for the plates is equivalent to 10000 electrons. Initially the single electron is directly in the...