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## Homework Statement

http://i.minus.com/jbxXiQZmmiOQtS.png [Broken]

## Homework Equations

Elecric flux = Integral of closed surface of E dotted into dA = q

_{enclosed}/ ε0

Rho = 50 x 10^-6 abs(z)

## The Attempt at a Solution

Integral of closed surface of E dotted into dA will end up being EA when E is constant.

as such,

EA =q

_{enclosed}/ ε0

I believe I have to use the function of the Rho in the integral

q

_{enclosed}= Intergal Rho * dV

As given we know to find dV we use the area (A) multiplied by dz.

Such as that

A * 50 x 10^-6 * Integral abs(z) dz

It here I believe that defining the limits tells us rather we are taking from within inside or outside. In my notes we are using a cylinder bar and Gaussian surface of a ring and going from 0 to R (R radius being that of cylinder bar) for inside and 0 to r (r being distance from center of cylinder bar to ring).

I have no idea what to put for my limits to get the inside and outside.

Moderator note: Moved to the introductory physics forum.

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