Q26P - Introduction to Electrodynamics

Question

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).

Step-by-Step Solution

Answer

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

Step 1: Define functions

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

Step 2: Determine value of potential at point

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${\mathrm{V}}_{\mathrm{a}}=\frac{\mathrm{\sigma h}}{2{\mathrm{\epsilon }}_0}$.

Step 3: Determine value of potential at center of the top

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

Step 4: Determine difference between and

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$V_a$and$\frac{\sigma h}{4{\epsilon }_0}In\left(1+\sqrt{2}\right)$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$V=\frac{\sigma h}{2{\epsilon }_0}\left(In\left(1+\sqrt{2}\right)-1\right)$.