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

 

[Hootenanny]

Hi Kolahal and welcome to PF,

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

HINT: Get your calculus books out.

 

[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]

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

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.

 

[Hootenanny]

So you have (just so I can see it more clearly);

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

You need to express you change in area (dA) in terms of radii.

Edit: Rashma got there before me.