Q2.52P - Introduction to Electrodynamics

Question

Two infinitely long wires running parallel to the x axis carry uniform

charge densities $\mathbf{+}\mathit{\lambda }$ and $\mathbf{-}\mathit{\lambda }$.

(a) Find the potential at any point $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$ using the origin as your reference.

(b) Show that the equipotential surfaces are circular cylinders, and locate the axis

and radius of the cylinder corresponding to a given potential ${\mathit{V}}_{\mathbf{0}}$.

Step-by-Step Solution

Answer

Answer

1. The potential at the point $\left(x,y,z\right)$ is $\mathit{V}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left[}\frac{{\left(y+a\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}{{\left(y-a\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}\mathbf{\right]}$.

2. The radius of the cylinder corresponding to given ${\mathit{V}}_{\mathbf{0}}$ is $\mathit{R}\mathbf{=}\mathit{a}\mathbf{ }\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathit{e}\mathit{c}\mathit{h}\mathbf{\left(}\frac{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{\right)}$.

Step 1: Define functions and determine the potential at any point

Write the expression for potential.

$\mathit{V}\mathbf{=}\frac{\mathbf{k}\mathbf{q}}{\mathbf{r}}$                                                                                  …… (1)

Here, V is the potential, k is Coulombs constant, q is the charge, and r is the distance.

a)

Write the expression for the potential at a distance s from an infinitely long straight wire that carries a uniform line charge density $\mathit{\lambda }$.

$\mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{\mathbf{s}}{\mathbf{a}}\mathbf{\right)}$                                                                 …… (2)

The two indefinitely long wires run parallel to the x-axis and carry uniform charge density $\mathbf{+}\mathit{\lambda }$ and $\mathbf{-}\mathit{\lambda }$.

The two infinitely long wire is shown in the figure below.

Write the expression for potential due to the long wire having charge density $\mathbf{+}\mathit{\lambda }$ at point P.

${\mathit{V}}_{\mathbf{+}}\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\left(\frac{s_{+}}{a}\right)$                                              …… (3)

Here, ${\mathit{s}}_{\mathbf{+}}$ is the distance of point p from the long wire having charge density $\mathbf{+}\mathit{\lambda }$.

Write the expression for potential due to the long wire having charge density $\mathbf{-}\mathit{\lambda }$ at point P.

${\mathit{V}}_{\mathbf{-}}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{s}}_{\mathbf{-}}}{\mathbf{a}}\mathbf{\right)}$                                                …… (4)

Here, ${\mathit{s}}_{\mathbf{-}}$ is the distance of point P from the long wire having charge density $\mathbf{-}\mathit{\lambda }$.

Hence, the expression for the potential at point P is calculated as follows:,

$$\begin{gathered} \mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{+}}{\mathit{V}}_{\mathbf{-}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{-}\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{s}}_{\mathbf{+}}}{\mathbf{a}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{S}}_{\mathbf{-}}}{\mathbf{a}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left(}\frac{{\mathbf{s}}_{\mathbf{-}}}{{\mathbf{s}}_{\mathbf{+}}}\mathbf{\right)} \end{gathered}$$                               …… (5)

From the above figure, the expression for ${\mathit{s}}_{\mathbf{+}}$ and ${\mathit{s}}_{\mathbf{-}}$ .is calculated as follows:

$$\begin{gathered} {\mathit{s}}_{\mathbf{+}}\mathbf{=}\sqrt{{\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{z}\mathbf{-}\mathbf{0}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{2}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\sqrt{{\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}} \\ {\mathit{s}}_{\mathbf{-}}\mathbf{=}\sqrt{{\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{z}\mathbf{-}\mathbf{0}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{2}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\sqrt{{\mathbf{(}\mathbf{y}\mathbf{+}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}} \end{gathered}$$

Substitute the values of the ${\mathit{s}}_{\mathbf{+}}$and ${\mathit{s}}_{\mathbf{-}}$ in equation (5).

$$\begin{gathered} \mathit{V}\left(x,y,z\right)\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\left[\frac{\sqrt{{\left(y+a\right)}^2+z^2}}{\sqrt{{\left(y-a\right)}^2+z^2}}\right] \\ \mathit{V}\left(x,y,z\right)\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\left[\frac{{\left(y+a\right)}^2+z^2}{{\left(y-a\right)}^2+z^2}\right] \end{gathered}$$

Therefore, the potential at the point $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$ is $\mathit{V}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\mathbf{\left[}\frac{{\mathbf{(}\mathbf{y}\mathbf{+}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}\mathbf{\right]}$.

Step 4: Determine the equipotential surfaces are circular cylinders , and locate the axis and radius of the cylinder corresponding to a given potential

b)

The value of potential is constant at all places the equipotential surface is constant.

Thus, the value of V is constant.

$\frac{{\left(y+a\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}{{\left(y+a\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}$  is constant.

Let’s consider that, 

$$\begin{gathered} \frac{{\mathbf{(}\mathbf{y}\mathbf{+}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}} \\ {\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{y}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathit{k}\mathbf{(}{\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{)} \\ {\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{y}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathit{k}\mathbf{[}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{ay}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{]} \end{gathered}$$.

Solve further,

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

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

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

Then further solve,

${\mathbf{[}\mathbf{y}\mathbf{-}\mathbf{a}\frac{\left(k+1\right)}{\left(k-1\right)}\mathbf{]}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{\left[}\frac{\mathbf{4}\mathbf{k}}{\left(k-1\right)}\mathbf{\right]}$

${\mathbf{[}\mathbf{y}\mathbf{-}\mathbf{a}\frac{(k+1)}{(k-1)}\mathbf{]}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}{\mathbf{\left[}\frac{\mathbf{2}\mathbf{a}\sqrt{\mathbf{k}}}{\mathbf{k}\mathbf{-}\mathbf{1}}\mathbf{\right]}}^{\mathbf{2}}$                     …… (6)

The above expression is written as follows:,

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

Comparing equations (6) and (7), we get the value of R.

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

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

$\mathbf{(}{\mathbf{y}}_{\mathbf{0}}\mathbf{,}{\mathbf{z}}_{\mathbf{0}}\mathbf{)}\mathbf{=}\mathbf{(}\frac{\mathbf{a}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{k}\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{0}\mathbf{)}$ and radius $\mathit{R}\mathbf{=}\frac{\mathbf{2}\mathbf{a}\sqrt{\mathbf{k}}}{\mathbf{k}\mathbf{-}\mathbf{1}}$.

Let’s assume that the , potential corresponding is ${\mathit{V}}_{\mathbf{0}}$.  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 $\mathit{I}\mathit{n}\mathit{k}$.

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

Consider that, $\mathit{P}\mathbf{=}\frac{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}$.  Then $\mathit{k}\mathbf{=}{\mathit{e}}^{\mathbf{P}}$.

Now, 

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

Then, 

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

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

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

Substitute ${\mathit{e}}^{\mathbf{P}}$ for k in $\frac{\mathbf{2}\mathbf{a}\sqrt{\mathbf{k}}}{\mathbf{k}\mathbf{-}\mathbf{1}}$ for R equation $\mathit{R}\mathbf{=}\mathit{a}\mathbf{ }\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathit{e}\mathit{c}\mathit{h}\mathbf{\left(}\frac{\mathbf{P}}{\mathbf{2}}\mathbf{\right)}$.,

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

Further solving,

$\mathit{R}\mathbf{=}\mathit{a}\mathbf{ }\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathit{e}\mathit{c}\mathit{h}\mathbf{\left(}\frac{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{\lambda }}\mathbf{\right)}$

Hence, the radius of the cylinder corresponding to the given ${\mathit{V}}_{\mathbf{0}}$ is $\mathit{R}\mathbf{=}\mathit{a}\mathbf{ }\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathit{e}\mathit{c}\mathit{h}\left(\frac{2{\mathrm{\pi \epsilon }}_0{V}_0}{\lambda }\right)$.