Elecric flux = Integral of closed surface of E dotted into dA = qenclosed / 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.
EA =qenclosed / ε0
I believe I have to use the function of the Rho in the integral
qenclosed = 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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