damn Gauss law is so confusing!

[stunner5000pt]

Positive charge is uniformly distributed throughout a non conducting cylindrical shell of inner radius R and outer radius 2R. What radial depth beneath outer surface of the charge distribution is the elctric field strength equal to one half the surface value?

errr

e0 = epsilon 0 the permittivity

e0 EA = Q enc

thus E = Qenc / A e0

to find where E = Qenc / 2Ae0 im totally stumped

your help would be greatly appreciated!

[BLaH!]

Start by finding the value for the electric field at the outer surface of the cylinder. We do this by making a cylindrical Gaussian surface around the cylinder. So Gauss' Law reads:

EA = \frac{Q_{enc}}{\epsilon}

So first off we need to know what the surface area is enclosed in our Gaussian surface. The surface area of a cylinder is

A = 2\pi r l

where l is the length of the cylinder and r is its radius. In our case the length is infinity but don't worry about that right now. The next step is to find the total charge that is included in our Gaussian surface. We know that the surface is UNIFORMLY distributed throughout the cylinder. We also know (obviously) that the cylinder is a three dimensional object. Therefore, we can define a uniform volume charge density \rhothat is equal to a constant.

\rho = \rho_{0} = \frac{Charge}{Volume}

Thus, to find the amount of charge enclosed we must multiply (or integrate for a non-uniform charge density) by the volume. Thus, the charge enclosed in our Gaussian surface is

Q_{enc} = \rho_{0}V = \rho_{0} [\pi (2R)^2 - \pi R^2]l = \rho_{0}3\pi R^2l

Thus. the electric field at the surface of the cylinder is

E(2R) =\frac{3\rho_{0}R}{2\epsilon}

Notice that the l in the above equations cancels out. You can get the second part in a similar way that I have shown. To get the electric field as a function of r just substitute r everywhere I had 2R above.

[stunner5000pt]

i wouldnt have thought of using the p idea. I was wondering how to get an explicit value for p

thank you very very much!!!!!!!