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Homework Statement
Find the total electric flux through the closed surface defined by [tex]p = 0.26, z = \pm 0.26[/tex] due to a point charge of [tex]60\mu C[/tex] located at the origin.
Note that in this question, p is defined to be what r is defined conventionally, and [itex]\phi[/itex] takes the place of [itex]\theta[/itex]. This is the book's notation, not mine and I'm using it here.
I'm trying to do the surface integral in terms of cylindrical basis vectors, henceforth denoted as [tex]\check{p}, \check{\phi}, \check{z}[/tex]
Homework Equations
I'm trying to do this question without using Gauss law or the divergence theorem.
The Attempt at a Solution
Now the surface can be seen to be that of a cylinder of radius 0.26, with the circular tops and bottom at z = -0.26 and 0.26.
The electric displacement flux density [itex]\vec{D}[/tex], I have worked out it to be [tex]\vec{D}(p,\phi,z) = \frac{Q}{4\pi} \frac{1}{\sqrt{p^2 + z^2}} \left ((\cos \phi + \sin \phi)\check{p} + (\cos \phi - \sin \phi) \check{\phi} + \check{z} \right)[/tex].
I hope I got it right.
Now the surface integral is to be performed piecewise, for the top and bottom pieces and the curved sides.
So for the curved side. the surface integral is given by [tex]\int_{-0.26}^{0.26} \int_0^{2\pi} \frac{Q}{4\pi} \frac{1}{\sqrt{0.26^2 + z^2} \cos \phi + \sin \phi d\phi dz[/tex].
I've omitted some steps such as the dot product of D with the normal vector (1,0,0) in cylindrical basis vectors.
This, strangely evaluates to 0.
For the other two tops, the surface integral is [tex]\int_0^{2\pi} \int_0^0.26 \frac{Q}{4\pi} \frac{1}{\sqrt{p^2 + 0.26^2} dp d\phi[/tex]. Multiplying this result by 2, because of the symmetrical configuration gives [tex]Q ln (\sqrt{2} +1)[/tex].
But of course this isn't the case, since we know that the end result should be just Q by Gauss law, the charge enclosed in the cylinder. Where is my mistake?