# Griffiths (electrostatics) problem

#### Kolahal Bhattacharya

A conicalsurface(an empty ice cream cone) carries a uniform surface charge density sigma.The height & radius of the cone are h & R.Find potential difference between apex & centre of the top

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#### Hootenanny

Staff Emeritus
Gold Member
Hi Kolahal and welcome to PF,

Could please show your thoughts and what you have attempted thus far?

HINT: Get your calculus books out.

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#### Kolahal Bhattacharya

I have started the problem considering a ring ll to circular base.Expressed its radius in terms of its height,R,h.Then, I used the integral(Griffiths:2.30):-(1/4 pi epsilon not) integral over (sigma/curly r)da
sigma= surface charge density,curly r=lr-r'l,da=area element

#### Reshma

Kolahal Bhattacharya said:
I have started the problem considering a ring ll to circular base.Expressed its radius in terms of its height,R,h.Then, I used the integral(Griffiths:2.30):-(1/4 pi epsilon not) :rofl: integral over (sigma/curly r)da
sigma= surface charge density,curly r=lr-r'l,da=area element
OK, you mean this:
$$V(r) = \frac{1}{4\pi \epsilon_0}\int \frac{\sigma}{r}da$$
Just a few hints:
Express $da$ in terms of "r" and evaluate the electric potential at two points, viz. the apex and the centre of its surface.

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#### Hootenanny

Staff Emeritus
$$V = \frac{1}{4\pi\epsilon_{0}} \int \frac{\sigma}{|r-r'|} \; da$$