# Conical Surface Potential Difference

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1. Mar 27, 2016

### Aroldo

Hey guys!
The question is related to problem 2.26 from Electrodynamics by Griffiths (3ed).

1. The problem statement, all variables and given/known data

A conical surface (an empty ice-cream cone) carries a uniform surface charge σ. The height of the cone is h, as the radius of the top. Find the potential difference between points a (the vertex) and b (the center of the top).

2. Relevant equations
Here I will call the potential V.

First of all, I assumed that at the vertex: V(a) = 0. (I can do that because I'm interested in V(b) - V(a), am I right?)

Then I calculated V(b). So:
V(a) - V(b) = - V(b) = -σh/(2ε) * ln (1 + (21/2/2))

But the book's solution didn't consider V(a) = 0, and found:
V(a) - V(b) = σh/(2ε) [1 - ln (1 + (21/2/2))]

Finally, my questions are:
Why is my assumption wrong?
How to calculate it assuming V(b) = 0?

2. Mar 27, 2016

### TSny

In deriving your expression for V(b), where did you choose V to be zero?

3. Mar 27, 2016

### Aroldo

At the point a = (0,0,0)

4. Mar 27, 2016

### TSny

Can you show an outline of your derivation of your expression for V(b)?

5. Mar 27, 2016

### Aroldo

Yes, of course.
r is the vector along the central axis and r' is the vector along the conical surface.

$$V(b) = \frac{1}{4\pi\epsilon_0}\int_0^b{\frac{\sigma\cdot da'}{|\textbf{r}-\textbf{r}'|}} = \frac{\sigma}{4\pi\epsilon_0}\sqrt{2}\pi\int_0^{\sqrt{2}h}\frac{r'dr'}{\sqrt{r'^2 + h^2 -\sqrt{2}hr'}} = \frac{\sigma h}{2\epsilon_0}\ln (1+\sqrt{2})$$

Last edited: Mar 27, 2016
6. Mar 27, 2016

### TSny

The integrand represents the potential at b due to a small element of charge $\sigma da$. It has the form $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$ for a point charge. Note that this expression assumes a particular place where V = 0. Where is that place?

7. Mar 27, 2016

### Aroldo

For a point charge, V = 0 at the infinity.
You helped me a lot to identify my own misunderstanding. In the previous exercise that I solved the potential wouldn't go to 0 at the infinity (it was due to a infinity distribution).

Thank you a lot!

8. Mar 27, 2016

### TSny

OK.

I'm not quite getting the result that you quoted as the given solution. I get that the argument of the log should be $1 + \sqrt{2}$ rather than $1 + \frac{\sqrt{2}}{2}$ . But I could be messing up somewhere.

9. Mar 27, 2016

### Aroldo

No no, you are right. My mistake. I already edited there.

Thanks again.

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