Question

Two long, straight copper pipes, each of radius R, are held a distance

2d apart. One is at potential ${\mathit{V}}_{\mathbf{0}}$, the other at $\mathbf{-}{\mathit{V}}_{\mathbf{0}}$(Fig. 3.16). Find the potential

everywhere. [Hint: Exploit the result of Prob. 2.52.]

Short answer

The potential at all the points are$V\left(x,y\right)=\frac{\lambda }{4\pi {\epsilon }_0}\ln \frac{z^2+{\left(y+a\right)}^2}{z^2+{\left(y-a\right)}^2}$ with $\lambda =\frac{2\pi {\epsilon }_0{V}_0}{\cosh \left(\frac{d}{R}\right)}$, $a=R\sqrt{{\cosh }^2\left({\cosh }^{-1}\left(\frac{d}{R}\right)\right)-1}$with $a=\sqrt{d^2-R^2}$.

Step-by-step solution

Given data

The following figure consists of two long straight copper pipes separated by distance 2d on x-

axis. One pipe is at potential $-V_0$and another pipe is at $+V_0$.

Write the expression for the charge due to the wire of charge density $+\lambda$at point $\left(x,y,z\right)$is,

${\mathbf{V}}_{\mathbf{+}}\mathbf{=}\mathbf{-}\left(\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{In}\frac{{\mathbf{s}}_{\mathbf{+}}}{\mathbf{d}}\right)$ …… (1)

Here, $S_{+}$is the distance from $+\lambda$to $\left(x,y,z\right)$.

Write the expression for the charge due to wire of charge density $-\lambda$at point (x, y, z) is,

${\mathbf{V}}_{\mathbf{-}}\mathbf{=}\left(\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{In}\frac{{\mathbf{s}}_{\mathbf{-}}}{\mathbf{d}}\right)$ …… (2)

Here, localid="1658318788283" $S_{-}$is the distance from $-\lambda$to $\left(x,y,z\right)$.

Determine total potential

Write the expression for the total potential.

$\mathit{V}\mathbf{=}\mathit{V}\mathbf{+}\mathit{V}$ ……. (3)

Here, V is the total potential.

Substitute the $-\left(\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\frac{S_{+}}{d}\right)$for $V_i$and $\left(\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\frac{S_{-}}{d}\right)$for V in above equation (3).

$$\begin{gathered} \mathit{V}\mathbf{=}\mathbf{-}\left(\frac{\lambda }{2\pi {\epsilon }_0}In\frac{S_{+}}{d}\right)\mathbf{+}\left(\frac{\lambda }{2\pi {\epsilon }_0}In\frac{S_{-}}{d}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\left[In\frac{S_{-}}{d}-In\frac{S_{+}}{d}\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\left(\frac{S_{-}}{S_{+}}\right) \end{gathered}$$ ……. (4)

From the figure, the distance ${\mathit{S}}_{\mathbf{+}}$and ${\mathit{S}}_{\mathbf{-}}$respectively are,

$$\begin{gathered} {\mathit{S}}_{\mathbf{+}}\mathbf{=}\sqrt{{\left(x-d\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}} \\ {\mathit{S}}_{\mathbf{-}}\mathbf{=}\sqrt{{\left(x+d\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}} \end{gathered}$$ ……. (5)

Substitute $\sqrt{{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$for ${\mathit{S}}_{\mathbf{+}}$and $\sqrt{{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$for ${\mathit{S}}_{\mathbf{-}}$in equation (5)

$$\begin{gathered} \mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\frac{\sqrt{{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}}{\sqrt{{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\frac{{\left({\left(x+d\right)}^2+y^2\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}}{{\left({\left(x-d\right)}^2+y^2\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}{\left(\frac{{(x+d)}^2+y^2)}{{(x-d)}^2+y^2}\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}}{{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)} \end{gathered}$$

Simplify the above equation, then the potential is,

$\mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}}{{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}$

Thus, the Potential at any point is $\mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}}{{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{d}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}$.

Determine charge density

The potential is constant at all the places on the equipotential surface. Hence from equation (1) $\frac{{\left(x+a\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}{{\left(x-a\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}$is constant (k) .

Therefore,

$$\begin{gathered} \frac{{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}\mathbf{=}\mathit{k} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}\mathbf{2}\mathit{a}\mathit{x}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathit{k}\left[x^2+a^2-2ax+z^2\right] \\ {\mathit{x}}^{\mathbf{2}}\left[k-1\right]\mathbf{+}{\mathit{a}}^{\mathbf{2}}\left[k-1\right]\mathbf{+}{\mathit{z}}^{\mathbf{2}}\left[k-1\right]\mathbf{-}\mathbf{2}\mathit{a}\mathit{x}\left[k+1\right]\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{a}\mathit{x}\frac{\left(k+1\right)}{\left(k-1\right)}\mathbf{=}\mathbf{0} \end{gathered}$$ …(6)

localid="1656933549284" ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{x}\mathit{a}\frac{(k+1)}{(k-1)}\mathbf{+}\left(a^2+z^2\right)\mathbf{=}\mathbf{0}$

Add the ${\left[a\frac{\left(k+1\right)}{\left(k-1\right)}\right]}^{\mathbf{2}}$on both sides.

$$\begin{gathered} {\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{x}\mathit{a}\frac{(k+1)}{(k-1)}\mathbf{+}{\mathbf{\left[}\mathbf{a}\frac{(k+1)}{(k-1)}\mathbf{\right]}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}{\mathbf{\left[}\mathbf{a}\frac{(k+1)}{(k-1)}\mathbf{\right]}}^{\mathbf{2}}\mathbf{-}{\mathit{a}}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{[}\mathbf{x}\mathbf{-}\mathbf{a}\frac{(k+1)}{(k-1)}\mathbf{]}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{[}{\mathbf{\left(}\frac{\mathbf{k}\mathbf{+}\mathbf{1}}{\mathbf{k}\mathbf{-}\mathbf{1}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{[}\mathbf{x}\mathbf{-}\mathbf{a}\frac{(k+1)}{(k-1)}\mathbf{]}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{\left[}\frac{\mathbf{4}\mathbf{k}}{{\mathbf{(}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{\right]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{[}\mathbf{x}\mathbf{-}\mathbf{a}\frac{(k+1)}{(k-1)}\mathbf{]}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}{\mathbf{\left[}\frac{\mathbf{2}\mathbf{a}\sqrt{\mathbf{k}}}{\mathbf{k}\mathbf{-}\mathbf{1}}\mathbf{\right]}}^{\mathbf{2}} \end{gathered}$$

The above expression is write as,

${\left[x-y_0\right]}^{\mathbf{2}}\mathbf{+}{\left(z-z_0\right)}^{\mathbf{2}}\mathbf{=}{\mathit{R}}^{\mathbf{2}}$ …… (7)

Here, ${\mathit{y}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{a}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}{(k-1)}$and ${\mathit{z}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$

Substitute the value of ${\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{z}}_{\mathbf{0}}$in equation (7).

Then the expression for R is

$\mathit{R}\mathbf{=}\frac{\mathbf{2}\mathbf{a}\sqrt{\mathbf{k}}}{\mathbf{k}\mathbf{-}\mathbf{1}}$

Thus, the represents circular cylinder with axis parallel to x-axis centered at

$\left(y_0,z_0\right)\mathbf{=}\left(\frac{a\left(k+1\right)}{k-1},0\right)$and radius $\mathit{R}\mathbf{=}\frac{\mathbf{2}\mathbf{a}\sqrt{\mathbf{k}}}{\mathbf{k}\mathbf{-}\mathbf{1}}$.

Let’s assume that, potential corresponds to ${\mathit{V}}_{\mathbf{0}}$, then

${\mathit{V}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathit{k}$

Rewrite the above equation for $\mathit{I}\mathit{n}\mathit{k}$

$$\begin{gathered} \frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{=}\mathit{I}\mathit{n}\mathit{k} \\ {\mathit{e}}^{\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}{\mathbf{\lambda }}}\mathbf{=}\mathit{k} \end{gathered}$$

Let’s consider that, $\mathit{P}\mathbf{=}\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}{\mathbf{\lambda }}$then $\mathit{k}\mathbf{=}{\mathit{e}}^{\mathbf{P}}$.

Now,

$$\begin{gathered} {\mathit{y}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{a}\left(k+1\right)}{\mathbf{k}\mathbf{-}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{a}\left(e^P+1\right)}{{\mathbf{e}}^{\mathbf{P}}\mathbf{-}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{a}\left(\frac{e^{P/2}+e^{-P/2}}{e^{P/2}-e^{-P/2}}\right) \end{gathered}$$

Then,

${\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathit{h}\left(\frac{P}{2}\right)$

Substitute $\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}$for P in above equation.

$$\begin{gathered} {\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathit{h}\mathbf{\left(}\frac{\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}}{\mathbf{2}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathit{h}\mathbf{\left(}\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{\right)} \end{gathered}$$ …… (8)

Substitute ${\mathit{e}}^{\mathbf{P}}$for k in $\mathit{R}\mathbf{=}\frac{\mathbf{2}\mathbf{a}\sqrt{\mathbf{k}}}{\mathbf{k}\mathbf{-}\mathbf{1}}$equation.

$$\begin{gathered} \mathit{R}\mathbf{=}\frac{\mathbf{2}\mathbf{a}\sqrt{{\mathbf{e}}^{\mathbf{P}}}}{{\mathbf{e}}^{\mathbf{P}}\mathbf{-}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathit{a}\frac{{\mathbf{e}}^{\mathbf{P}\mathbf{/}\mathbf{2}}}{{\mathbf{e}}^{\mathbf{P}}\mathbf{-}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathit{a}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{P}\mathbf{/}\mathbf{2}}\mathbf{-}{\mathbf{e}}^{\mathbf{P}\mathbf{/}\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}}{{\mathbf{e}}^{\mathbf{P}\mathbf{/}\mathbf{2}}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{P}\mathbf{/}\mathbf{2}}} \end{gathered}$$

So,

$\mathit{R}\mathbf{=}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{}\mathit{e}\mathit{c}\mathit{h}\mathbf{\left(}\frac{\mathbf{P}}{\mathbf{2}}\mathbf{\right)}$

Substitute the value $\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}$for P in above equation.

$\mathit{R}\mathbf{=}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{e}\mathit{c}\mathit{h}\mathbf{\left(}\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{\right)}$ ….... (9)

Hence, the radius of the cylinder corresponding to given ${\mathit{V}}_{\mathbf{0}}$is $\mathit{R}\mathbf{=}\mathbf{}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{e}\mathit{c}\mathit{h}\mathbf{\left(}\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{\right)}$.

As, ${\mathit{y}}_{\mathbf{0}}\mathbf{\to }\mathit{d}$equation (8) becomes,

$\mathit{d}\mathbf{=}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathit{h}\mathbf{\left(}\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{\right)}$ ......(10)

And from equation (9),

$\mathit{R}\mathbf{=}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{e}\mathit{c}\mathbf{}\mathit{h}\mathbf{\left(}\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{\right)}$ ......(11)

Divide the equation (10) with (11)

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{d}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{a}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{h}\left({\displaystyle \frac{2{\mathrm{\pi \epsilon }}_0{V}_0}{\lambda }}\right)}{\mathbf{a}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{e}\mathbf{c}\mathbf{}\mathbf{h}\left(\frac{2\pi {\epsilon }_0{V}_0}{\lambda }\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{d}}{\mathbf{R}}\mathbf{=}\left[\frac{cosh\left({\displaystyle \frac{2{\mathrm{\pi \epsilon }}_0{V}_0}{\lambda }}\right)}{sinh\left(\frac{2\pi {\epsilon }_0{V}_0}{\lambda }\right)}\right]\left[sinh\left(\frac{2{\mathrm{\pi \epsilon }}_0{V}_0}{\lambda }\right)\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{d}}{\mathbf{R}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{h}\left(\frac{2{\mathrm{\pi \epsilon }}_0{V}_0}{\lambda }\right) \\ \frac{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}{\mathit{h}}^{\mathbf{-}\mathbf{1}}\left(\frac{d}{R}\right) \end{gathered}$$

Thus,

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{{\mathbf{cosh}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}{\displaystyle \frac{d}{R}}\mathbf{\right)}}$

Hence, the linear charge density is $\mathit{\lambda }\mathbf{=}\frac{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{c}\mathbf{o}\mathbf{s}{\mathbf{h}}^{\mathbf{-}\mathbf{1}}\left({\displaystyle \frac{d}{R}}\right)}$.