[YMP/EM-03001] Potential Due to Conical Surface

Question
A conical surface ( an empty ice cream cone ) carries a uniform surface charge density $\sigma_0$. The height of the cone and the radius of the base are both equal to $a$. Find the potentials at the vertex of the cone and at the center of the base.

Solution
We divide the conical surface into thin rings by planes parallel to the base. Let $2 \alpha$ be the angle of the cone.

\FigNoNumX{50,10}{60}{60}{MidSemAnsQ1}{}

The radius of the ring $ = x \tan\alpha=x.$
The slant height of the ring $= \sqrt{2}dx$. (width along
the curved surface)
$\therefore$ area of the ring $= 2\pi x \sqrt{2} dx$
and the total charge on the ring =$2\pi x\sqrt{2} dx \sigma_0$.
Potential at the vertex, $P$, due to this ring shaped element of area, is
\begin{eqnarray}
d\phi(P)&=& \frac{1}{4\pi\epsilon_0}\times\sigma_0 2\pi x\surd{2}dx\,
\times \frac{1}{\surd{2}x}
=\frac{\sigma}{2\varepsilon_0}\, dx.
\end{eqnarray}

Hence the potential due to the cone at the vertex, P, is
\begin{eqnarray}
\phi(P) &=& \int_0^a\
\Big(\frac{\sigma}{2\epsilon_0}\Big) dx
= \left(\frac{\sigma a}{2 \varepsilon_0}\right).
\end{eqnarray}
The potential at the center of the base, Q, is due to the ring element at $x$ is
\begin{eqnarray}
d\phi(Q) &=& \Big(\frac{1}{4\pi\varepsilon_0} \Big)
{2\pi x \surd 2 \, dx\sigma_0 }\times
\frac{1}{\sqrt{x^2+(a-x)^2}}\\
&=& \frac{\surd 2\sigma_0}{2\varepsilon_0} \frac{x \,dx}
{\sqrt{2x^2 -2ax +a^2}}
\end{eqnarray}
$\therefore$ the potential at $Q$ is
\begin{eqnarray}
\phi(Q) &=& \frac{\surd2\sigma_0}{2\varepsilon_0}\int_0^a
\frac{x\,dx}{\sqrt{2x^2-2ax +a^2}}\
\equiv \frac{\sigma}{2\varepsilon_0} I \\
\text{where } &I& =\int_0^a \frac{x\,dx}{\sqrt{x^2-ax +a^2/2} }
\end{eqnarray}

Computing the integral $I$

\begin{eqnarray}
I &=& \int_0^a \frac{x\,dx}{\sqrt{x^2-ax +a^2/2} }\\
&=& \int_0^a \frac{(x-a/2)\,dx}{\sqrt{x^2-ax +a^2/2} }
+ \frac{a}{2}\int_0^a \frac{dx}{\sqrt{x^2-ax +a^2/2} }\\
&=& \sqrt{x^2-ax +a^2/2} \Big|_0^a + \frac{a}{2}
\int_0^a\frac{dx}{\sqrt{(x-a/2)^2 + a^2/4)}} \Label{E1}\\
&=&\frac{a}{2} \sinh^{-1}\left.\Big(\frac{(x-a/2)}{a/2}\Big) \right|_0^a\\
&=& a \sinh^{-1}(1)
\end{eqnarray}

The integral appearing in \EqRef{E1}) can also be written as
\begin{eqnarray}
I=\frac{a}{2}\int_0^a\frac{dx}{\sqrt{(x-a/2)^2 + a^2/4)}}& = &
\int_{-a/2}^{a/2}\frac{dt}{\sqrt{t^2+a^2/4}} \\
&=& \frac{a}{2}\log\left(t+
\sqrt{t^2+a^2/4}\right)\Big|_{-a/2}^{a/2}\\
&=&\frac{a}{2} \log\Big(\frac{\surd2+1}{\surd 2 -1}\Big) \\
&=& a \log(\surd2 +1)
\end{eqnarray}
Hence the final answer for the potential at point \(Q\) is
\begin{eqnarray}
\phi(Q) = \frac{\sigma}{2\varepsilon_0} I=\frac{a}{2\varepsilon_0}\log(\surd2
+1)
\end{eqnarray}

Comments
The curved area of a cone is \(2\pi a \ell\) where \(\ell\) is the slant height. Written in terms of height \(h\) it becomes \(2\pi a h \sec \alpha\) The curved area with height \(h\) in the range \(x\) and \(x+dx\) is \\
\[ 2\pi a (x+dx)\sec\alpha - 2\pi x \sec \alpha = 2\pi a \sec\alpha\, dx\]
{Thus it is correct to take the slant height in computation of the area of the strip.}