91影视

The following figure consists of two long straight copper pipes separated by distance $2\mathrm{d}$on -axis. One pipe is at potential $-{\mathrm{V}}_0$and another pipe is at $+{\mathrm{V}}_0$.

Write the expression for the charge due to the wire of charge density $+\mathrm{位}$at point (x,y,z) is,

鈥︹�� (1)

Here, $\mathrm{s}$is the distance from $+\mathrm{位}$to $(\mathrm{x},\mathrm{y},\mathrm{z})$.

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

localid="1657421100992" $\mathbf{V}\mathbf{=}\left(\frac{位}{\mathrm{鈭�}{蔚}_0}\ln 鈦�\frac{s_{鈭�}}{d}\right)$鈥︹�� (2)

Here, $\mathrm{s}$is the distance from $-\mathrm{位}$to $(\mathrm{x},\mathrm{y},\mathrm{z})$.

Write the expression for the total potential.

$\mathrm{V}={\mathrm{V}}_{+}+{\mathrm{V}}_{-}$鈥︹��. (3)

Here, $\mathrm{V}$is the total potential.

Substitute the $鈭�\left(\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0}\ln 鈦�\frac{{\mathrm{s}}_{+}}{\mathrm{d}}\right)$for ${\mathrm{V}}_{+}$and $\left(\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0}\ln 鈦�\frac{{\mathrm{s}}_{鈭�}}{\mathrm{d}}\right)$for ${\mathrm{V}}_{-}$in above equation (3).

$\mathrm{V}=鈭�\left(\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0}\ln 鈦�\frac{{\mathrm{s}}_{鈭�}}{\mathrm{d}}\right)+\left(\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0}\ln 鈦�\frac{{\mathrm{s}}_{鈭�}}{\mathrm{d}}\right)$

$=\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0}\left[\ln 鈦�\frac{{\mathrm{s}}_{鈭�}}{\mathrm{d}}鈭�\ln 鈦�\frac{{\mathrm{s}}_{+}}{\mathrm{d}}\right]$

From the figure, the distance and respectively are,

${\mathrm{s}}_{+}=\sqrt{(\mathrm{x}鈭�\mathrm{d}{)}^2+{\mathrm{y}}^2}$

Substitute $\sqrt{(\mathrm{x}鈭�\mathrm{d}{)}^2+{\mathrm{y}}^2}$for role="math" localid="1657421577606" ${\mathrm{s}}_{+}$and $\sqrt{(\mathrm{x}+\mathrm{d}{)}^2+{\mathrm{y}}^2}$for $s_{-}$in equation (5)

$\mathrm{V}=\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0}\ln 鈦�\frac{\sqrt{(\mathrm{x}+\mathrm{d}{)}^2+{\mathrm{y}}^2}}{\sqrt{(\mathrm{x}鈭�\mathrm{d}{)}^2+{\mathrm{y}}^2}}$

$=\frac{\mathrm{i}鈰�}{2{\mathrm{蟺锄}}_0}\ln 鈦�\frac{{\left((\mathrm{x}+\mathrm{d}{)}^2+{\mathrm{y}}^2\right)}^{1/2}}{{\left((\mathrm{x}鈭�\mathrm{d}{)}^2+{\mathrm{y}}^2\right)}^{1/2}}$

$=\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0}\ln 鈦�{\left(\frac{(\mathrm{x}+\mathrm{d}{)}^2+{\mathrm{y}}^2}{(\mathrm{x}鈭�\mathrm{d}{)}^2+{\mathrm{y}}^2}\right)}^{1/2}$

$=\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0}\frac{1}{2}\ln 鈦�\left(\frac{(\mathrm{x}+\mathrm{d}{)}^2+{\mathrm{y}}^2}{(\mathrm{x}鈭�\mathrm{d}{)}^2+{\mathrm{y}}^2}\right)$

Simplify the above equation, then the potential is,

$\mathrm{V}=\frac{\mathrm{位}}{4{\mathrm{蟺蔚}}_0}\ln 鈦�\left(\frac{(\mathrm{x}+\mathrm{a}{)}^2+{\mathrm{y}}^2}{(\mathrm{x}鈭�\mathrm{a}{)}^2+{\mathrm{y}}^2}\right)$

Thus, the Potential at any point is $\mathrm{V}=\frac{\mathrm{位}}{4{\mathrm{蟺蔚}}_0}\ln 鈦�\left(\frac{(\mathrm{x}+\mathrm{a}{)}^2+{\mathrm{y}}^2}{(\mathrm{x}鈭�\mathrm{a}{)}^2+{\mathrm{y}}^2}\right)$.

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

Therefore,

$\frac{(\mathrm{x}+\mathrm{a}{)}^2+{\mathrm{z}}^2}{(\mathrm{x}鈭�\mathrm{a}{)}^2+{\mathrm{z}}^2}=\mathrm{k}$

${\mathrm{x}}^2+{\mathrm{a}}^2+2\mathrm{ax}+{\mathrm{z}}^2=\mathrm{k}\left[{\mathrm{x}}^2+{\mathrm{a}}^2鈭�2\mathrm{ax}+{\mathrm{z}}^2\right]$

${\mathrm{x}}^2[\mathrm{k}鈭�1]+{\mathrm{a}}^2[\mathrm{k}鈭�1]+{\mathrm{z}}^2[\mathrm{k}鈭�1]鈭�2\mathrm{ax}[\mathrm{k}+1]=0$

${\mathrm{x}}^2+{\mathrm{a}}^2+{\mathrm{z}}^2鈭�2\mathrm{ax}\frac{(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}=0$鈥︹�� (6)

$x^2鈭�2\mathrm{xa}\frac{(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}+\left({\mathrm{a}}^2+{\mathrm{z}}^2\right)=0$

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

$x^2鈭�2\mathrm{xa}\frac{(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}+{\left[\mathrm{a}\frac{(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}\right]}^2+{\mathrm{z}}^2={\left[\mathrm{a}\frac{(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}\right]}^2鈭�{\mathrm{a}}^2$

${\left[\mathrm{x}鈭�\mathrm{a}\frac{(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}\right]}^2+{\mathrm{z}}^2={\mathrm{a}}^2\left[{\left(\frac{\mathrm{k}+1}{\mathrm{k}鈭�1}\right)}^2鈭�1\right]$

${\left[\mathrm{x}鈭�\mathrm{a}\frac{(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}\right]}^2+{\mathrm{z}}^2={\mathrm{a}}^2\left[\frac{4\mathrm{k}}{(\mathrm{k}鈭�1{)}^2}\right]$

${\left[\mathrm{x}鈭�\mathrm{a}\frac{(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}\right]}^2+{\mathrm{z}}^2={\left[\frac{2\mathrm{a}\sqrt{\mathrm{k}}}{\mathrm{k}鈭�1}\right]}^2$

The above expression is write as,

${\left[\mathrm{x}鈭�{\mathrm{y}}_0\right]}^2+{\left(\mathrm{z}鈭�{\mathrm{z}}_0\right)}^2={\mathrm{R}}^2$ 鈥︹�� (7)

Here, ${\mathrm{y}}_0=\frac{\mathrm{a}(\mathrm{k}+1)}{(\mathrm{k}鈭�1)}$and ${\mathrm{Z}}_{0=}0$

Substitute the value of ${\mathrm{y}}_{0,}{\mathrm{z}}_0$in equation (7).

Then the expression for localid="1657459125250" $\mathrm{R}$is

$\mathrm{R}=\frac{2\mathrm{a}\sqrt{\mathrm{k}}}{\mathrm{k}鈭�1}$

Thus, the represents circular cylinder with axis parallel to x-axis centered at $\left({\mathrm{y}}_0,{\mathrm{z}}_0\right)=\left(\frac{\mathrm{a}(\mathrm{k}+1)}{\mathrm{k}鈭�1},0\right)$and radius $\mathrm{R}=\frac{2\mathrm{a}\sqrt{\mathrm{k}}}{\mathrm{k}鈭�1}$.

Let鈥檚 assume that, potential corresponds to ${\mathrm{v}}_0$, then

${\mathrm{V}}_0=\frac{\mathrm{位}}{4{\mathrm{蟺蔚}}_0}\ln 鈦�\mathrm{k}$

Rewrite the above equation for $\mathrm{lnk}$$\mathrm{lnk}$

$\frac{4{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}=\ln 鈦�\mathrm{k}$

${\mathrm{e}}^{4{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}=\mathrm{k}$

Let鈥檚 consider that, $\mathrm{P}=\frac{4{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\widehat{\mathrm{位}}}$then $\mathrm{k}={\mathrm{e}}^{\mathrm{渭}}$.

Now,

${\mathrm{y}}_0=\frac{\mathrm{a}(\mathrm{k}+1)}{\mathrm{k}鈭�1}$

$=\frac{\mathrm{a}\left({\mathrm{e}}^{\mathrm{p}}+1\right)}{{\mathrm{e}}^{\mathrm{p}}鈭�1}$

$=\mathrm{a}\left(\frac{{\mathrm{e}}^{\mathrm{P}/2}+{\mathrm{e}}^{\mathrm{P}/2}}{{\mathrm{e}}^{\mathrm{P}/2}鈭�{\mathrm{e}}^{鈭�\mathrm{P}/2}}\right)$

Then,

${\mathrm{y}}_0=\mathrm{acoth}鈦�\left(\frac{\mathrm{P}}{2}\right)$

Substitute $\frac{4{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}$for $\mathrm{P}$in above equation.

${\mathrm{y}}_0=\mathrm{acoth}鈦�\left(\frac{\frac{4{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}}{2}\right)$

$=\mathrm{acoth}鈦�\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)$ 鈥︹�� (8)

Substitute $e^{\mathrm{p}}$for $\mathrm{k}$in $\mathrm{R}=\frac{2\mathrm{a}\sqrt{\mathrm{k}}}{\mathrm{k}鈭�1}$equation.

$\mathrm{R}=\frac{2\mathrm{a}\sqrt{{\mathrm{e}}^{\mathrm{p}}}}{{\mathrm{e}}^{\mathrm{p}}鈭�1}$

$=2\mathrm{a}\frac{{\mathrm{e}}^{\mathrm{蚁}/2}}{{\mathrm{e}}^{\mathrm{p}}鈭�1}$

$=2\mathrm{a}\frac{1}{{\mathrm{e}}^{\mathrm{p}/2}鈭�{\mathrm{e}}^{鈭�\mathrm{p}/2}}$

$=\mathrm{a}\frac{2}{{\mathrm{e}}^{\mathrm{P}/2}鈭�{\mathrm{e}}^{鈭�\mathrm{P}/2}}$

So,

$\mathrm{R}=\mathrm{acosech}鈦�\left(\frac{\mathrm{P}}{2}\right)$

Substitute the value $\frac{4{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}$for $\mathrm{P}$in above equation.

$\mathrm{R}=\mathrm{acosech}鈦�\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)$ 鈥�.... (9)

Hence, the radius of the cylinder corresponding to given ${\mathrm{v}}_0$is $\mathrm{R}=\mathrm{acosech}鈦�\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)$.

As, ${\mathrm{y}}_0鈫�\mathrm{d}$equation (8) becomes,

$\mathrm{d}=\mathrm{acoth}鈦�\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)$......(10)

And from equation (9),

$\mathrm{R}=\mathrm{acosec}鈦�\mathrm{h}\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)$ ......(11)

Divide the equation (10) with (11)

$\frac{\mathrm{d}}{\mathrm{R}}=\frac{\mathrm{acoth}鈦�\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)}{\mathrm{acosec}鈦�\mathrm{h}\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)}$

role="math" localid="1657460122368" $\frac{\mathrm{d}}{\mathrm{R}}=\left[\frac{\cos 鈦�\mathrm{h}\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)}{\sin 鈦�\mathrm{h}\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)}\right]\left[\sin 鈦�\mathrm{h}\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)\right]$

$\frac{\mathrm{d}}{\mathrm{R}}=\cos 鈦�\mathrm{h}\left(\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}\right)$

$\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{\mathrm{位}}={\cosh }^{鈭�1}鈦�\left(\frac{\mathrm{d}}{\mathrm{R}}\right)$

Thus,

$\mathrm{位}=\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{{\cosh }^{鈭�1}鈦�\left(\begin{array}{l}\mathrm{d}\\ \mathrm{R}\end{array}\right)}$

Hence, the linear charge density is $\mathrm{位}=\frac{2{\mathrm{蟺蔚}}_0{\mathrm{V}}_0}{{\cosh }^{鈭�1}鈦�\left(\begin{array}{l}\mathrm{d}\\ \mathrm{R}\end{array}\right)}$.