# Electric flux for a rectangular surface? [closed]

I have the following homework problem:

A line of charge $\lambda$ is located on the z-axis. Determine the electric flux for a rectangular surface with corners at coordinates: $(0, R, 0)$, $(w, R, 0)$, $(0,R, L)$, and $(w, R, L)$.

This is what I have come up with so far:

The line of charge, is located on the $z$-axis. We can recall that $\Phi = \int_{S}\vec{E}~\mathrm{d}A$. We initially note that this is parallel to the $xz$-plane, ergo we will integrate in respect to $x$ and $z$. Due to the fact that this is an infinite line of charge, there is no change in the field as we vary the distance along the $z$-axis. Our integral is $$\int_0^L\int_0^w\vec{E}~\mathrm{d}x\mathrm{d}z$$ We can recall that $$E=\frac{\lambda}{2\pi\varepsilon_0r}$$ We can see that $r=\sqrt{x^2+R^2}$ by the Pythagorean Theorem. By substitution we have the following integral: $$\int_0^L\int_0^w\frac{\lambda}{2\pi\varepsilon_0\sqrt{x^2+R^2}}\mathrm{d}x\mathrm{d}z = \frac{\lambda L}{2\pi\varepsilon_0}\int_0^w\frac{1}{\sqrt{x^2+R^2}}\mathrm{d}x$$

When I solve this I get:

$$\Phi=\frac{\lambda L}{2\pi\varepsilon_0}\sinh^{-1}\left(\frac{w}{R}\right)$$

I am not sure where I am going wrong. I may be doing something conceptually incorrect.

## closed as off-topic by David Z♦Nov 14 '13 at 3:38

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Here is the solution in my opinion: One thing you forgot is that flux involves the force perpendicular to the area. F.dA remember? So the solution integral would be $\frac{\lambda}{L2\pi\epsilon_0} \int\frac{1}{\sqrt{x^2 + R^2}}\sin\theta\ \mathrm{d}x\mathrm{d}z$. If you look at the figure attached

\begin{align}F &= \int_0^L \int_0^w \frac{\lambda}{2\pi\epsilon_0}\frac{1}{\sqrt{x^2 + R^2}}\sin\theta\ \mathrm{d}x\mathrm{d}z \\ &= \int_0^L \int_0^w \frac{\lambda}{2\pi\epsilon_0}\frac{R}{(\sqrt{x^2 + R^2})^2}\sin\theta\ \mathrm{d}x\mathrm{d}z\qquad\biggl(\sin\theta = \frac{R}{\sqrt{x^2 + R^2}}\biggr) \\ &= \frac{\lambda R}{2\pi\epsilon_0}\int_0^L \int_0^w \frac{1}{(x^2 + R^2)}\sin\theta\ \mathrm{d}x\mathrm{d}z \\ &= \frac{\lambda L}{2\pi\epsilon_0 R} \int_0^w \frac{1}{1 + (x/R)^2}\ \mathrm{d}x \\ &= \frac{\lambda L}{2\pi\epsilon_0}\biggl[\tan^{-1}\biggl(\frac{x}{R}\biggr)\biggr]_0^w \\ &= \frac{\lambda L}{2\pi\epsilon_0}\tan^{-1}\biggl(\frac{w}{R}\biggr) \end{align}

• actually i find this way of posting answer more readable than the MathJax however if you want to learn loo MathJax in a easy way just see different question and answers and click edit to see how they are written it is lot easy – Dimensionless Nov 13 '13 at 14:53
• @Akash if you find that easier to read you're in the distinct minority. Equations (and text) should always be transcribed on this site; we strongly discourage members from posting screenshots or photos of handwritten equations. big_biceps, I'll transcribe it for you this time, but in the future, see e.g. this. But more importantly, please don't post complete solutions to homework problems! We usually delete them. – David Z Nov 14 '13 at 3:41
• @DavidZ got it. – big_biceps Nov 14 '13 at 5:35
• @DavidZ i just find it more readable but yeah it is wrong i just said i also never post answer like this but is said it so sorry not gonna be again – Dimensionless Nov 14 '13 at 9:35

flux is not a vector. hence you have to calculate E.dA . write the electric field with direction and then take its dot product with area vector and then do the double integral.

I think the integral would simply be $Φ=(λ/L2πε0)\sin^{−1}(w/R)$ and not $\sinh^{-1}$.

• Is that correct? The answer my professor provided is $\tan^{-1}(w/R)$, but I do not know how he got there. – yousuf soliman Nov 13 '13 at 9:31
• First of all, I apologize. the integral sinh-1 is correct. Let me write the solution in a fresh answer box. – big_biceps Nov 13 '13 at 12:18