A cone with apex at the origin has a height $h$ and a top radius $h$, a uniform charge density with no charge on the top face.

I need to find the potential $V$ at a position $z$ on the cone's axis using spherical coordinates. I'm told I won't be able to evaluate the $r'$ integral analytically so my answer will have the integral expression.

I've done the steps in the photo (please see) but I don't see why I can't evaluate the integral at the end (could use wolframalpha.com for example). Please assist me - what did I do wrong? Or is my work correct?

my attempt at the problem


For the final integral, first complete the square in the denominator to get ${r'^2+z^2-\sqrt 2 zr'}=(r'-\frac{\sqrt2}{2}z)^2+\frac{z^2}{2}$. The integral then is: $$\int_0^h dr' \frac{r'}{\sqrt{(r'-\frac{\sqrt2}{2}z)^2+\frac{z^2}{2}}}$$ Now use the substitution $r' - \frac{z}{\sqrt2} = \frac{z}{\sqrt 2}\tan u$ to get: $$\int_{-\pi/4}^{\tan^{-1}(\sqrt 2h/z - 1)}\frac{\frac {z^2} {2}(\tan u + 1)(1+\tan^2u)}{\frac{z}{\sqrt 2}(1+\tan^2u)^{1/2}}du$$ $$=\frac{z}{\sqrt 2}\int_{-\pi/4}^{\tan^{-1}(\sqrt 2h/z - 1)}{(\tan u + 1)(1+\tan^2u)^{1/2}}du$$ $$=\frac{z}{\sqrt 2}\int_{-\pi/4}^{\tan^{-1}(\sqrt 2h/z - 1)}{\frac{(\tan u + 1)}{|\cos u|}}du$$ $$=\frac{z}{\sqrt 2}\int_{-\pi/4}^{\tan^{-1}(\sqrt 2h/z - 1)}{\frac{(\tan u + 1)}{\cos u}}du$$ $$=\frac{z}{\sqrt 2}\int_{-\pi/4}^{\tan^{-1}(\sqrt 2h/z - 1)}{\frac{\sin u}{\cos^2 u}}du+\frac{z}{\sqrt 2}\int_{-\pi/4}^{\tan^{-1}(\sqrt 2h/z - 1)}{\sec{u} \ }du$$ The first integral is easily solved by the substitution $x=\cos u$, turning it into a $\int \frac{dx}{x}=\ln x +c$ type integral. The second integral is a simple secant integral $\int\sec u du = \ln|\sec u + \tan u|+c$.

I've left the final calculations for you to go through. Also note that you've written $z = r \cos \theta = r \cos 0 = r$ on the top of the page which is incorrect ($\theta = \pi/4$). I would just write the potential as a function of $z$, so you just have to replace $r$ with $z$ in your calculations (which is what I've done above), since you've incorrectly assumed that $r=z$.

  • $\begingroup$ Thank you very much. But does this make sense based on the physics problem as its stated? $\endgroup$ – Arthur Alex Karapetov Nov 5 '18 at 3:12
  • $\begingroup$ Check my edited answer. Other than that, it looks OK to me. $\endgroup$ – Sahand Tabatabaei Nov 5 '18 at 3:38

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