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$\left(x,y,z\right)$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 .

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$\lambda$.

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

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

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

Write the expression for potential due to the long wire having charge density$+\lambda$at poin P.

localid="1657515619518" $V_{+}=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\left(\frac{s_{+}}{a}\right)$ …… (3)

Here,localid="1657515432949" $s_{+}$is the distance of point $P$from the long wire having charge density .

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

$V_{-}=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\left(\frac{s_{-}}{a}\right)$ …… (4)

Here,$s_{-}$is the distance of point$P$from the long wire having charge densitylocalid="1657515687607" $-\lambda$.

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

$V=V_{+}{V}_{-}$

$$\begin{gathered} =\frac{-\lambda }{2{\mathrm{\pi \epsilon }}_0}In\left(\frac{s_{+}}{a}\right)+\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\left(\frac{s_{-}}{a}\right) \\ =\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\left(\frac{s_{-}}{s_{+}}\right) \end{gathered}$$ …… (5)

From the above figure, the expression for $s_{+}$and$s_{-}$is calculated as follows:

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

Substitute the values of the $s_{+}$and $s_{-}$in equation (5).

$$\begin{gathered} V=\left(x,y,z\right)=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\left[\frac{\sqrt{{\left(y+a\right)}^2+z^2}}{\sqrt{{\left(y-a\right)}^2+z^2}}\right] \\ V=\left(x,y,z\right)=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\left[\frac{{\left(y+a\right)}^2+z^2}{{\left(y-a\right)}^2+z^2}\right] \end{gathered}$$

Therefore, the potential at the point$\left(x,y,z\right)$is$V\left(x,y,z\right)=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}In\left[\frac{{\left(y+a\right)}^2+z^2}{{\left(y-a\right)}^2+z^2}\right]$.

b)

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

Thus, the value of $V$is constant.

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

Let’s consider that

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

Solve further

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

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

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

Then further solve

$$\begin{gathered} {\left[y-a\frac{\left(k+1\right)}{\left(k-1\right)}\right]}^2+z^2=a^2\left[\frac{4k}{{\left(k-1\right)}^2}\right] \\ {\left[y-a\frac{\left(k+1\right)}{\left(k-1\right)}\right]}^2+z^2=\left[\frac{2a\sqrt{k}}{k-1}\right]......\left(6\right) \end{gathered}$$

The above expression is written as follows:

${\left[y-y_0\right]}^2+{\left(z-z_0\right)}^2=R.......\left(7\right)$

Here,role="math" localid="1657520201223" $Y_0=\frac{a\left(k+1\right)}{\left(k-1\right)}\mathrm{and}{\mathrm{z}}_0=0$.

Substitute the value ofrole="math" localid="1657518779174" $y_0,z_0$in equation (7).

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

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

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

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

Let’s assume that the potential corresponding is$V_0$. Then,

$V_0=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}Ink$

Rewrite the above equation for $Ink$.

$$\begin{gathered} \frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }=Ink \\ {e}^{\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }}=k \end{gathered}$$

Consider that$P=\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }$. Then$k={\mathrm{e}}^{\mathrm{p}}$.

Now,

$$\begin{gathered} y_0=\frac{a\left(k+1\right)}{k-1} \\ =\frac{a\left(e^p+1\right)}{e^p-1} \\ =a\left(\frac{e^{p/2}+e^{-p/2}}{e^{p/2}-e^{-p/2}}\right) \end{gathered}$$

Then,

Here,$y_0=\frac{a\left(k+1\right)}{\left(k-1\right)}$, and $z_0=0$.

Substitute the value of$y_0,z_0$in equation (7).

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

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

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

$\left(v_0,z_0\right)=\left(\frac{a\left(k+1\right)}{k-1},0\right)\mathrm{and}\mathrm{radius}R=\frac{2a\sqrt{k}}{k-1}$.

Let’s assume that the potential corresponding is .$V_0$Then,

$$\begin{gathered} V_0=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}Ink \\ \mathrm{Rewrite}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{for}Ink. \\ \frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }=Ink \\ {e}^{\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }}=k \end{gathered}$$

Consider that $P=\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }$.Then$k=e^P$.

Now,

$$\begin{gathered} y_0=\frac{a\left(k+1\right)}{k-1} \\ =\frac{a\left(e^P+1\right)}{e^P+1} \\ =a\left(\frac{e^{P/2}+e^{-P/2}}{e^{P/2}-e^{-P/2}}\right) \end{gathered}$$

Then,

$Y_0=acoth\left(\frac{p}{2}\right)$

Substitute the$\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }$for $P$in the above equation.

$$\begin{gathered} y_0=acoth\left(\frac{\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }}{2}\right) \\ =acoth\left(\frac{2{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }\right) \end{gathered}$$

Substitute $e^P$for $k$in $\frac{2a\sqrt{k}}{k-1}$for $R$equation $R=a\cos ech\left(\frac{P}{2}\right)$.

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

Further solving

$R=a\cos ech\left(\frac{2{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }\right)$

Hence, the radius of the cylinder corresponding to the given $V_0$is

$R=a\cos ech\left(\frac{2{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }\right)$.