1
$\begingroup$

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 $\sqrt{2}h$

$\endgroup$
1
  • $\begingroup$ 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

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.