# Spherical integration for cone to find electric potiential

I am integrating a hollow cone (point at origin) to get the electric potiential at a point $$b$$ at the center of the cone (at a height $$h$$ --also assume the radius of the circular top to be length h at this height). Assume a uniform surface charge density $$\sigma$$ I want to find the electric potiential at point $$b$$.

I am aware of the "ring slice method" and my issue is not with what is the final value but why is it that using spherical coordinates for integration does not work? I have tried converting the position vector to using cartesion base vectors and using still spherical coordinate variables but I still do not get the right answer. To make things clear I have tried utilizing the following integrals using spherical coordinates;

$$\frac{\sigma \sin (\theta ) }{2 \epsilon }\left(\int \frac{r'}{r-r'} \, dr\right)$$

where $$\theta$$ is $$45^\circ$$ and $$r=h$$ (the point $$b$$)

*note some simplification took place, I thought this would be fine since the integrand is not a vector but it appears not to be so, since the answer does not come out right. So next I thought maybe the vector nature of " $$R=r-r'$$ " is somehow affecting the so that's when I tried,

putting things in terms of the cartesian base vector form, that is,

$$r \sin (\theta ) \sin (\phi ) \hat{x}+r \sin (\theta ) \cos (\phi ) \hat{y}+r \cos (\theta ) \hat{z}$$

So that my integral becomes

$$\frac{\sigma \sin (\theta ) }{\sqrt{2} \epsilon }\left(\int_0^{\sqrt{2}h} \frac{r}{\sqrt{\left(r-r'\right)^2+\left(r'\right)^2}} \, dr\right)$$

which is pretty close to a step in the solution manual however not quite, the funny thing is that if I use the substitution $$r'=l/\sqrt{2}$$ I get the same form as the solution value. This does not make sense though because I am already integrating the hypotenuse variable which is the longest side of the triangle, so what could $$l$$ represent!

I am very confused and could use some help. I hope I explained things clearly enough but if need be I can attach pictures.

P.S. sorry I did not include the limits of integration, my knowledge on latex is not very good and this is the best I could do. The first integral I already have calculated (the $$2\pi$$ result) and simplified, the limits in the remaining integral should be from $$0$$ to $$\sqrt{2}h$$

• meta.math.stackexchange.com/q/5020 i am still learning it myself and I am pretty sure the integral font can be enlarged
– user108787
Commented Sep 12, 2016 at 22:10

In spherical coordinates, consider a point $\vec{r}$ on the cone, at distance/radius r from the tip and azimuthal angle $\phi$. The cone surface element at $\vec{r}$, of length $dr$ and width $r\sin\theta \; d\phi$, gives a corresponding charge element $$d\sigma = \sigma r\sin\theta \; d\phi dr$$ while the distance from $\vec{r}$ to the point $b$ at the cone's center reads $$d = \sqrt{r^2\sin^2\theta + (h - r\sin\theta)^2}$$ For $\theta=\pi/4$ these become $$d\sigma = \sigma \sqrt{2} \;(r/\sqrt{2}) \; d\phi \;d(r/\sqrt{2})$$ $$d = \sqrt{(r/\sqrt{2})^2 + (h - r/\sqrt{2})^2}$$ Substitute into the expression for the potential and you should retrieve what you are looking for.