Recent content by jasonleroy

I don't think the charge can move from the balloon to the wall.

Imagine bringing a negatively charged particle close to a neutral one. The electrons in the neutral particle will be repulsed by the negative particle and try to move away leaving the protons on the near side of the neutral particle and the electrons on the far side of the neutral particle...

Charge objects can induce a dipole moment on neutral objects, and thereby have a weak attraction. Is that what you're referring too?

Once it's lying on the equator, you can measure it's weight.

They are not weightless. In fact, they are falling towards earth. It just so happens that their orbit velocity is such that they fall "around" the Earth rather than straight towards the center. If they were stationary, they'd fall immediately into earth.

That's it, but the cross product is AB*sin(x). I'm not following what you meant by the line below, but you got the main point.

You have created two vectors with zero cross product. Do you understand what can be said about two vectors whose cross product is zero?

Are you asking if the number you plug in for Vmax in problem "a" should have a phase angle associated with it? I would assume no. It would be explicitly stated in the problem.

You can use a fet as a simple switch.

From what I understand, if you were to track an electrons journey, it would never cross the gap (provided a perfect dielectric between the plates). If A were connected to the positive side of a battery, and B to the negative side, a negative charge would build on plate B, repelling the electrons...

I think the proximity of the magnet to the metal helps the cutting of the magnetic flux. If you think of the extreme, a pea sized magnetic falling through a tube with a diameter of 1 meter would need to be VERY strong to see any effect. I am by far no expert though.

I suppose the simplest example would be measuring a perfect short. If there's no distance between you and the short, S11 is 1∠180°. Now, if you add a length equivalent to 30 degrees for example, you have a new S11 of 1∠120°. In my simplistic way of thinking about it, why isn't it 1∠240°? You...

Thanks for the response. What seems to be counter-intuitive to me is that a line of longer physical length shows up as having a smaller phase angle than the original. In my mind, longer length = larger phase angle. However, I must be thinking of it from the wrong perspective. The phase angle...

Hopefully I can word this correctly. Consider viewing S11 of a short on a smith chart. It starts at the left and moves clockwise around the edge. If you then add a small length of line between the reference plane and the short, the phase angle wraps further around the outside in the clockwise...

Thanks! I'll just have to get out there and interview. Since I'm on educational leave, I expect to get an offer to return to my current company. Just want to make sure the offer is fair.