# Electric flux through closed surface

by Defennder
Tags: electric, flux, surface
 HW Helper P: 2,616 1. The problem statement, all variables and given/known data Find the total electric flux through the closed surface defined by $$p = 0.26, z = \pm 0.26$$ due to a point charge of $$60\mu C$$ located at the origin. Note that in this question, p is defined to be what r is defined conventionally, and $\phi$ takes the place of $\theta$. 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 $$\check{p}, \check{\phi}, \check{z}$$ 2. Relevant equations I'm trying to do this question without using Gauss law or the divergence theorem. 3. 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 $$\vec{D}(p,\phi,z) = \frac{Q}{4\pi} \frac{1}{\sqrt{p^2 + z^2}} \left ((\cos \phi + \sin \phi)\check{p} + (\cos \phi - \sin \phi) \check{\phi} + \check{z} \right)$$. 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 $$\int_{-0.26}^{0.26} \int_0^{2\pi} \frac{Q}{4\pi} \frac{1}{\sqrt{0.26^2 + z^2} \cos \phi + \sin \phi d\phi dz$$. 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 $$\int_0^{2\pi} \int_0^0.26 \frac{Q}{4\pi} \frac{1}{\sqrt{p^2 + 0.26^2} dp d\phi$$. Multiplying this result by 2, because of the symmetrical configuration gives $$Q ln (\sqrt{2} +1)$$. 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?
 HW Helper P: 2,616 I think there's something wrong with the latex I entered earlier. These should be the correct last two expressions: $$\int_{-0.26}^{0.26} \int_0^{2\pi} \frac{Q}{4\pi} \frac{1}{\sqrt{0.26^2 + z^2}} (\cos \phi + \sin \phi) d\phi dz$$ $$\int_0^{2\pi} \int_0^{0.26} \frac{Q}{4\pi} \frac{1}{\sqrt{p^2 + 0.26^2}} dp d\phi$$
 HW Helper P: 2,616 Ok I worked through the expression fo D again and this time I got this: $$\vec{D}_{(p,\phi,z)} = \frac{Q}{4\pi} \left ( \frac{p}{p^2+z^2} \check{p} + \frac{z}{(p^2+z^2)^{3/2}} \check{z} \right)$$ This time, D doesn't depend on phi at all. Before I go any further, is this correct? Because I worked out the surface integral using this expression and didn't manage to get Q. EDIT: I don't know if the Latex display is working but it doesn't seem to reflect the changes I made to the code. I used http://rogercortesi.com/eqn/index.php to check the syntax and it seemed all right.
 HW Helper P: 2,616 Ok I managed to get that expression for D in cylindrical basis vectors, but somehow I still can't get the answer. Here's the flux through the curve surface: $$\frac{Q}{4\pi} \int_{0.26}^{-0.26} \int^{2\pi}_0 \frac{p}{(p^2+z^2)^{3/2}} d\check{\phi}dz$$ This works out to be $$\frac{Q}{0.26\sqrt{2} }$$ Flux through the cylindrical tops: $$\frac{Q}{4\pi} \int_{0}^{2\pi} \int^{0.26}_0 \frac{z}{(p^2+z^2)^{3/2}} dpd\phi$$ Multiplying this by 2 gives $$\frac{Q}{0.26\sqrt{2}}$$ Adding them up doesn't give Q, so did I setup the integrals correctly?