Recent content by Physicsman69

A gaussian surface just around the point at the center wouldn't yield anything. The charge at the center is unknown.

Wait, for: For #2, this would mean that q0 would be actually be zero, correct? A charge uniformly distributed over the surface, can act like a charge concentrated at the center, therefore Q0 + q0 = Q0, so q0 = 0?

Yes, you can draw a gaussian surface outside of the sphere which would yield $$E4\pi r^2 = Q_0/\epsilon$$ And, then a gaussian surface drawn inside the conducting shell would yield $$E4\pi r^2 = 0/\epsilon$$ Wouldn't this mean the charge in the center equals $$-Q_0$$ To offset the positive...

Edit: Forgot to type "stumped" at the end of the title 1. Homework Statement Instead of typing it out, a link to a scanned document of the problem is here: http://imgur.com/Be3jSLp. Homework Equations The equations to use are stated in the problem here: http://imgur.com/Be3jSLp The Attempt...

The question is: You drop a coin down a well. After 3.2 seconds you hear the sound of the coin splashing into the water surface below. How far below lies the water surface in the well? After doing all the work the answer comes out to d = 46.0 m. However, when solving for "d" I had to solve a...