anderkve
commented
3 years ago Hi @giammy00,
Thanks for spotting this! This is an inconsistency that crept in when I updated the project description a little while back. I wanted to state a definition of \omega_z rather than \omega_z^2, because I thought it would be less confusing. But then at the same time I wanted to keep everything in terms of real numbers. But that was a mistake, because as you point out this amounts to implicitly making the assumption that q > 0.
The equations of motion stated in the project description are entirely general -- i.e. they work for both signs of q -- as long as we define \omega_z^2 to be
\omega_z^2 = \frac{2 q V_0}{m d^2}
i.e. without any absolute value on q. This way the stated electric potential will give an oscillating solution for z(t) when q > 0, and a diverging solution z(t) when q < 0, exactly as you would expect by just looking at the z-component of the E-field. (But the latter case will require you to think about complex arguments in z(t).)
I'll correct the project description now, and add that you are free to assume q > 0 when writing down the solution for z(t) if you prefer sticking to the case with guaranteed real numbers.
edit: I noticed it is stated in problem 5 that we are using Ca+ ions as charge particles. However, since problem 1 is more general, I think that my question could still be relevant.
My question is : are we always going to assume that the particle has a positive charge? While working out the equations of motion , I was wondering what would happen if q is a negative charge. If I define omega_z = sqrt ( (2|q|V_0)/(md^2) ) (as suggested) in the case of the negative charge I would get the following equation for motion along z: \ddot z - \omega_z^2 * z = 0 (the sign in front of the \omega_z^2 term is switched; the same would happen in all of the other equations of motion). Could you please tell me if I am right, or , if I am wrong, where my mistake is, please? Thank you very much