91Ó°ÊÓ

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

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 ,

$\mathrm{V}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}∫\frac{\mathrm{σ}}{\mathrm{r}\text{'}}\mathrm{da}$

Here,$\mathrm{σ}$is the surface charge density, is the surface integral,${\mathrm{Î\mu }}_0$is the permittivity

for free space.

Consider the surface area is calculated as,$2\mathrm{Ï€°ù}$, therefore the value of is,

$da=2\mathrm{Ï€°ùdr}$

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 $2\mathrm{Ï€°ùdr}$with limits 0 to$\sqrt{2h}$in the equation for the volume.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{a}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}{∫}_0^{\sqrt{2\mathrm{h}}}\frac{\mathrm{σ}\left(2\mathrm{Ï€}\right)}{\mathrm{r}\text{'}}\left(\frac{\mathrm{r}\text{'}}{\sqrt{2}}\right)\mathrm{dr} \\ =\frac{2\mathrm{πσ}}{4{\mathrm{Ï€Î\mu }}_0\sqrt{2}}\left(\sqrt{2\mathrm{h}}\right) \\ =\frac{\mathrm{σ³ó}}{2{\mathrm{Î\mu }}_0} \end{gathered}$$

The value of potential at point is${\mathrm{V}}_{\mathrm{a}}=\frac{\mathrm{σ³ó}}{2{\mathrm{Î\mu }}_0}$.

Consider the following equation,

${\mathrm{V}}_{\mathrm{b}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}{∫}_0^{\sqrt{2\mathrm{h}}}\left(\frac{\mathrm{σ}2\mathrm{Ï€°ù}}{{\mathrm{r}}^{\mathrm{m}}}\right)\mathrm{dr}$

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$\frac{1}{4{\mathrm{Ï€Î\mu }}_0}{∫}_0^{\sqrt{2\mathrm{h}}}\left(\frac{\mathrm{σ}2\mathrm{Ï€°ù}}{{\mathrm{r}}^{\mathrm{m}}}\right)\mathrm{dr}$for$V_b$and integrate with

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

$$\begin{gathered} {\mathrm{V}}_{\mathrm{b}}=\frac{2\mathrm{πσ}}{4{\mathrm{Ï€Î\mu }}_0}\frac{1}{\sqrt{2}}{∫}_0^{\sqrt{2\mathrm{h}}}\frac{\mathrm{r}}{\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2-\sqrt{2}\mathrm{hr}}}\mathrm{dr} \\ =\frac{\mathrm{σ}}{2\sqrt{2}{\mathrm{Î\mu }}_0}{\left[\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2-\sqrt{2}\mathrm{hr}}+\frac{\mathrm{h}}{\sqrt{2}}\mathrm{In}\left[2\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2-\sqrt{2}\mathrm{hr}}+2\mathrm{r}-\sqrt{2\mathrm{h}}\right]\right]}_0^{\sqrt{2\mathrm{h}}} \\ =\frac{\mathrm{σ}}{2\sqrt{2}{\mathrm{Î\mu }}_0}\left[\mathrm{h}+\frac{\mathrm{h}}{\sqrt{2}}\mathrm{In}\left(2\mathrm{h}+2\sqrt{2\mathrm{h}}-\sqrt{2}\mathrm{h}\right)-\mathrm{h}-\frac{\mathrm{h}}{\sqrt{2}}\left(2\mathrm{h}-\sqrt{2}\mathrm{h}\right)\right] \\ =\frac{\mathrm{σ}}{2\sqrt{2}{\mathrm{Î\mu }}_0}\frac{\mathrm{h}}{\sqrt{2}}\left(\mathrm{In}\left(2\mathrm{h}+\sqrt{2}\mathrm{h}\right)\mathrm{In}\left(2\mathrm{h}-\sqrt{2}\mathrm{h}\right)\right) \end{gathered}$$

Solve further as,

$$\begin{gathered} V_b=\frac{\mathrm{σ}h}{4\mathrm{Î\mu }}In\left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right) \\ =\frac{\mathrm{σ}h}{4{\mathrm{Î\mu }}_0}In\left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)\left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right) \\ =\frac{\mathrm{σ}h}{4{\mathrm{Î\mu }}_0}In\left(\frac{2+\sqrt{2}}{\sqrt{2}}\right) \\ =\frac{\mathrm{σ}h}{4{\mathrm{Î\mu }}_0}In\left(1+\sqrt{2}\right) \end{gathered}$$

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

The potential difference between a and b is,

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

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

$$\begin{gathered} V=\frac{\mathrm{σ}h}{4{\mathrm{Î\mu }}_0}In\left(1+\sqrt{2}\right)-\frac{\mathrm{σ}h}{2{\mathrm{Î\mu }}_0} \\ =\frac{\mathrm{σ}h}{2{\mathrm{Î\mu }}_0}\left(In\left(1+\sqrt{2}\right)-1\right) \end{gathered}$$

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