Recent content by chw42

Ok, so your equation E = 2k(lambda)/r must be integrated over the right distance to find potential. So the potential at point R is 2klambda(ln(r)), where r is equal to .58 meters. Potential at point P is the same thing, but with a different radius since it is at (.58, .58) (hint: use...

To get the E field, you need to manipulate the volume density. Say p = volume density. We need to get two lambdas so we can use the equation E = 2k(lambda cylinder + lambda shell)/r Where r is equal to the distance of the point. To get the linear density of the cylinder, whose p is 40, we...

I figured it out. You have to account for it, not totally sure why, but to find out the potential of the ring it's KQ(1/b - 1/c), then you add that to the potential at the location you want.

I'm also interested in this question. I got the answer for the potential at v(b), but not at v(c), the outside of the insulating sphere. Thing is, I don't know how the outside shell can have any effect since it has zero charge. I know that the inside of the shell will be negatively charged while...