Recent content by Aero6

The shape I'm using is a gaussian cylinder. Right, so the electric field for s> b is 0 because the coaxial cable is neutral, so we do not see any charge outside. The electric field for s<a can be solved using: E-field=Qencl/2*pi*r*L and Qencl can be found by using: Qencl=integral of...

Homework Statement A long coaxial cable carries a volume charge density rho=alpha*s on the inner cylinder (radius a) and a uniform surface charge density on the outer cylindrical shell (radius b). This surface charge is negative and of just the right magnitude so that the cable as a whole is...

The problem word for word is: "A solid hemisphere of radius R has its center of curvature at the origin, with the positive z axis pointing out the 'dome' of the hemisphere. it has a charge density rho=alpha*r, with r measured from the origin. a) find the electric field (magnitude and...

Homework Statement Hi, I'm wondering if this is the proper way to approach this problem. The question says to: a)find the electric field at the center of curvature of the hemisphere (center of the flat bottom). Homework Equations Gauss's law: integral E*da = Qencl/epsilon...

Hi I've been working on this problem repeatedly and thought I understood how to solve separable equations problems but I keep getting the wrong answer. y' = (1-2x)y^2 Here's what I got for the problem: y'y^2 = (1-2x) \int(1\y^2) dy = \int(1-2x)dx ln | y^2| = x-x^2 + C e^ln|y^2|...