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 $\sqrt{2}h$

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

1 Answer 1


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.

  • $\begingroup$ Haven't had time yet to confirm your answer but I know where I went wrong in my 2nd integral, I treated the cos in the field vector to be 45 degrees when I should have had it set to zero, that gives me the correct answer. Still unsure why my initial integral does not work out though. $\endgroup$
    – skyfire
    Sep 13, 2016 at 14:24

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