Two long straight wires, carrying opposite uniform line charges,$\pm A$are situated on either side of a long conducting cylinder (Fig. 3.39). The cylinder(Which carries no net charge) has radius ,and the wires are a distance from the axis. Find the potential.

The potential at a distancefrom an infinitely long straight wire that carries a uniform line charge density $\lambda$is,

$V=-\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s}{a}\right)$ …… (1)

Here,${\epsilon }_0$ is permittivity of the free space and is the distance.

The following figure shows that long straight wires are situated on either side of a long conducting cylinder.

From the above figure, the expression for the distance $y_0$is,

$\begin{array}{rcl}{y}_0& =& b+\frac{a-b}{2}\\ & =& \frac{a+b}{2}\\ & & \end{array}$

If goes to $\frac{a-b}{2}$then there is following condition,

$\begin{array}{rcl}{\left(\frac{a-b}{2}\right)}^2& =& {\left(\frac{a+b}{2}\right)}^2-R\\ {\left(a-b\right)}^2& =& {\left(a+b\right)}^2-4R^2\\ {\left(a+b\right)}^2-{\left(a-b\right)}^2& =& 4R^2\\ 4ab& =& 4R^2\\ b& =& \frac{R^2}{a}\\ & & \end{array}$

From the figure, write the expression for$s_1^2,s_2^2,s_3^2$ and$s_4^2$ are

$$\begin{gathered} s_1^2={\left(y+a\right)}^2+z^2 \\ {s}_2^2={\left(y+b\right)}^2+z^2 \\ {s}_3^2={\left(y-b\right)}^2+z^2 \\ {s}_4^2={\left(y-a\right)}^2+z^2 \end{gathered}$$

Write the expression for electric potential at point P due to $+\lambda$.

$V_1=-\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s_4}{a}\right)$

Write the expression for electric potential at point P due to $-\lambda$.

$V_2=\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s_1}{a}\right)$

Write the expression for electric potential at point P due to$+\lambda$.

$V_3=-\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s_2}{b}\right)$

Write the expression for electric potential at point P due to $-\lambda$.

$V_4=\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s_3}{b}\right)$

Write the expression for the total electric potential.

$$\begin{gathered} V=-\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s_4}{a}\right)+\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s_1}{a}\right)-\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s_2}{b}\right)+\frac{\lambda }{2\pi {\epsilon }_0}In\left(\frac{s_3}{b}\right) \\ =\frac{\lambda }{2\pi {\epsilon }_0}\left[In\left(\frac{s_1}{a}\right)-In\left(\frac{s_4}{a}\right)\right]+\frac{\lambda }{2\pi {\epsilon }_0}\left[In\left(\frac{s_3}{b}\right)-In\left(\frac{s_2}{b}\right)\right] \\ =\frac{\lambda }{4\pi {\epsilon }_0}\left[In{\left(\frac{s_1}{a}\frac{a}{s_4}\right)}^2+In{\left(\frac{s_3}{b}\frac{b}{s_2}\right)}^2\right] \\ =\frac{\lambda }{4\pi {\epsilon }_0}\left[In{\left(\frac{s_1}{s_4}\right)}^2+In{\left(\frac{s_3}{s_2}\right)}^2\right] \end{gathered}$$

Thus,

$V=\frac{\lambda }{4\pi {\epsilon }_0}In\left[\frac{s_1^2{s}_3^2}{s_4^2{s}_2^2}\right]$

Substitute ${\left(y+a\right)}^2+z^2$for$s_1^2$, ${\left(y+b\right)}^2+z^2$for$s_2^2$,${\left(y-b\right)}^2+z^2$for $s_3^2$and ${\left(y-a\right)}^2+z^2$for $s_4^2$.

$V=\frac{\lambda }{4\pi {\epsilon }_0}In\frac{\left[{\left(y+a\right)}^2+z^2\right]}{\left[{\left(y-a\right)}^2+z^2\right]}\frac{\left[{\left(y-b\right)}^2+z^2\right]}{\left[{\left(y+b\right)}^2+z^2\right]}$

Substitute$s\cos \varphi$for $y$and$s\sin \varphi$for Z and $\frac{R^2}{a}$for b in above equation.