(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

"A conical surface (an empty ice-cream cone) carries a uniform surface charge [tex]\sigma[/tex]. The height of the cone ish, as is the radius of the top. Find the potential at the centre of the top, taking infinity as reference point." - Griffiths

My result for the potential differs from the answer's. Can someone please check my solution

2. Relevant equations

[tex]V = \iint_S \frac{1}{4\pi \epsilon_0}\frac{\sigma}{x} dS[/tex], where [tex]x[/tex] is the distance from the source to the point.

3. The attempt at a solution

My diagram is in the attached picture.

First off, I used cylindrical coordinates with the equation z = r, z > 0 to graph the cone

I found [tex]dS = \sqrt{2}zdA[/tex]

The distance from any point on the cone to the origin is [tex]\sqrt{2} z[/tex], so using the cosine law,

[tex]x = \sqrt{2z^2-2hz+h^2}[/tex]

So we have

[tex]V = \int_0^h\int_0^{2\pi} \frac{1}{4\pi \epsilon_0} \frac{\sigma}{\sqrt{2z^2-2hz+h^2}} \sqrt{2}z d\theta dz=\frac{\sigma h}{4\epsilon}\ln(2\sqrt{2}+3)[/tex]

Answer: [tex]V = \frac{\sigma h}{2\epsilon}\ln(1+\sqrt{2})[/tex]

thanks!

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# Homework Help: The potential at points around a cone

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