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

Consider the conical surface, this surface carrying uniform charge density as $\mathit{\sigma }$.

The required diagram is shown below.

Here, h is the height of the cone, r is the radius of the cone at the distance r', P is the small infitesimal where the potential determines

Write the formula for find the potential at point a ,

$\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \frac{\mathrm{\sigma }}{\mathrm{r}\text{'}}\mathrm{da}$

Here,$\sigma$is the surface charge density, is the surface integral,${\mathrm{\epsilon }}_0$is the permittivity

for free space.

Consider the surface area is calculated as,$2\mathrm{\pi r}$, therefore the value of is,

$da=2\mathrm{\pi rdr}$

By using Pythagoras theorem, the value of r in terms of r' is determined.

$$\begin{gathered} r{\text{'}}^2=2r^2 \\ r=\frac{r\text{'}}{\sqrt{2}} \end{gathered}$$

Substitute $\frac{r\text{'}}{\sqrt{2}}$for r and $2\mathrm{\pi rdr}$with limits 0 to$\sqrt{2h}$in the equation for the volume.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{a}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}{\int }_0^{\sqrt{2\mathrm{h}}}\frac{\mathrm{\sigma }\left(2\mathrm{\pi }\right)}{\mathrm{r}\text{'}}\left(\frac{\mathrm{r}\text{'}}{\sqrt{2}}\right)\mathrm{dr} \\ =\frac{2\mathrm{\pi \sigma }}{4{\mathrm{\pi \epsilon }}_0\sqrt{2}}\left(\sqrt{2\mathrm{h}}\right) \\ =\frac{\mathrm{\sigma h}}{2{\mathrm{\epsilon }}_0} \end{gathered}$$

The value of potential at point is${\mathrm{V}}_{\mathrm{a}}=\frac{\mathrm{\sigma h}}{2{\mathrm{\epsilon }}_0}$.

Consider the following equation,

${\mathrm{V}}_{\mathrm{b}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}{\int }_0^{\sqrt{2\mathrm{h}}}\left(\frac{\mathrm{\sigma }2\mathrm{\pi r}}{{\mathrm{r}}^{\mathrm{m}}}\right)\mathrm{dr}$

By using Pythagoras theorem the value of$r^m$is determined.

$r^m=\sqrt{h^2+r^2-\sqrt{2hr}}$

Substitute $\sqrt{h^2+r^2-\sqrt{2hr}}$for$r^m$in$\frac{1}{4{\mathrm{\pi \epsilon }}_0}{\int }_0^{\sqrt{2\mathrm{h}}}\left(\frac{\mathrm{\sigma }2\mathrm{\pi r}}{{\mathrm{r}}^{\mathrm{m}}}\right)\mathrm{dr}$for$V_b$and integrate with

the limits 0 to $\sqrt{2h}$.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{b}}=\frac{2\mathrm{\pi \sigma }}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{\sqrt{2}}{\int }_0^{\sqrt{2\mathrm{h}}}\frac{\mathrm{r}}{\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2-\sqrt{2}\mathrm{hr}}}\mathrm{dr} \\ =\frac{\mathrm{\sigma }}{2\sqrt{2}{\mathrm{\epsilon }}_0}{\left[\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2-\sqrt{2}\mathrm{hr}}+\frac{\mathrm{h}}{\sqrt{2}}\mathrm{In}\left[2\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2-\sqrt{2}\mathrm{hr}}+2\mathrm{r}-\sqrt{2\mathrm{h}}\right]\right]}_0^{\sqrt{2\mathrm{h}}} \\ =\frac{\mathrm{\sigma }}{2\sqrt{2}{\mathrm{\epsilon }}_0}\left[\mathrm{h}+\frac{\mathrm{h}}{\sqrt{2}}\mathrm{In}\left(2\mathrm{h}+2\sqrt{2\mathrm{h}}-\sqrt{2}\mathrm{h}\right)-\mathrm{h}-\frac{\mathrm{h}}{\sqrt{2}}\left(2\mathrm{h}-\sqrt{2}\mathrm{h}\right)\right] \\ =\frac{\mathrm{\sigma }}{2\sqrt{2}{\mathrm{\epsilon }}_0}\frac{\mathrm{h}}{\sqrt{2}}\left(\mathrm{In}\left(2\mathrm{h}+\sqrt{2}\mathrm{h}\right)\mathrm{In}\left(2\mathrm{h}-\sqrt{2}\mathrm{h}\right)\right) \end{gathered}$$

Solve further as,

$$\begin{gathered} V_b=\frac{\sigma h}{4\epsilon }In\left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right) \\ =\frac{\sigma h}{4{\epsilon }_0}In\left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)\left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right) \\ =\frac{\sigma h}{4{\epsilon }_0}In\left(\frac{2+\sqrt{2}}{\sqrt{2}}\right) \\ =\frac{\sigma h}{4{\epsilon }_0}In\left(1+\sqrt{2}\right) \end{gathered}$$

The value of potential at point$V_b=\frac{\sigma h}{4{\epsilon }_0}In\left(1+\sqrt{2}\right)$

The potential difference between a and b is,

$\mathrm{V}={\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}$

Substitute$\frac{\sigma h}{2{\epsilon }_0}$for$V_a$andlocalid="1656137828545" $\frac{\sigma h}{4{\epsilon }_0}In\left(1+\sqrt{2}\right)$for$V_b$in above equation.

$$\begin{gathered} V=\frac{\sigma h}{4{\epsilon }_0}In\left(1+\sqrt{2}\right)-\frac{\sigma h}{2{\epsilon }_0} \\ =\frac{\sigma h}{2{\epsilon }_0}\left(In\left(1+\sqrt{2}\right)-1\right) \end{gathered}$$

Therefore, the potential difference between a and b is$V=\frac{\sigma h}{2{\epsilon }_0}\left(In\left(1+\sqrt{2}\right)-1\right)$.